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CONTENTS PHYSICAL CHEMISTRY-3

Pages



Chemical Equilibrium – Theory

1 – 11



Ionic Equilibrium – Theory

12 – 21



Some Basic Concepts of Chemistry – Theory

22 – 39



Chemical Equilibrium – Exercise – 1

A1 – A9



Ionic Equilibrium – Exercise – 1

A10 – A18



Some Basic Concepts of Chemistry – Exercise – 1

A19 – A26



Chemical Equilibrium – Exercise – 2

B1 – B3



Ionic Equilibrium – Exercise – 2

B4 – B6



Some Basic Concepts of Chemistry – Exercise – 2

B7 – B9



Chemical Equilibrium – Exercise – 3

C1 – C13



Ionic Equilibrium – Exercise – 3

C14 – C27



Some Basic Concepts of Chemistry – Exercise – 3

C28 – C39



Chemical Equilibrium – Exercise – 4

D1 – D4



Ionic Equilibrium – Exercise – 4

D5 – D11



Some Basic Concepts of Chemistry – Exercise – 4

D12 – D14

CHEMICAL EQUILIBRIUM

CHEMICAL EQUILIBRIUM

1. INTRODUCTION

2.3 Concept of Chemical Equilibrium

Equilibrium represents the state of a process in which the proprties like temperature, pressure, concentration of the system do not show any change with the passage of time. If the opposing processes involve only physical changes, the equilibrium is called physical equilibrium. If the opposing processes involve chemical changes, i.e., the opposing processes are chemical reactions, the equilibrium is called chemical equilibrium.

2. EQUILIBRIA IN CHEMICAL PROCESSES 2.1 Reversible Reactions A reaction in which not only the reactants react to form the products under certain conditions but also the products react to form reactants under the same conditions is called a reversible reaction. In other words a reaction which takes place not only in the forward direction but also in the backward direction under the same conditions is called a reversible reaction.

x

It is the state of minimum Gibb’s energy

x

dG = 0 and 'G = 0 at this state

x

Rate of forward reaction = Rate of backward reaction

x

This equilibrium is dynamic and stable in nature Dynamic Nature of Chemical Equilibrium In the Haber’s process, starting with definite amounts of N2 and H2 and carrying out the reaction at a particular temperature, when equilibrium is attained, the concentrations of N2, H2 and NH3 become constant. If the experiment is repeated by taking deuterium (D2) in place of H2 but with the same amounts and exactly similar conditions as before, equilibrium is attained containing D2 and ND3 in place of H2 and NH3 but in the same amounts. Now, if the two reaction mixtures are mixed, then after some time, it is found that the concentrations of ammonia and hydrogen are same except that now all forms of ammonia (i.e., NH3, NH2D, NHD2, ND3) and all forms of hydrogen (i.e., H2, HD, ) are present. This shows that at equilibrium, the reaction 2 is still going on, i.e., equilibrium is dynamic in nature.

A reversible reaction between A and B to form C and D is represented as : ZZX C + D A + B YZZ

2.2 Irreversible Reactions If a reaction cannot take place in the reverse direction, i.e., the products formed do not react to give back the reactants under the same conditions, it is called an irreversible reaction. A + B  o C+D

1

CHEMICAL EQUILIBRIUM 't o0 Units of rate : conc/time or mol/Ls 3.2 Overall rate of a reaction aA + bB lcC l + dD Overall Rate = – (1/a) 'A/'t = – (1/b) 'B/'t = + (1/c) 'C/'t = + (1/d) 'D/'t Overall rate : Rate of forward reaction – Rate of backward reaction “At equilibrium the overall rate of a reversible reaction becomes zero”.

4. EQUILIBRIUM CONSTANT (K) 2.4 Characteristics of Chemical Equilibrium (i)

4.1 Law of Mass Action

At equilibrium, the concentration of each of the reactants and the products becomes constant.

(ii)

The rate of a reaction is proportional to the product of the active masses of the reactants, each raised to the power equal to its stoichiometric coefficient as represented by the balanced chemical equation.

At equilibrium, the rate of forward reaction becomes equal to the rate of backward reaction and hence the equilibrium is dynamic in nature.

aA + bB + cC + ......  o Products

(iii) A chemical equlibrium can be established only if none of the products is allowed to escape out or separate out as a solid.

Rate of reaction v [A]a [B]b [C]c ..... Law of Chemical Equilibrium is a result obtained by applying the Law of Mass Action to a reversible reaction in equilibrium.

e.g. if CO2 gas escapes out in case of decomposition of CaCO , the reaction will no longer remain reversible. Similarly, the reaction is irreversible if one of the products separates out as solid,

ZZX C + D A + B YZZ

AgNO3 + KCl  o AgCl p + KNO

Rate of the forward reaction v [A] [B] = kf [A] [B] Rate of the backward reaction v [C] [D] = kb [C] [D]

(iv) Chemical equilibrium can be attained from either direction, i.e., from the direction of the reactants as well as from the direction of the products.

At equilibrium, Rate of forward reaction = Rate of backward reaction

3. RATE OF A REACTION k f [A] [B] k b [C] [D] or Average Rate = Change in concentration/Time taken = 'c/'t

[C][D] [A][B]

kf kb

K

At constant temperature, as kf and kb are constant, therefore,

'c = Final Concentration – Initial concentration

kf k b = K is also constant at constant temperature and is

3.1 Instantaneous rate

called ‘Equilibrium constant’.

Lim 'c/'t = dc/dt

2

CHEMICAL EQUILIBRIUM The product of the molar concentrations of the products, each raised to the power equal to its stoichiometric coefficient divided by the product of the molar concentrations of the reactants, each raised to the power equal to its stoichiometric coefficient is constant at constant temperature and is called Equilibrium constant.

4.4 Reaction Quotient (Q) aA + bB lcC l + dD Q = [C]c [D]d/ [A]a [B]b Q > Kc : Reaction will tend towards backward direction

4.2 Characteristics of Equilibrium Constant (i)

(ii)

Q > Kc : Reaction will tend towards forward direction Q > Kc : Reaction will be at equilibrium

The value of the equilibrium constant for a particular reaction is always constant depending only upon the temperature of the reaction and is independent of the concentrations of the reactants with which we start or the direction from which the equilibrium is approached.

Note : Q is a variable which always approaches Keq which is a constant.

If the reaction is reversed, the value of the equilibrium constant in inversed. 4.5 Calculating Equilibrium Concentrations

(iii) If the equation (having equilibrium constant K) is divided by 2, the equilibrium constant for the new equation is the square root of K (i.e.,

Suppose we are given the following data :

K ).

(iv) If the equation (having equilibrium constant K) is multiplied by 2, the equilibrium constant for the new equation is the square of K (i.e., K2) (v)

1.

The balanced reaction and value of Kc

2.

The initial concentration of the reactants, or the initial moles

3.

Volume of the container And we need to find the final equilibrium concentration of the reactants and products. Then we can follow the following steps :

If the equation (having equilibrium constant K) is written in two steps (having equilibrium constant K1 and K2) then K1 × K2 = K.

Step-1

(vi) The value of the equilibrium constant is not affected by the addition of a catalyst to the reaction.

Write down the balanced chemical equation for the reaction Step-2

This is because the catalyst increases the speed of the forward reaction and the backward reaction to the same extent.

Under every reactant and product, write down the initial moles/concentration Step-3

4.3 Extent of Reaction

Subtract the amount reacting and add the amount produced in terms of a variable x and note down the equilibrium concentration in terms of x. If we are dealing in moles then we will need to divide the moles by volume to obtain concentrations.

High value of Kc (Kc > 103) o At equilibrium reaction is forward dominant Low value of Kc (Kc < 10–3) o At equilibrium reaction is backward dominant

Step-4

Moderate value of Kc (between 10 and 10 ) o At equilibrium neither direction dominates 3

–3

Substitute the equilibrium concentration in the expression of Kc and equate it to the value of Kc. Step-5 Solve the above equation and calculate the value of x and in case of multiple solutions select the value which is sensible from reaction point of view. Then back substitute

3

CHEMICAL EQUILIBRIUM the value of x in the equilibrium concentration expression and obtain the actual value of the same.

POINTS TO REMEBER

4.6 Equilibrium constant Kp

Whether the reaction is exothermic or endothermic on increasing the temperature, it will tend towards endothermic direction.

aA + bB l cC + dD Kp = (PCc × PDd)/ (PAa × PBb)

5. HOMOGENOUS EQUILIBRIA

where PC, PD, PA, PB are partial pressures of A, B, C and D respectively.

Reactions in which all reactants and products are in the same phase Homogenous Reactions can further be divided into three sub-categoreis :

Kp = Kc (RT)'n R = 0.0821 L atm/mol–K

5.1 Those reactions where gaseous moles increase ('n = +ve)

4.7 Relationship of Keq and 'G

PCl5 (g) lPCl3 (g) + Cl2 (g) 5.2 Those reactions where gaseous moles remain the same ('n = 0)

For any reaction : 'G = 'Gº + RT InQ

Where Q is the reaction quotient.

H2 (g) + I2 (g) l 2HI (g)

At equilibrium 'G = 0

5.3 Those reactions where gaseous moles decrease ('n = – ve)

'Gº = –RT lnKeq = – 2.303RT log10Keq

2NO2 (g) l N2O4 (g)

where K eq is generally taken as Kp. It depends on the definition of standard values to define 'Gº.

6. DEGREE OF DISSOCIATION, D

If the standard active masses are taken as 1 M each then we will take Keq = KC and if they are taken as 1 atm each then we will take Keq = Kp.

It is defined as the fraction of molecules dissociating. For example, if 100 molecules are present and only 40 dissociate then the degree of dissociation is 0.4 or 40%.

4.8 Dependence of Keq on temperature

7. HETEROGENEOUS EQUILIBRIA

Van’t Hoff Equation : ln (K2/K1) = ('H/R) (1/T1–1/T 2)

or

Category of reactions where various phases/states exist in the same reaction.

log10 (K2/K1) = ('H/2.303R) (1/T1–1/T2)

In presence of gases the activity of solids and excess liquids is constant. Therefore we can assume the active masses of solids and excess liquids as constant.

Exothermic Reactions 'H = negative On increasing the temperature the equilibrium constant will decrease i.e. The reaction will become more backward dominant.

CaCO3(s) l CaO(s) + CO2(g) Kc = [CO2]

Endothermic Reaction

Kp = PCO 2

'H = possitive

8. LE CHATELIER’S PRINCIPLE

On increasing the temperature the equilibrium constant will increase. The reaction will become more forward dominant.

If a distrubance is introduced in an equilibrium mixture it will behave so as to undo the distrubance and re-establish equilibrium.

4

CHEMICAL EQUILIBRIUM 8.5 Effect of a Catalyst

8.1 Effect of Change of Concentration

A catalyst increases the rate of the chemical reaction by making available a new low energy pathway for the conversion of reactants to products. It increases the rate of forward and reverse reactions that pass through the same transition state and does not affect equilibrium. Catalyst lowers the activation energy for the forward and reverse reactions by exactly the same amount. Catalyst does not affect the equilibrium composition of a reaction mixture. It does not appear in the balanced chemical equation or in the equilibrium constant expression.

If in a reaction in equilibrium, the concentration of any reactant is increased, the equilibrium shifts in the forward direction. On the other hand, if the concentration of any product is increased, the equilibrium shifts in the backward direction. The reverse happens if the concentrations are decreased. 8.2 Effect of change of temperature Exothermic reactions are favoured by low temperature whereas endothermic reations are favoured by high temperature.

eg contact process,

8.3 Effect of change of pressure

2SO2 (g)  O2 (g) U 2SO3 (g); Kc

1.7 u1026

Low pressure favours those reactions which are accompanied by increase in total number of moles and high pressure favours those reactions which take place with decrease in total number of moles. However, pressure has no effect on an equilibrium reaction which proceeds with no change in total number of moles.

Practically the oxidation of SO2 to SO3 is very slow. Thus, platinum or divanadium penta-oxide (V2 O5) is used as catalyst to increase the rate of the reaction. 8.6 Effect of adding an inert gas to a reaction mixture in equilibrium.

8.4 Effect of change in volume

(a)

If the reaction takes place at constant volume addition of an inert gas will not change the molar concentrations of the reactants and products. Hence, the state of equilibrium will remain unaffected.

(b)

If the reaction takes place at constant pressure, addition of inert gas must accompany in increase in the total volume to keep pressure constant. Thus, reaction shifts towards larger number of moles.

The effect of decrease of volume is equivalent to the effect of increase of pressure. hence the effect of decrease in volume will be to shift the equilibrium in the direction in which the number of moles decreases.

5

CHEMICAL EQUILIBRIUM

SOLVED EXAMPLES Example - 3

Example - 1

One mole of H2O and one mole of CO are taken in a 10 litre vessel and heated at 725 K. At equilibrium 40% of water (by mass) reacts with CO according to the equation:

The following reaction has attained equilibrium CO(g)  2H 2 (g)

CH 3 OH(g), 'H o

92.0kJ mol1

What will happen if (i) volume of the reaction vessel is suddenly reduced to half? (ii) the partial pressure of hydrogen is suddenly doubled? (iii) an inert gas is added to the system? CH 3 OH

Kc

(i)

When volume of the vessel is reduced to half, the concentration of each reactant or product becomes double.

CO H 2

Thus,

, Kp

Calculate the equilibrium constant for the reaction.

PCO u p2H 2

2 CH3OH

Qc

2[CO] u (2[H2 ])

2

1 Kc 4

Sol.

CH 3OH to make Q c

(ii)

QP

p CO

u (2 p H2 )

As Q c  K c , equilibrium

2

K

1 Kp . 4

2HI(g) Initial pressure 0.2 atm 0.04 atm

H 2 (g)

 I2 (g)

0

0

0.16 atm 2

H 2 CO2 H 2 O CO

H 2 (g)  I2 (g)

p

2 HI

0.08atm u 0.08atm (0.04atm)2

2HI(g) is 54.8. If 0.5 mol L–1 of HI

(g) is present at equilibrium at 700K, what are the concentrations of H2(g) and I2(g) assuming that we initially started with HI (g) and allowed it to reach equilibrium at 700 K.

Sol.

2HI(g)

1 54.8

H 2 (g)  I2 (g), K

At equilibrium [HI]= 0.5 mol L–1,

0.16 atm 2

H2

(Decrease in the pressure of HI = 0.2–0.04=0.16 atm.) p H 2 u pI 2

0.444

At 700 K, equilibrium constant for the reaction

[I 2 ]

= 0.08 atm

Kp

0.04 u 0.04 0.06 u 0.06

Example - 4

A sample of HI (g) is placed in a flask at a pressure of 0.2 atm. At equilibrium, the partial pressure of HI (g) is 0.04 atm. What is Kp for the given equilibrium?

?

0.04 mol L1 ,

Kc .

Example - 2

At eqm.

1  0.40 mol L1 = 0.06 mol L–1, 10

[CO2] = 0.04 mol L–1,

As volume remains constant, molar concentration will not change. Hence there is no effect on the state of equilibrium.

Sol.

0.4 mol L1 10

H2

Again Qp < Kp, equilibrium will shift in the forward direction to make Qp= Kp. (iii)

At equilibrium H 2 O [CO]= 0.06 mol L–1.

will shift in the forward direction, producing more of

PCH3 OH

H 2 O(g)  CO2 (g)

PCH3 OH

Sol.

2

H 2 O(g)  CO(g)

This gives x

4.0.

6

x mol L1 ? K

xu x (0.5) 2

1 54.8

0.068. i.e., [H2]= [I2] = 0.068 mol L–1

CHEMICAL EQUILIBRIUM

What is the equilibrium concentration of each of substances in the equilibrium when the initial concentration of ICI was 0.78 M?

At eqm.1–0.171 0.180–0.171 0.171 mol

Initial conc.

At eqm.

Kc =

0.78-2x I 2 Cl2 ICl

2

Molar concs 8.29/V

I 2 (g) + Cl 2 (g)

0.78M

0

0

x

? 0.14

or or x

0.292  0.748 x or 1.748 x

Kc

0.171/V

CH 3 COOC 2 H 5 H 2 O

0.171/ V 0.171/ V

CH 3COOH C 2 H5 OH

0.829 / V 0.009 / V

3.92

CH3COOC2 H5  H2O

Initial 1.000 mol 0.500 mol After time t 1–0.214

0.500–0.214

0.214 mol 0.214 mol

0.14 0.374

0.292 or x

= 0.786 mol = 0.286 mol 0.167

Reaction quotient (Qc) =

I 2 = Cl2 =0.167M, ICI =0.78-2×0.167M=0.446M

0.214 / V (0.214 / V) (0.786 / V)(0.286 / V)

0.204

As Q c z K c , equilibrium has not been attained

Example - 6

Example - 7

Ethyl acetate is formed by the reaction between ethanol and acetic acid and the equilibrium represented as: CH 3 COOH(l )  C2 H 5 OH(l )

A sample of pure PCl5 was introduced into an evacuated vessel at 473 K. After equilibrium was attained, concentration of PCl5 was found to be 0.5×10–1 mol L–1. If value of Kc is 8.3 × 10–3, what are the concentrations of PCl3 and Cl2 at equilibrium?

CH 3COOC 2 H5 ( )  H 2 O(l )

(i) Write the concentration ratio (reaction quotient), Qc for this reaction (Note: water is not in excess and is not a solvent in this reaction)

PCl3 (g)

PCl5 (g)

Sol.

(ii) At 293 K, if one starts with 1.00 mol of acetic acid and 0.18 mol of ethanol, there is 0.171 mol of ethyl acetate in the final equilibrium mixture. Calculate the equilibrium constant.

 Cl2(g)

At eqm 0.5× 10–1 mol L–1 x mol L–1 x mol L–1 (suppose)

x2 0.5 u10 1

? Kc

(iii) Starting with 0.5 mol of ethanol and 1.0 mol of acetic acid and maintaining it at 293 K, 0.214 mol of ethyl acetate is found after some time. Has equilibrium been reached? (i) Qc

0.171/V

(iii) CH3COOH  C2 H5OH

xu x (0.78  2 x) 2

Hence, at equilibrium,

Sol.

0.009/V

x

x 0.78  2x

x2 0.14(0.78  2x)2 or

0.171 mol

= 0.829 mol = 0.009 mol

Suppose at equilibrium, I2 = Cl2 =xmolL-1 . Then 2ICl

CH3COOC2 H5  H2O

Initial 1.00 mol 0.180 mol

I 2 (g)+Cl 2 (g),K c =0.14

2I Cl(g) Sol.

CH3COOH  C2 H5OH

(ii)

Example - 5

or

x2

x

CH 3 COOC2 H5 H 2 O CH 3 COOH C2 H5 OH

(8.3 u 10 3 )(0.5 u 101 )

4.15 u104 Hence, PCl3

7

8.3 u103 (Given)

4.15 u 104 or

2.04 u102 M

eq

Cl2

eq

0.02M

0.02M

CHEMICAL EQUILIBRIUM Example - 8

Example - 10 Dihydrogen gasused inHaber’sprocessisproduced by reacting methane fromnatural gas with high temperature steam. The first stage of two stage reaction involves the formation of CO and H2. In second stage, CO formed in first state is reacted with more steam in water gas shift reaction,

Equilibrium constant, K c for the reaction, N 2 (g)  3H 2 (g)

2NH 3 (g) at 500 K is 0.061. At a

particular time, the analysis shows that composition of the reaction mixture is 3.0 mol L–1 N2, 2.0 mol L–1 H2 and 0.5 mol L–1 NH3. Is the reaction at equilibrium? If not, in which direction does the reaction tend to proceed to reach equilibrium?

Sol.

NH 3

Qc

CO2 (g)  H2 (g) . If a reaction

vessel at 400ºC is charged with an equimolar mixture of CO and steam such that pCO

2

N2 H2

CO(g)  H2 O(g)

0.5 3

2

3.0 2.0

3

pH2O = 4.0 bar, what

will be the partial pressure of H2 at equilibrium? Kp = 0.1 at 400ºC.

0.0104

As Q c z K c , reaction is not in equilibrium

Sol.

Suppose the partial pressure of H2 at equilibrium= p bar CO(g)  H 2 O(g)

As Q c  K c , reaction will proceed in the forward direction.

CO2(g) + H2 (g)

Initial pressure 4.0 bar 4.0 bar Example - 9

At eqm

(4–p)

(4–p)

p

p

The equilibrium constant for the reaction H 2 (g)  Br2 (g)

2HBr(g) at 1024 K is 1.6 ×105. Find

Kp

the equilibrium pressure of all gases if 10.0 bar of HBr is introduced into a sealed container at 1024 K.

Sol.

H2 (g)  Br2 (g). K

2HBr(g)

p2 (4  p)2

0.1(given) ?

p 4p

0.1 0.316

p = 1.264–0.316p or 1.316p=1.264 or p=0.96 bar. Hence,

1 (1.6 u105 )

(pH2 )eq = 0.96 bar Example - 11

Initial 10 bar At eqm. 10–p

Kp

p/2

(p / 2)(p / 2) (10  p) 2

Predict which of the following reaction will have appreciable concentration of reactants and products:

p/2 2

1 p 5 1.6 u10 4(10  p)2

(a) Cl 2 (g) 1 1.6 u105

(b) Cl2 (g)  2NO(g)

402p

1 or 4 u 10 2 p 4 u 10 2

20 or p

20 402

2(10  p) or

Sol.

= p/2 = 2.5×10–2 bar, p HBr

2NO2 Cl(g) Kc

1.8

For reaction (c), as Kc is neither high nor very low, reactants and products will be present in comparable amounts.

Example - 12

4.98 u10 2 bar

Hence, at equilibrium pH2

5 u 10 39

2NOCl(g),Kc 3.7 u108

(c) Cl2 (g)  2NO2 (g)

Taking square root of both sides, we get p 2(10  p)

2Cl(g), K c

2O3 (g), is The value of Kc for reaction, 3O2(g) –50 2.0 × 10 at 25ºC. If the equilibrium concentration of O2 in air at 25ºC is 1.6 × 10–2, what is the concentration of O3?

pBr2 10  p | 10 bar Sol.

8

Kc

O3 O2

2 3

? 2.0 u 10

50

O3

2

1.6 u102

3

CHEMICAL EQUILIBRIUM or [O3]2 = (2.0 × 10–50) (1.6×10–2)3 = 8.192×10–56 or

Example - 15

–28

[O3]= 2.86×10 M

Prove that the pressure necessary to obtain 50% dissociation of PCl5 at 500 K is numerically equal to three times the value of the equilibrium constant, Kp.

Example - 13 CH4(g) + H2O (g), is at The reaction, CO(g)+3H2(g) equilibrium at 1300 K in a 1 L flask. It also contains 0.30 mol of CO, 0.10 mol of H2 and 0.02 mol of H2O and an unknown amount of CH4 in the flask. Determine the concentration of CH4 in the mixture. The equilibrium constant, Kc, for the reaction at the given temperature is 3.90.

Sol.

Initial moles 1

CO H 2

3

0.30 0.10

3

p PCl5

0.5 uP 1.5

P ,PPCl3 3

p Cl2

0.5 uP 1.5

P 3

? Kp

p PCl3 u pCl2

–2

or [CH4]= 0.0585 M=5.85×10 M

p PCl5

(P / 3)(p / 3) (p / 3)

2C(s)  O 2 (g)

2AB(g) + B 2 (g)

2AB 2

with a degree of dissociation, x, which is small compared with unity. Deduce the expression for x in terms of the equilibrium constant Kp and the total pressure, P.

Sol.

2AB(g)  B2 (g)

2AB 2 1 mole

At equilibrium 1 – x

0

0

P 3

P or P 3

1 x u P, p AB 1 2 x 2 p AB u p B2

p 2AB2

2x

x

2x u P, p B2 1  2x 2

2CO(g)

On reducing the volume, the pressure will increase. By Le Chatelier’s principle, equilibrium will shift to the side accompanied by decrease of pressure, i.e., decrease in the number of gaseous moles, i.e., backward direction.

The degree of dissociation is 0.4 at 400 K and 1.0 atm for the gaseous reaction PCl5

x uP 1  2x

§ 2x u P · § x u P ·§ 1  x · uP¸ ¨ ¸ u¨ ¸¨ © 1  2x ¹ © 1  2x ¹© 1  2x ¹

PCl3 + Cl2.

Assuming ideal behaviour of all the gases, calculate the density of the equilibrium mixture at 400 K and 1.0 atmosphere. (Atomic mass of P = 31.0 and Cl = 35.5)

2

PCl5

Sol.

PCl3 + Cl2

3

4x P 1  x 1  2x

or K p

Initial At eqm.

Neglection x in comparison to unity,

3

4x P or x

1 mole 0.6

0 0.4

0 0.4 moles

Average molecular mass of mixture

1/ 3

Kp

3K p.

Example - 17

Total number of moles at equilibrium = 1–x+2x+x=1+2x

? Kp

0.5 uP 1.5

What is the effect of reducing the volume on the system described below?

At emperature T, a compound AB 2(g) dissociates according to the reaction

pAB2

0.5 Total=1.5 moles

Example - 16

Example - 14

Initial

0.5

(molar conc= No. of moles

because volume of flask = 1L)

Sol.

0

If P is the total required pressure, then

CH 4 0.02

? 3.90

0

Moles at eqm. 1–0.5=0.5

CH 4 H 2 O

Kc

PCl3  Cl2

PCl5

Sol.

§ Kp · ¨ ¸ © 4P ¹

9

CHEMICAL EQUILIBRIUM Step 2. To calculate extra CO2 to be added

0.6(31  5 u 35.5)  0.4(31  3u 35.5)  0.4(2 u35.5) 0.6  0.4  0.4 125.1  55  42.6 1.4

Suppose extra CO2 to be added = x mole. Then writing the reverse reaction, we have

148.92

addition

w RT or PM M

nRT

+ H2

0.2 + x

0.6

CO + H2O

initial moles after

For ideal gases,

PV

CO2

w RT V

0.4

Moles (Molar

d RT

Conc.)at new (0.2+x –0.2) (0.6–0.2)

?d

PM RT

1u148.92 = 4.5374 g L–1 0.08205 u 400

equilibrium

=x

Kc '

At 540 K, 0.10 mol of PCl5 are heated in a 8.0 L flask. The pressure of the equilibrium mixture is found to be 1.0 atm. Calculate Kp and Kc for the reaction.

Initial

x

x

Total no. of moles at eqm. = 0.1 + x PV

nRT, i.e., n

? 0.1  x

1u 8 0.0821u 540

PV RT

0.18 or x

0.6 u 0.5 x u 0.4

1 Kc

K c (RT)

0.75 moles

For the dissociation

0.04

N 2 O 4 (g)

Initial moles Moles at eqm.

Kp

x

0.08

(0.08 / 8)(0.08 / 8) (0.02 / 8) 'n

1 or

Sol.

0.18

reaction ? Kc

= 0.5

The degree of dissociation of N2O4 into NO2 at one atmospheric pressure and 313 K is 0–310. Calculate Kp of the dissociation reaction at this temperature. What will be the degree of dissociation at 10 atmospheric pressure at the same temparture?

0.1 mol

At eqm. 0.1 – x

= 0.4

Example - 20

PCl3  Cl2

PCl5

0.6 (0.3+0.2)

(V=1L) (given)

Example - 18

Sol.

0.3

0.04(0.0821u 540)

1

1

2NO2 (g) 0

1–0.310=0.69 2×0.310=0.62 (D 0.310,Given)

1.77

Total = 0.69+0.62=1.31 moles

Example - 19

At one atmospheric

An equilibrium mixture, CO(g)  H 2 O(g)

pressure,

CO 2 (g)  H 2 (g), present in a

vessel of one litre capacity at 1000 K was found to contain 0.4 mole of CO, 0.3 mole of H2O, 0.2 mole of CO2 and 0.6 mole of H2. If it is desired to increase the concentration of Co to 0.6 mole by adding CO2 into the vessel, how many moles of it must be added into equilibrium mixture at constant temperature in order to get this change. Sol.

Kp

CO(g) H 2 (g) CO(g) H 2 O(g)

p 2NO2 p N 2O 4

0.69 u 1 atm, p NO2 1.31 (0.62 /1.31atm) 2 (0.69 /1.31atm)

0.62 u1atm 1.31

0.425atm

At 10 atmoshperic pressure, suppose degree of dissociation = D . Then

Step 1. To calculate Kc of the reaction.

Kc

p N 2 O4

Initial moles

0.2 u 0.6 1 0.4 u 0.3

N 2 O4

2NO 2

1

0

Moles at eqm 1  D

10

2D ,

Total = 1+ D

CHEMICAL EQUILIBRIUM

1 D u 10atm, p NO2 1 D

p N 2 O4

Kp

[20D /(1  D )]2 [10(1  D ) /(1  D)]

?

40D 2 1  D2

2D u 10atm 1 D

40D 2 1  D (1  D )

Example - 23 0.1 mole of N2O4 (g) was sealed in a tube under atmospheric condition at 25ºC. Calculate the number of moles of NO2(g) present if the equilibrium N2O4(g) is reached after some time (Kp = 0.14).

40D 2 1  D2

0.425

Neglecting D2 in comparison to 1, 40 D2 = 0.425

Sol.

N 2 O 4 (g)

2NO2(g)

Initial amount

0.1 mole

0

At equilibrium

(0.1–x)

2x, Total=0.1+x moles

As P=1 atm, p N2O4

or D = 0.103 = 10.3 % Example - 21

Kp

When α - D glucose is dissolved in water, it undergoes mutarotation to form an equilibrium mixture of α - D glucose and β - D glucose containing 63.6% of the latter. Calculate Kc for the mutarotation.

? At equilibrium

Kc

36.4%

63.6 36.4

63.6% ?

1.747

Example - 22 At 77ºC and one atmoshpheric pressure, N2O4 is 70% dissociated into NO2. What will be the volume occupied by the mixture under these conditions if we start with 10g of N2O4? Sol.

Molar mass of

N 2 O4

28  64

92g mol 1

N 2 O4

2NO2

Initial moles

10 92

0

After dissociation

10 70 10  u 92 100 92

p N 2O 4

2×0.076

= 0.109–0.076=0.033 = 0.152

11

4x 2 0.01  x 2

2x atm 0.1  x

(2x /(0.1  x)) 2 (0.1  x) /(0.1  x)

4x 2 (0.1  x)(0.1  x)

E - D glucose

α-D glucose

Sol.

p 2NO2

0.1  x , p NO2 0.1  x

4x 2 0.01  x 2

0.14 or 4.14 x2 = 0.0014 or x=0.018

No. of moles of NO2 at equilibrium = 2x=2×0.018= 0.036 mole

IONIC EQUILIBRIUM

IONIC EQUILIBRIUM Ionic equilibrium is the study of equilibrium in the reactions where formation of ions take place in aqueous solution.

The major drawback of this theory is that the basis of all definitions is water. BOH l B+ + OH–

1. ELECTROLYTES

1.2.2 Bronsted-Lowry concept Acids : Those compounds which can transfer protons that + is H to other compounds

Electrolytes are those compounds which on dissolving in polar solvents like water break into ions. The solution of electrolytes conducts electricity because of the presence of ions.

Bases: Those compounds which can accept protons. That is, Bronsted acids are “proton donors” and Bronsted bases are “Proton acceptors”.

1.1. Classification of electrolytes

HCl

+ H2 O l

H 3O + +

Cl

1.1.1 Electrolytes can be classified on the basis of their strength into two categories:

Acid-1

Strong electrolytes : Those electrolytes which easily break into ions and give almost complete dissociation. Eg. HCl, NaOH, NaCl, HNO3, HClO4, H2SO4, CaCl2 etc

Conjugate Acid-Base pairs: Pairs which are separated by a proton and exhibit opposite behaviours in the two directions of the same reaction. Eg. In the above reaction HCl and Cl are conjugate acid-base pairs.

Weak electrolytes : Those electrolytes which dissociate partially. Eg. CH3COOH, NH4OH, HCN, H2C2O4, and all organic acids and bases etc.

Note : In a conjugate pair if acid is strong the base is weak and vice-versa.

Base-1

Acid-2

Base-2

1.1.2 Electrolytes can be further classified on the basis of the kind of compound they are.

1.2.3 Lewis Concept



Acids



Bases

Base: is a compound which can transfer its lone pair of electrons.



Salts

Eg. BF3 + NH3 l H3N oBF3

1.2 Acids and Bases

In the above reaction BF3 is a lewis acid and NH3 is a lewis base.

Acid: A compound which can accept a pair of electrons.

The definition of acids and bases varies from theory to theory:

2. DISSOCIATION OF WEAK ACIDS AND BASES

1.2.1 Arrhenius theory

HA l H+ + A–

This theory defines acids and bases from the perspective of water as a solvent.

Ka = cD2/(1 – D) Ka is called “ionisation constant” or the “Dissociation constant” of the acid.

Arrhenius acids: Those compounds which will increase + H ion concentration in water. Eg HCl, H2SO4, CH3COOH etc

For low dissociation : 1 – D | 1

HA l H+ + A–

Ka = cD2

Arrehenius bases: Those compounds which will increase – OH ion concentration in water. Eg. NaOH, Ca(OH)2, NH4OH etc

ŸD  ( K a / C) This expression will be valid only when D< 0.05 (5%)

12

IONIC EQUILIBRIUM [H+] = cD =

cK a = [A–]

6. BUFFER SOLUTIONS

Similarly for a Weak base Ka = cD/(1 – D)

Solutions which can resist any change in pH on addition of small amount of acid or base.

K b is called “ionisation constant” or “dissociation constant” for the base. D 

Buffer solutions are of three types:

( K b / C This expression will be valid only when

6.1 Acidic Buffer Eg.

D< 0.05 (5%) [B+] = [OH–] =

CH3COOH + CH3COONa

cK b

Henderson-Hasselbalch equation

Note :

pH = pKa + log [salt]/[acid]

D weak electrolytes increase on dilution

6.1.1 Buffer Range: pKa– 1 dpH d pK + 1 a

3. SELF­IONISATION OF WATER

6.2 Basic Buffer Eg.

H2O lH+ + OH– +

NH4OH and NH4Cl

–

Kw = [H ] [OH ]

pOH = pKb + log [salt]/[base]

Kw is called ionic product of water.

6.2.1 Buffer range:

For pure water: [H+] = [OH–] +

pKb – 1 d pOH d pKb+ 1 –

At 25º C we know that [H ][OH ] = 10 [H+] = [OH–] =

Kw

–14

6.3 Mixed Buffer Eg.

10 7 M

CH3COONH4

Acidic: [H3O+] > [OH–] Neutral: [H3O+] = [OH–]

Note :

Basic: [H3O+] < [OH–]



pH of a buffer solution does not change on dilution.

Note: Kw increases as temperature increases.



For any acid-conjugate base pair Ka.Kb=Kw

4. pH SCALE pH = – log10[H+]

7. POLYPROTIC ACIDS

pH + pOH = pKw = 14 (at 25º)

pH < 7 Acidic

Those acids which can furnish more than one H + permolecule. Eg. H2SO4, H2CO3, H3PO4

pH > 7 Basic

For any polyprotic acid : K1 > K2 > K3 ....so on

pH = 7 Neutral

For dissociation of H2A

At 25ºC

i.e. The concentration of the second ion of a polyprotic acid is almost equal to the second dissociation constant.

5. MIXTURE OF TWO WEAK ACIDS [H+] = (x + y) =

[A2–] | K

c1K 1  c 2 K 2

13

IONIC EQUILIBRIUM

8. SALT HYDROLYSIS

9. SPARINGLY SOLUBLE SALTS & PRECIPITATION

Depending on the nature of the parent acid and base there can be 4 type of salts:

When a salt is dissolved in water then it starts breaking into ions and after sometime the solubility process attains equilibrium.

8.1 Salt of strong acid and strong base (NaCl)

AgCl(s) l Ag+(aq) + Cl–(aq)

This type of salt do not get hydrolysed.

Ksp = [Ag+] [Cl–] = Q = I.P.

Neutral solution with pH = 7

I.P. < KspŸ forward reaction, more salt can be dissolved

8.2 Salt of weak acid and strong base (CH3COONa)

I.P. = K Ÿ saturation, no more salt can be dissolved sp

This type of salt give acidic solution on hydrolysis.

I.P. > Ksp Ÿ backward reaction, precipitation of solid salt will take place

Kh = CDh2/ (1 – Dh) = Kw/Ka ; pH = 7 + ½ pKa – ½ log C 8.3 Salt of Strong acid and weak base (NH4Cl) This type of salt give basic solution on hydrolysis. Kh = CDh2/ (1 – Dh) = Kw/Kb ; pH = 7 – ½ pKb – ½ log C 8.4 Salt of weak acid and weak base (CH3COONH4) This type of salt may give acidic,basic or neutral solution. Kh = Dh2/ (1 – Dh)2 = Kw/(Ka × Kb) ; pH = 7 + ½ pKb – ½ pKb

14

IONIC EQUILIBRIUM

SOLVED EXAMPLES Example - 1

(d)

Calculate the pH of the following solutions:

? M2 = 1.36 × 10–2 M

(a) 2g of TlOH dissolved in water to give 2 litre of the solution

ª¬ H  º¼

(b) 0.3 g of Ca(OH)2 dissolved in water to give 500 mL of the solution

pH

Sol.

2g 1 Molar conc. of TlOH = (204 16 1)gmol1 u 2L

Ka

2.21u 10 12 M

pK a

 log(2.21 u 10 12 ) 12  (0.3424) 11.66

pH

0.3g

1

pH

ª¬ OH ¼º

2 Ca(OH) 2

2 u (8.11 u 10 ) M

Sol.

 log(16.22 u 103 )



ª¬ OH ¼º

pOH

 log(1.74 u 10 3 ) CD

0.1u (0.132)2 1.74 u103 2.76

0.1u 0.132 1.32 u 10 2 M 2  0.1206 1.88

ª¬ OH  ¼º

3.75 u 10 2 M

Kb

2

3.75 u 10 M

2  0.0574 1.43

CodH   OH

9.95 ? pOH 14  9.95

 or log ª¬ OH ¼º

 log(3.75 u10 2 )

CD

3  0.2405

 log(1.32 u 102 )

 log ª¬ OH  ¼º

Molar conc. of NaOH 0.3g 1 u 40g mol 1 0.2L

CD

CD.CD CD2 2 C(1  D) 1  D  CD

Cod  H 2 O pH

3  1.2101 1.79

pH 14  1.79 12.21 (c)

C – CD

0

The pH of 0.005 M codeine (C18H21NO3) solution is 9.95. Calculate the ionization constant and pKb.

3

= 16.22×10–3 M pOH

0

Example - 3

Ca(OH) 2 o Ca 2   2 OH  ?

C

ª¬ H  º¼

3 Molar conc. of Ca(OH)2 = 40  34 gmol1 u 0.5L 8.11u10 M



2  0.1335 1.87

CH 2 (Br)COO   H 

CH 2 (Br)COOH

Conc at eqm

4.52 u10 3 M

ª¬ H  ¼º 1014 /(4.52 u 10 3 )

 log(1.36 u 102 )

Intial Conc

= 4.52 ×10–3 M TlOH

1.36 u10 2 M.

HCl

The degree of ionization of a 0.1 M bromoacetic acid solution is 0.132. Calculate the pH of the solution and the pKa bromoacetic acid.

(d) 1 mL of 13.6 M HCl is diluted with water to give 1 litre of the solution.

? ª¬ OH  º¼

M 2 u 1000mL

Example - 2

(c) 0.3 g of NaOH dissolved in water to give 200 mL of the solution

Sol.

M 2 V2 ?13.6M u 1mL

M1V1

4.05i.e.

4.05 4.05

5.95

or

8.913 u 105 M

ª¬CodH º¼ ª¬OH º¼ Cod

ª¬

 2

º¼ Cod

8.91u105 5 u103

= 1.588 × 10–6

? pH 14  1.43 12.57

pK b

15

 log 1.588 u 10 6

6  0.1987

5.8

2

IONIC EQUILIBRIUM Example - 7

Example - 4

The solubility of Sr(OH)2 at 298 K is 19.23 g/L of solution. Calculate the concentration of strontium and hydroxyl ions and the pH of the solution. (Atomic mass of Sr = 87.6)

Calculate the hydrogen ion concentration in the following biological fluids whose pH are given: (a) Human muscle–fluid, 6.83 (b) Human stomach fluid, 1.2

Sol.

(c) Human blood, 7.38

Solubility of Sr(OH)2 in moles

(d) Human saliva, 6.4 Sol.

(a) log[H  ] ?[H  ]

pH



? [H ]

 pH

7.17

L–1 =

2.8

Sr(OH)2 o Sr2+ + 2 OH– 2

6.31u 10 M

pH

8.62

(c) log[H ]

7.38

Antilog 8.62

(d) log[H  ]

pH

pOH

4.169 u 10 8 M

6.4

Antilog 7.60

? [Sr 2  ] 0.1581M.[OH  ]

The pH of milk, black coffee, tomato juice, lemon juice and egg white are 6.8, 5.0, 4.2, 2.2 and 7.8 respectively. Calculate corresponding hydrogen ion concentration in each.

Sol.

ª¬ H  º¼

or

–5

Similar to Q. 7.55 [Ans 1.5×10 M, 10 M,

If 0.561 g KOH is dissolved in water to give 200 mL of solution at 298 K, calculate the concentration of potassium hydrogen and hydroxyl ions. What is its pH? 0.561 1000 M u 56 200 



0.050M 

As KOH o K  OH ,? ª¬ K ¼º ª¬ H  ¼º

 ¬ª H ¼º

4.57 u 10 3 M

4.57 u 10 3 M

Ka

(4.57 u10 3 )(4.57 u 103 ) 0.1

D

Ka / C

2.09 u 104

2.09 u104 / 0.1 0.0457

Example - 9 

¬ªOH ¼º

The ionic product of water at 310 K is 2.7 ×10–14. What is the pH of neutral water at this temperature?

0.05M

K w / ¬ª OH  ¼º 10 14 / 0.05 1014 /(5 u 102 )

Sol.

= 2.0×10–13 M. +

Anti log 3.66

ª¬ CNO  ¼º

Example - 6

KOH

H   CNO 

HCNO

pH=2.34 mean –log[H+]= 2.34 or log [H+]=–2.34= 3.66

6.31×10–3 M, 1.58×10–8 M]

Sol.

0.5, ? pH 14  0.5 13.5

The pH of 0.1M solution of cyanic acid (HCNO) is 2.34. Calculate the ionization constant of the acid and its degree of ionization in the solution.

3.981 u10 7 M

–7

 log 0.3162

2 u 0.1581 0.3162M

Example - 8

7.60

Example - 5

Sol.

19.23gL1 0.1581M 121.6gmol1

Assuming complete dissociation,

1.2

Antilog 2.8 

?[H  ]

6.83

Antilog 7.17 1.479 u10 7 M

(b) log[H  ]

?[H  ]

Molar mass of Sr(OH)2= 87.6+34=121.6g mol–1

ª¬ H  º¼ pH

–13

pH =–log [H ]=– log (2.0×10 )= 13–0.3010= 12.699

 log ª¬ H  ¼º

7  0.2156

16

2.7 u10 14

Kw

1.643 u107 M

 log(1.643 u 107 )

6.78

IONIC EQUILIBRIUM Example - 13

Example - 10

Prove that the degree of dissociation of a weak monoprotic acid is given by

+

One the basis of the equation pH =– log [H ], the pH of 10–8 mol dm–3 solution of HCl should be 8. However, it is observed to be less than 7.0. Explain the reason. Sol.

D

pH of 10 –8 M HCl solution is not 8 because this concentration is so low that H+ ion produced from H2O in the solution (viz. 10–7M) cannot be neglected. Total [H+] = 10–8 + 10–7 M. Solving and calculating pH, we get the value close to 7 but less than 7 as the solution is acidic.

1 where Ka is the dissociation constant 1  10(pK a  pH)

of the acid. Sol.

Suppose we start with C mol L–1 of the weak monoprotic acid HA. Then H+ + A–

HA Example - 11

Initial molar conc. C

pH of a solution of a strong acid is 5.0. What will be the pH of the solution obtained after diluting the given solution 100 times? Sol.

+

pH = 5 means [H ]= 10 5

10 100

 ¬ª H º¼

–5

M. On diluting 100 times,

 Also, ª¬ H º¼

7  0.3010

ª¬ H  ¼º

6.699.

?

or or log [H+] = – 2.85 = 3.15 or

or [H+] = antilog 3.15 = 1.413 × 10–3 M

Ka

ª¬ H º¼ HA

2

1.413 u10 0.08

K a (1  D) uD D2

H+ + A–

K a (1  D ) D

[log Ka  log(1  D)  log D]

pK a  log(1  D )  log D

log

or

pH of HOCl=2.85 i.e., –log [ H+] = – 2.85



Ka (1  D) ..(i) D2

CD

 log ¬ªH ¼º

or pH

pH of 0.08 mol dm–3 HOCl solution is 2.85. Calculate its Ionisation constant.

For weak monobasic acid HA

CD 2 or C 1 D

Substituning the value of C from eqn. (i), we get

Example - 12

Sol.

CD.CD C(1  D)

Thus, Ka

107 M

 log(2 u 10 7 )

CD CD

after dissociation= C(1– D )

This should give pH =7 but it cannot be so because solution is acidic and pH should be less than 7. The reason is that [H+] from H2O cannot be neglected. Thus, total [H+] = 10–7 M (from HCl) + 10–7 M (from H2O) = 2 × 10–7 M ? pH

C– CD

Molar conc

1 D D

pK a  pH or

1 D 10pK a pH D

1  1 10 pK a pH D 1 1 1  10pK a pH or D D 1  10pK a pH

Example - 14 Arrange the given compounds in the decreasing order of basicity on the basis of Bronsted–Lowry concept: BaO, CO2, SO3, B2O3, Cl2O7

3 2

= 2.4957 × 10–5. Sol

BaO + H2O CO 2  H 2 O

17

Ba(OH)2 (Basic) H 2 CO3 (weakly acidic)

IONIC EQUILIBRIUM SO 3  H 2 O

H 2SO 4 (Strongly acidic)

B2 O3  3H 2 O

2H3 BO3 (Very weaky acidic)

Cl 2 O 7  H 2 O

2HClO4 (Very strongly acidic)

Example - 18 Why PO 43- ion is not amphiprotic? Sol.

Hence, in the decreasing order of basicity, we gave

well as accept proton. PO 34 ion can accept proton(s) but

BaO ! B 2 O 3 ! CO 2 ! SO 3 ! Cl 2 O7 .

cannot donate any proton. Hence, PO34 is not amphiprotic.

Example - 15

Example - 19

What are the conjugate bases of the following?

In the reaction between BF3 and C2H5OC2H5 which one of them will act as an acid? Justify your anwer.

3+

CH3OH, HN3, [Al(H2O)6] . Sol.

An amphiprotic ion is one which can donate proton as

Sol.

CH 3O  N 3 [Al(H 2 O)5 OH]2 (H+ ion has been removed

The reaction between BF3 and C2H5OC2H5 is

methoxideion , azideion

from one H2O molecule). Example - 16 Glycine is an D –amino acid which exists in the form of +

Zwitter ion as N H 3CH 2COO- . Write the formula of

As BF3 accepts a pair of electrons, hence BF3 is the Lewis acid.

its conjugate base. Sol.

Example - 20

Conjugate acid = 

Which is a stronger base in each of the following pairs and why?



N H 3 CH 2 COO   H 

N H3 CH 2 COOH

(i) H2O, Cl– (ii) CH3COO–, OH– Conjugate base = 

N H 3 CH 2 COO   H 

Sol

NH 2 CH 2 COO 

(i) H2O (ii) OH– Refer to Bronsted – Lowry concept for relative strengths.

Example - 17

Example - 21

Write reaction for autoprotolysis of water. How is ionic product of water related to ionization constant of water? Derive the relationship. Sol.

Classify the following species as Lewis acids and Lewis bases NH3, BF3, SnCl4, C5H5N, CO, Ni2+

Autoprotolysis of H2O takes place as: Sol. 

H2O  H2O

H 3 O  OH

For ionization of H2O, H 2 O 



Ki

[H ][OH ] [H 2 O]

Kw

K i u 55.55.

K2 [H 2 O]

Lewis acids: BF3, SnCl4, Ni2+



Lewis bases: NH3, C5H5N, CO. Example - 22

H   OH 

Explain why pH of 0.1 molar solution of acetic acid will be higher than that of 0.1 molar solution of HCl?

Kw or 55.55mol L1

Sol.

18

Acetic acid is a weak electrolyte. It is not completely ionized and hence gives less H+ ion concentration. HCl is a strong acid. It is completely ionized giving more H + ion concentration. As pH =–log [H+]; less the [H+], greater will be the pH.

IONIC EQUILIBRIUM HA + OH–

Example - 23 Calculate the approximate pH of a 0.100 M aqueous H2S solution. K1 and K2 for H2S are 1.00 ×10–7 and 1.30 ×10–13 respectively at 25ºC,

A– + H2O

[A  ] [HA][OH  ]

K

.......(i)

Further, for the weak acid, HA Sol.

K2 < < K1. Hence H+ ions are mainly from 1st dissociation, H+ +HS– i.e. H2S ª¬ H º¼ ª¬ HS º¼ H 2S 

K1



[H  ][A  ] [HA]

......(ii)

Also Kw = [H+] [ OH–]

.....(iii)

Ka

2

ª¬ H º¼ or ª¬ H  º¼ H 2S



K1[H 2S]

H+ + A–,

From eqns. (i), (ii) and (iii), ? [H  ]

107 u 10 1

10 4

Ka Kw

K Hence, pH=4 Example - 24

The pH of 0.05 M aqueous solution of diethylamine is 12.0. Calculate its Kb Sol.

At the equivalence point, CH3COONa is formed and its 0.1 M 0.05M . It is a salt of weak acid 2 and strong base. The formula for finding the pH of such a salt is

(C 2 H 5 ) 2 NH  H 2 O

[(C 2 H 5 ) 2 NH] 0.05  0.01 0.04M [(C 2 H 5 ) 2 N H 2 ][O H  ] [(C 2 H 5 ) 2 N H ]

K

1  [log K 2  log K a  log c] 2

An aqueous solution contains 10% ammonia by mass and has a densityof 0.99 g cm–3. Calculate hydroxyl and hydrogen ion concentration in this solution. Ka for NH 4+ = 5.0×10-10 M

1 (14  5  0.2788  2  0.6990) 2

Sol.

10 1 u u 1000 = 5.82 M 17 100 / 0.99

Example - 25 A certain weak acid has Ka = 1.0×10–4. Calculate the equilibrium constant for its reaction with a strong base.

weak

10% ammonia by mass means 10g NH3 are present in 100g of the solution. ? Molarity of the solution

8.71

HA  BOH strong

+

NH 3  H 2 o NH 4 OH U NH 4  OH 

BA + H2O

or HA + B + OH

–

+

2.5 u 10  3

Example - 27

1  [14  (5  0.2788)  (2  0.6990)] 2

Sol.

10 2 u 10  2 0.04

[(C 2 H 5 ) 2 NH 2 ] [OH  ]

1 ? pH  [log1014  log(1.9 u105 ) –log (5×10–2)] 2

17.42 2

(C 2 H 5 ) 2 NH 2  OH 

As pH = 12, ? [H  ] 1012 M or [OH – ] = 10 –2 M,

concentration =

pH

1010

Example - 26

Calculate the pH at equivalence point when a solution of 0.10 M acetic acid is titrated with a solution of 0.10 M NaOH solution, Ka for acetic acid = 1.9×10–5 Sol.

104 1014

–

B + A +H2 O or

Initial conc.

C mol L–1

After dissociation

C – C D = C (1 –D)

19

C D

CD

IONIC EQUILIBRIUM

 ? [OH ] C D

Total volume after mixing = 500 ml KhC

? In the final solution, after mixing , [OH – ]=

102 5

10 14 u 5.82 = 1.079 × 10–2 M 5.0 u10 10

Kw uC Ka

?[H  ]

Kh C

C

10 14 1.079 u10 2

= 0.9268 × 10

–12

Example - 30 –13

M = 9.268 × 10

Calculate the degree of dissociation of 0.5 M NH3 at 25ºC in a solution of pH=12.

M

The ionization constant of NH 4+ in water is

Initial conc. C mol L–1

0

0

5.6 × 10–10 at 25ºC. The rate constant for reaction of

After disso. C  CD

CD

CD

–

10

NH and OH to form NH3 and H2O at 25ºC is 3.4×10 s

pH 12 means[H  ] 10 12 or [OH  ] 10 2

litre mol–1 sec–1. Calculate the rate constant for prton transfer from water of NH3. Sol.

NH 4  OH 

NH 4 OH

Sol.

Example - 28

+ 4

NH +4  H 2 O

? [OH  ] CD 102 or D

NH4OH+H+, Ka = 5.6 ×10–10 NH 4  OH  , k

NH3 + H2O

5 u 10 12

 log[H  ]  log(5 u10 12 ) 12  0.69 11.31

? pH Kw [OH  ]

1014 2 u 10 3

2 u 103 M or[H  ]

102 0.5

3.4 u 1010

102 C

2 u 10 2 or 2%.

Aim. To find kf Example - 31

We know that for a conjugate acid–base pair K acid u K base

?K

Kw Ka

K w ,i.e., K a u K b

Calculate the ratio of pH of a solution containing 1 mole of CH3COONa + 1 mole of HCl per litre to that of a solution containing 1 mole of CH3COONa + 1 mole of CH3 COOH per litre.

Kw

10 14 5.6 u10 10 Sol.

But K

kf ?k f k

Kb u k

10 14 u3.4 u1010 5.6 u10 10

Case I. Calculation of pH of solution containing 1 mole of CH3COONa + 1 mole of HCl per litre CH 3COONa  HCl o CH3 COOH  NaCl

Initial moles 1 mole

0.607 u106

6.07 u105

0

0

Moles after reaction

Example - 29

0

0

1

1

i.e [CH 3 COOH] 1mol L1

What will be the resultant pH when 200 ml of an aqueous solution of HCl (pH=2.0) is mixed with 300 ml of an aqueous solution of NaOH (pH=12.0)? Sol.

1 mole

CH 3 COOH

pH = 2 means [H+] = 10–2M

Initial conc.

pH = 12 means [H+]= 10–12or [OH–]= 10–2 M

After dissociation

Thus, 200 ml of 10–2 M HCl are mixed with 300 ml of 10–2 M NaOH. After neutralisation NaOH left = 100 ml of 10–2 M

+

H+

C mol L–1 C  CD

? [H  ] CD . But D

20

CH3COO–

CD Ka C

CD

IONIC EQUILIBRIUM

? [H ] C

Ka C

(ii) When 3/4th of the acid has been neutralized Ka C

Ka

K1/a 2

1 ?  log[H  ]  log K a ,i.e.(pH)1 2

1

('C 1molL )

CH 3COOH  NaOH o CH 3 COONa  H 2 O

1  log K a ...(i) 2

Initial conc. 0.1M

Case II. Calculation of pH of solution containing 1 mole of CH3COONa + 1 mole of CH3 COOH per litre Applying Henderson equation,

(pH)2 = pKa + log

[Salt] [Acid]

pK a

 log K a ...(ii)

1 After 3/4th 0.1 u M 4

3 0.1 u M 4

neutralization = 0.025 M

= 0.075 M

? pH = – log 10–5 + log

0.075 = 5 + 0.4771 = 5.4771. 0.025

Example - 33

[Salt] [Acid] 1mol L1 ]

Calculate the amount of (NH4)2 SO4 in g which must be added to 500 mL of 0.2 M NH3 to yield a solution of pH = 9.35. Kb for NH3 = 1.78×10–5.

Example - 32 0.1 M CH3 COOH (pH=3) is titrated with 0.05 M NaOH solution. Calculate the pH when (i) 1/4th of the acid has been neutralized. (ii)3/4th of acid has been neutralized. Sol.

Sol.

As it is a basic buffer, pOH = pK b + log

Calculation of dissociation constant of the acid  log K  log

As pH = 3, ?[H ] 103 M, [CH3COO ] [H ] 103 M

Ka

[CH 3COO  ][H  ] [CH3 COOH]

103 u103 0.1

[NH 4 ] [NH 4 OH]

As pH = 9.35, ?pOH 14  9.35 10

[Salt] [Base]

4.65

5

Millimoles of NH4OH is solution = 0.2×500 = 100 Suppose millimoles of NH 4 to be added=x

(i) When 1/4th of the acid has been neutralized ?

CH 3COOH  NaOH o CH 3COONa  H 2 O Initial conc. 0.1 M

4.65

 log(1.78 u 105 )  log

(5  0.2504)  log

After 1/4th 0.1u

3 4

0.1u

0.075M ?pH

pK a  log

1 4

x 100

x / 500 100 / 500

x 100

0.0996 1.004  0.1

or

log

or

log x = 2.1 or x = 125.9

?

Millimoles of (NH4)2SO4 to be added =

0.025M [Salt] [Acid]

 log10 5  log

0.025 0.075

125.9 62.95 2

( '1 millimole of (NH 4 )2 SO 4 { 2 millimoles of NH 4+ ) ?

5  0.4771 4.5229

21

Mass of (NH4)2SO4 to be added = (62.95 × 10–3 moles) (132 g mol–1) = 8.3094 g.

SOME BASIC CONCEPTS OF CHEMISTRY

SOME BASIC CONCEPTS OF CHEMISTRY

2. MATTER

1. CHEMISTRY

Matter is defined as any thing that occupies space possesses mass and the presence of which can be felt by any one or more of our five senses.

Chemistry is defined as the study of the composition, properties and interaction of matter. Chemistry is often called the central science because of its role in connecting the physical sciences, which include chemistry, with the life sciences and applied sciences such as medicine and engineering.

Matter can exist in 3 physical states viz. solid, liquid, gas. Solid - a substance is said to be solid if it possesses a definite volume and a definite shape, e.g., sugar, iron, gold, wood etc.

Various branches of chemistry are

Liquid- A substance is said to be liquid, if it possesses a definite volume but no definite shape. They take up the shape of the vessel in which they are put, e.g., water, milk, oil, mercury, alcohol etc.

1.1 Physical chemistry The branch of chemistry concerned with the way in which the physical properties of substances depend on and influence their chemical structure, properties, and reactions.

Gas- a substance is said to be gaseous if it neither possesses definite volume nor a definite shape. This is because they fill up the whole vessel in which they are put, e.g., hydrogen, oxygen etc.

1.2 Inorganic chemistry

The three states are interconvertible by changing the conditions of temperature and pressure as follows

The branch of chemistry which deals with the structure, composition and behavior of inorganic compounds. All the substances other than the carbon-hydrogen compounds are classified under the group of inorganic substances. 1.3 Organic chemistry The discipline which deals with the study of the structure, composition and the chemical properties of organic compounds is known as organic chemistry. 1.4 Biochemistry The discipline which deals with the structure and behavior of the components of cells and the chemical processes in living beings is known as biochemistry.

3. CLASSIFICATION OF MATTER AT

1.5 Analytical chemistry

MACROSCOPIC LEVELL

The branch of chemistry dealing with separation, identification and quantitative determination of the compositions of different substances.

At the macroscopic or bulk level, matter can be classified as (a) mixtures (b) pure substances.

22

SOME BASIC CONCEPTS OF CHEMISTRY These can be further sub-divided as shown below Pure substances can be further classified into elements and compounds. Element- An element is defined as a pure substance that contains only one kind of particles. Depending upon the physical and chemical properties, the elements are further subdivided into three classes, namely (1) Metals (2) Nonmetals and (3) Metalloids. Compound- A compound is a pure substance containing two or more than two elements combined together in a fixed proportion by mass. Further, the properties of a compound are completely different from those of its constituent elements. Moreover, the constituents of a compound cannot be separated into simpler substances by physical methods. They can be separated by chemical methods.

(a) Mixtures : A mixture contains two or more substances present in it (in any ratio) which are called its components. A mixture may be homogeneous or heterogeneous.

4. PROPERTIES OF MATTER Every substance has unique or characteristic properties. These properties can be classified into two categories – physical properties and chemical properties.

Homogeneous mixture- in homogeneous mixture the components completely mix with each other and its composition is uniform throughout i.e it consist of only one phase. Sugar solution and air are thus, the examples of homogeneous mixtures.

4.1 Physical Properties

Heterogeneous mixtures- In heterogeneous mixture the composition is not uniform throughout and sometimes the different phases can be observed. For example, grains and pulses along with some dirt (often stone) pieces, are heterogeneous mixtures.

Physical properties are those properties which can be measured or observed without changing the identity or the composition of the substance. Some examples of physical properties are color, odor, melting point, boiling point, density etc. 4.2 Chemical properties

Any distinct portion of matter that is uniform throughout in composition and properties is called a Phase.

Chemical properties are those in which a chemical change in the substance occurs. The examples of chemical properties are characteristic reactions of different substances; these include acidity or basicity, combustibility etc.

(b) Pure substances :- A material containing only one substance is called a pure substance.

5. MEASUREMENT 5.1 Physical quantities In chemistry, a substance is a form of matter that has constant chemical composition and characteristic properties. It cannot be separated into components by physical separation methods, i.e. without breaking chemical bonds. They can be solids, liquids or gases.

All such quantities which we come across during our scientific studies are called Physical quantities. Evidently, the measurement of any physical quantity consists of two parts

23

SOME BASIC CONCEPTS OF CHEMISTRY International d’Unités – abbreviated as SI) was established by the 11th General Conference on Weights and Measures (CGPM from Conference Generale des Poids at Measures). The CGPM is an inter governmental treaty organization created by a diplomatic treaty known as Meter Convention which was signed in Paris in 1875.

(1) The number, and (2) The unit A unit is defined as the standard of reference chosen to measure any physical quantity. 5.2 S.I. UNITS

The SI system has seven base units and they are listed in table given below.

The International System of Units (in French Le Systeme

These units pertain to the seven fundamental scientific quantities. The other physical quantities such as speed, volume, density etc. can be derived from these quantities. The definitions of the SI base units are given below : Definitions of SI Base Units

Unit of length

metre

The metre is the length of the path travelled by light in vacuum during a time interval of 1/299 792 458 of a second.

Unit of mass

Kilogram

The kilogram is the unit of mass; it is equal to the mass of the internationl prototype of the kilogram.

Unit of time

second

The second is the duration of 9 192 631 770 periods of the radiation corresponding to the transition between the two hyperfine levels of the ground state of the caesium-133 atom.

Unit of electric current

ampere

The ampere is that constant current which, if maintained in two straight parallel conductors of infinite length, of negligible circular cross-section, and placed 1 metre apart in vacuum, would produce between these conductors a force equal to 2 × 10–7 newton per metre of length.

Unit of thermodynanic temperature

kelvin

The kelvin, unit of thermodynamic temperature, is the fraction 1/273. 16 of the thermodynamic temperature of the triple point of water.

Unit of amount of substance

mole

1.

The mole is the amount of substance of a system which contains as many elementary entities as there are atoms in 0.012 kilogram of carbon-12; its symbol is “mol.”.

2.

When the mole is used, the elementary entities must be specified and may be atoms, molecules, ions, electrons, other particles, or specified groups of such particles.

Unit of luminous intensity

The candela is the luminous intensity, in a given direction, of a source that emits monochromatic radiation of frequency 540 × 1012 hertz and that has a radiant intensity in that direction of 1/683 watt per steradian.

candela

24

SOME BASIC CONCEPTS OF CHEMISTRY

7. LAW OF CHEMICAL COMBINATION The mass standard is the kilogram since 1889. It has been defined as the mass of platinum-iridium (Pt-Ir) cylinder that is stored in an airtight jar at International Bureau of Weights and Measures in Sevres, France. Pt-Ir was chosen for this standard because it is highly resistant to chemical attack and its mass will not change for an extremely long time.

7.1 Law of conservation of mass “In a chemical reaction the mass of reactants consumed and mass of the products formed is same, that is mass is conserved.” This is a direct consequence of law of conservation of atoms. This law was put forth by Antoine Lavoisier in 1789.

6. SOME IMPORTANT DEFINITION

7.2 Law of Constant / Definite Proportions The ratio in which two or more elements combine to form a compound remains fixed and is independent of the source of the compound. This law was given by, a French chemist, Joseph Proust.

6.1 Mass and Weight Mass of a substance is the amount of matter present in it while weight is the force exerted by gravity on an object. The mass of a substance is constant whereas its weight may vary from one place to another due to change in gravity. The SI unit of mass is the kilogram (kg). The SI derived unit (unit derived from SI base units) of weight is newton.

7.3 Law of Multiple Proportions When two elements combine to form two or more compounds then the ratio of masses of one element that combines with a fixed mass of the other element in the two compounds is a simple whole number ratio. This law was proposed by Dalton in 1803.

6.2 Volume Volume is the quantity of three-dimensional space enclosed by some closed boundary, for example, the space that a substance (solid, liquid, gas, or plasma) or shape occupies or contains. Volume is often quantified numerically using the SI derived unit, the cubic meter.

7.4 Law of Reciprocal Proportions When three elements combine with each other in combination of two and form three compounds then the ratio of masses of two elements combining with fixed mass of the third and the ratio in which they combine with each other bear a simple whole number ratio to each other. This Law was given by Richter in 1792.

6.3 Density The mass density or density of a material is defined as its mass per unit volume. The symbol most often used for density is U (the lower case Greek letter rho). SI unit of density is kg m–3.

7.5 Gay Lussac’s Law of Gaseous Volumes

6.4 Temperature

This law was given by Gay Lussac in 1808. He observed that when gases combine or are produced in a chemical reaction they do so in a simple ratio by volume provided all gases are at same temperature and pressure.

Temperature is a physical property of matter that quantitatively expresses the common notions of hot and cold. There are three common scales to measure temperature — °C (degree celsius), °F (degree fahrenheit) and K (kelvin). The temperature on two scales is related to each other by the following relationship:

7.6 Avogadro Law In 1811, Avogadro proposed that equal volumes of gases at the same temperature and pressure should contain equal number of molecules.

°F = 9/5 (°C) + 32 K = °C + 273.15

25

SOME BASIC CONCEPTS OF CHEMISTRY

8. DALTON’S ATOMIC THEORY

9.3 Method 2 Mass of 6.022 × 10 23 atoms of that element taken in grams.

In 1808, Dalton published ‘A New System of Chemical Philosophy’ in which he proposed the following: 1.

Matter consists of indivisible atoms.

2.

All the atoms of a given element have identical properties including identical mass. Atoms of different elements differ in mass.

3. 4.

This is also known as molar atomic mass.

“ Mass of 1 atom in amu and mass of 6.022 × 1023 atoms in grams are numerically equal.

Compounds are formed when atoms of different elements combine in a fixed ratio.

“ When atomic mass is taken in grams it is also called the molar atomic mass.

Chemical reactions involve reorganization of atoms. These are neither created nor destroyed in a chemical reaction.

“ 6.022 × 1023 is also called 1 mole of atoms and this number is also called the Avogadro’s Number.

9. ATOM

“ Mole is just a number. As 1 dozen = 12; Atom is the smallest part of an element that can participate in a chemical reaction. {Note : This definition holds true only for non-radioactive reactions}

1 million = 10 6; 1 mole = 6.022 × 1023.

10. MOLECULES

9.1 Mass of an Atom

A group of similar or dissimilar atoms which exist together in nature is known as a molecule. e.g. H2, NH3.

There are two ways to denote the mass of atoms. 9.2 Method 1

The mass of molecules is measured by adding the masses of the atoms which constitute the molecule. Thus, the mass of a molecule can also be represented by the two methods used for measuring the mass of an atom viz. amu and g/mol.

Atomic mass can be defined as a mass of a single atom which is measured in atomic mass unit (amu) or unified mass (u) where 1 a.m.u. = 1/12th of the mass of one C12 atom

26

SOME BASIC CONCEPTS OF CHEMISTRY is called the Excess Reagent. e.g. if we burn carbon in air (which has an infinite supply of oxygen) then the amount

11. CHEMICAL REACTIONS

of CO2 being produced will be governed by the amount of

A chemical reaction is only rearrangement of atoms. Atoms

carbon taken. In this case, Carbon is the LR and O2 is the

from different molecules (may be even same molecule) rearrange

ER.

themselves to form new molecules.

13. PERCENT YIELD Points to remember :

As discussed earlier, due to practical reasons the amount of

“ Always balance chemical equations before doing any calculations

product formed by a chemical reaction is less than the amount

“ The number of molecules in a reaction need not to be conserved e.g.

of product formed to the amount predicted when multiplied

predicted by theoretical calculations. The ratio of the amount by 100 gives the percentage yield.

N2 + 3 H2 Æ 2 NH3. The number of molecules is not conserved

Actual Yield Percentage Yield = Theoretical Yield × 100

If we talk about only rearrangement of atoms in a balanced chemical reaction then it is evident that the mass of the atoms in the reactants side is equal to the sum of the masses of the atoms on the products side. This is the Law of Conservation of Atoms and Law of Conservation of Mass.

14. REACTIONS IN AQUEOUS MEDIA Two solids cannot react with each other in solid phase and hence need to be dissolved in a liquid. When a solute is dissolved in a solvent, they co-exist in a single phase called the solution. Various parameters are used to measure the strength of a solution.

12. STOICHIOMETRY

The strength of a solution denotes the amount of solute which is contained in the solution. The parameters used to denote the strength of a solution are:

The study of chemical reactions and calculations related to it is called Stoichiometry. The coefficients used to balance the reaction are called Stoichiometric Coefficients.

“ Mole fraction X : moles of a component / Total moles of solution.

Points to remember :

“ Mass% : Mass of solute (in g) present in 100g of solution.

“ The stoichiometric coefficients give the ratio of molecules or moles that react and not the ratio of masses.

“ Mass/Vol : Mass of solute (in g) present in 100mL of solution

“ Stoichiometric ratios can be used to predict the moles of product formed only if all the reactants are present in the stoichiometric ratios.

“ v/v : Volume of solute/volume of solution {only for liq-liq solutions}

Practically the amount of products formed is always less than the amount predicted by theoretical calculations

“ g/L : Wt. of solute (g) in 1L of solution

mass of solute 6 “ ppm : mass of solution u 10

12.1 Limiting Reagent (LR) and Excess Reagent (ER)

moles of solute “ Molarity (M) : volume of solution (L)

If the reactants are not taken in the stoichiometric ratios then the reactant which is less than the required amount determines how much product will be formed and is known

moles of solute “ Molality (m) : mass of solvent (kg)

as the Limiting Reagent and the reactant present in excess

27

SOME BASIC CONCEPTS OF CHEMISTRY

IMPORTANT RELATIONS 1.

e.g. molality remains unchanged with temperature. Formulae involving volume are altered by temperature e.g. Molarity.

Relation between molality (m) Molarity (M), density (d) of solution and molar mass of solute (MO)

17. INTRODUCTION TO EQUIVALENT CONCEPT

d : density in g/mL

Equivalent concept is a way of understanding reactions and processes in chemistry which are often made simple by the use of Equivalent concept.

MO : molar mass in g mol–1 Molality, m

2.

M u 1000 1000d  MM O

17.1 Equivalent Mass “The mass of an acid which furnishes 1 mol H+ is called its Equivalent mass.”

Relationship between molality (m) and mole fraction (XB) of the solute m

XB 1000 u 1  XB MA

m

“The mass of the base which furnishes 1 mol OH– is called its Equivalent mass.”

1  X A 1000 u XA MA

17.2 Valency Factor (Z) Points to remember : Valency factor is the number of H+ ions supplied by 1

“ Molarity is the most common unit of measuring strength of solution.

molecule or mole of an acid or the number of OH- ions supplied by 1 molecule or 1 mole of the base.

“ The product of Molarity and Volume of the solution gives the number of moles of the solute, n = M × V

Equivalent mass, E

“ All the formulae of strength have amount of solute. (weight or moles) in the numerator.

Molecular Mass Z

17.3 Equivalents

“ All the formulae have amount of solution in the denominator except for molality (m). No. of equivalents =

15. DILUTION LAW

wt. of acid / base taken Eq. wt.

When a solution is diluted, more solvent is added, the moles of solute remains unchanged. If the volume of a solution having a Molarity of M1 is changed from V1 to V2 we can

It should be always remembered that 1 equivalent of an acid reacts with 1 equivalent of a base.

write that: M1V1 = moles of solute in the solution = M2V2

16. EFFECT OF TEMPERATURE

18. MIXTURE OF ACIDS AND BASES

Volume of the solvent increases on increasing the

Whenever we have a mixture of multiple acids and bases we can find whether the resultant solution would be acidic or basic by using the equivalent concept. For a mixture of multiple acids and bases find out the equivalents of acids

temperature. But it shows no effect on the mass of solute in the solution assuming the system to be closed i.e. no loss of mass.

and bases taken and find which one of them is in excess. The formulae of strength of solutions which do not involve volume of solution are unaffected by changes in temperature.

28

SOME BASIC CONCEPTS OF CHEMISTRY

22. EQUIVALENT VOLUME OF GASES

19. LAW OF CHEMICAL EQUIVALENCE

Equivalent volume of gas is the volume occupied by 1 equivalent of a gas at STP.

According to this law, one equivalent of a reactant combines with one equivalent of the other reactant to give one equivalent of each product . For, example in a reaction aA + bB o cC + dD irrespective of the stoichiometric coefficients, 1 eq. of A reacts with 1 eq. of B to give 1 eq. each of C and 1eq of D

Equivalent mass of gas = molecular mass /Z. Since 1 mole of gas occupies 22.4L at STP therefore 1 equivalent of a gas will occupy 22.4/Z L at STP. e.g. Oxygen occupies 5.6L, Chlorine and Hydrogen occupy 11.2L.

20. EQUIVALENT WEIGHTS OF SALTS

23. NORMALITY

To calculate the equivalent weights of compounds which are neither acids nor bases, we need to know the charge on the cation or the anion. The mass of the cation divided by the charge on it is called the equivalent mass of the cation and the mass of the anion divided by the charge on it is called the equivalent mass of the anion. When we add the equivalent masses of the anion and the cation, it gives us the equivalent mass of the salt. For salts, Z in the total amount of positive or negative charge furnished by 1 mol of the salt.

The normality of a solution is the number of equivalents of solute present in 1L of the solution.

N

equivalents of solute volume of solution (L)

The number of equivalents of solute present in a solution is given by Normality × Volume (L). On dilution of the solution the number of equivalents of the solute is conserved and thus, we can apply the

21. ORIGIN OF EQUIVALENT CONCEPT

formula : N1V1 = N2V2

Equivalent weight of an element was initially defined as weight of an element which combines with 1g of hydrogen. Later the definition wad modified to : Equivalent weight of an element is that weight of the element which combines with 8g of Oxygen.

Caution : Please note that the above equation gives rise to a lot of confusion and is a common mistake that students make. This is the equation of dilution where the number of equivalents are conserved. Now, since one equivalent of a reactant always reacts with 1 equivalent of another reactant a similar equation is used in problems involving titration of acids and bases. Please do not extend the same logic to molarity.

Same element can have multiple equivalent weights depending upon the charge on it. e.g. Fe2+ and Fe3+.

Relationship between Normality and Molarity N = M × Z ; where ‘Z’ is the Valency factor

29

SOME BASIC CONCEPTS OF CHEMISTRY

SOLVED EXAMPLES Example - 1

Example - 5

Classify the following substances into elements, compounds and mixtures.

What is the S.I. unit of mass ? Sol. S.I. unit of mass is kilogram (kg).

(i) Air (ii) Diamond (iii) LPG (iv) Dry ice (v) Graphite (vi) Steel (vii) Marble (viii) Smoke (ix) Glucose (x) Laughing gas.

Example - 6 In the reaction, A + B2 o AB2, identify the limiting reagent, if any, in the following mixtures

Sol. Elements

: Diamond; Graphite

(i) 300 atoms of A + 200 molecules B2

Compounds : Marble; Glucose; Laughing gas; Dry ice Mixtures

(ii) 2 mol A + 3 mol B2

: Air; LPG; Steel; Smoke

(iii) 100 atoms of A + 100 molecules of B2

Example - 2

(iv) 5 mol A + 2.5 mol B2

Classify the following mixtures as homogeneous and heterogeneous.

(v) 2.5 mol A + 5 mol B2

(i) Air (ii) Smoke (iii) Petrol (iv) Sea water (v) lodized table salt (vi) Aerated water (vii) Mixture of sand and common salt (viii) Gun powder (ix) Milk (x) Muddy water.

Sol. (i) According to the given reaction, 1 atom of A reacts with 1 molecule of B2.

Sol. Homogeneous : Air; Petrol; Iodized table salt; Sea water; Aerated water; Milk.

? 200 molecules of B2 will react with 200 atoms of A and 100 atoms of A will be left unreacted. Hence, B2 is the limiting reagent while A is the excess reagent.

Heterogeneous : Smoke; Gun powder; Mixture of sand common salt; Muddy water.

(ii) According to the given reaction, 1 mol of A reacts with 1 mol of B2 . Hence A is limiting reagent. (iii) No limiting reagent.

Example - 3 Why Law of conservation of mass should better be called as Law of conservation of mass and energy ?

(iv) 2.5 mol of B2 will react with 2.5 mol of A. Hence, B2 is the limiting reagent.

Sol. In nuclear reactions, it is observed that the mass of the products is less than the mass of the reactants. The difference of mass, called the mass defect, is converted into energy according to Einstein equation, E = ' m c2. Hence, we better call it as a low of conservation of mass and energy.

(v) 2.5 mol of A will react with 2.5 mol of B2. Hence, A is the limiting reagent. Example - 7 Is the law of constant composition true for all types of compounds ? Explain why or why not.

Example - 4

Sol. No, law of constant composition is not true for all types of compounds. It is true only for the compounds obtained from one isotope. For example, carbon exists in two common isotopes, 12C and 14C.

If the speed of light is 3.0 × 108 m s–1, calculate the distance covered by light in 2.00 ns. Sol. Distance covered = Speed × Time = 3.0 × 108m s–1 × 2.00 ns = 3.0 × 108m s–1 × 2.00 ns ×

10 9 s = 6.00 × 10–1m 1 ns = 0.600 m

30

SOME BASIC CONCEPTS OF CHEMISTRY Example - 8

Example - 11

Why atomic masses are the average values ?

Carbon and oxygen are known to form two compounds. The carbon content in one of these is 42.9% while in the other it is 27.3%. Show that this data is in agreement with the law of multiple proportions.

Sol. Most of the elements exist in different isotopes, i.e., atoms with different masses, e.g., Cl has two isotopes with mass numbers 35 and 37 existing in the ratio 3 : 1 Hence, average value is taken.

Sol. Oxide 1 Example - 9 What mass of sodium chloride would be decomposed by 9.8 g of sulphuric acid, if 12 g of sodium bisulphate and 2.75 g of hydrogen chloride were produced in a reaction assuming that the law of conservation of mass is true ?

Carbon

Oxygen

42.9%

57.1%

? Amount of oxygen that combines with 1 g carbon

57.1 = 1.33 g 42.9

Sol. NaCl + H2SO4 = NaHSO4 + HCl

Oxide 1

According to law of conservations of mass, Total masses of reactants = Total masses of products

Carbon

Oxygen

27.3%

72.7%

? Amount of oxygen that combines with 1 g carbon

Let the mass of NaCl decomposed be x g; so

72.7 = 2.66 g 27.3

x + 9.8 = 12.0 + 2.75 = 14.75

Ratio of oxygen in oxide (1) and (2) = 1 : 2

x = 4.95 g

Thus, Law of multiple proportion is verified.

Example - 10 In an experiment, 2.4 g of iron oxide on reduction with hydrogen yield 1.68 g of iron. In another experiment 2.9 g of iron oxide given 2.03 g of iron on reduction with hydrogen. Show that the above data illustrate the law of constant proportions.

Example - 12 In three moles of ethane (C2H6), calculate : (i) Number of moles of carbon atoms (ii) Number of moles of hydrogen atoms

Sol. In the first experiment

(iii) Number of molecules of ethane

The mass of iron oxide = 2.4 g The mass of iron after reduction = 1.68 g

Sol. (i) 1 mole of C2H6 contains 2 moles of carbon atoms

The mass of oxygen = Mass of iron oxide – Mass of iron

? 3 moles of C2H6 will C-atoms = 6 moles

= (2.4 – 1.68) = 0.72 g

(ii) 1 mole of C2H6 contains 6 moles of hydrogen atoms

Ration of oxygen and iron = 0.72 : 1.68

? 3 moles of C2H6 will contain H-atoms = 18 moles

= 1 : 2.33

(iii) 1 mole of C2H6 contains Avogadro’s no., i.e.,

In the second experiment

6.02 × 1023 molecules

The mass of iron oxide = 2.9 g

? 3 moles of C2H6 will contain ethane molecules

The mass of iron after reduction = 2.03 g

= 3 × 6.02 × 1023

The mass of oxygen = (2.9 – 2.03) = 0.87 g Ratio of oxygen and iron = 0.87 : 2.03

= 18.06 × 1023 molecules

= 1 : 2.33

31

SOME BASIC CONCEPTS OF CHEMISTRY

Example - 16

Example - 13

Calculate the mass per cent of different elements present in sodium sulphate (Na2SO4).

Zinc sulphate crystals contain 22.6% of zinc and 43.9% of water. Assuming the law of constant proportions to be true, how much zinc should be used to produce 13.7 g of zinc sulphate and how much water will they contain ?

Sol. Mass % of an element

Sol. 100 g of zinc sulphate crystals are obtained from

Mass of that element in the compound u 100 Molar mass of the compound

= 22.6 g zinc 1 g of zinc sulphate crystals will be obtained from

Now, molar mass of Na2SO4 = 2 (23.0) + 32.0 + 4 × 16.0

= 22.6/100 g zinc

= 142 g mol–1

13.7 g of zinc sulphate crystals will be obtained from Mass percent of sodium =

22.6 × 13.7 = 3.0962 g of zinc 100 100 g of zinc sulphate crystals contain water = 43.9 g

Mass per cent of sulphur =

32 u 100 = 22.54 % 142

Mass per cent of oxygen =

64 u 100 = 45.07% 142

1 g of zinc sulphate crystals contain water = 43.9/100 g 13.7 g of zinc sulphate crystals shall contain water

43.9 × 13.7 = 6.0143 g 100

46 u 100 = 32.39 % 142

Example - 17

Example - 14

Calculate the amount of carbon dioxide that could be produced when

What will be the mass of one 12C atom in g ?

(i) 1 mole of carbon is burnt in air.

Sol. 1 mol of 12C atoms = 6.022 × 1023 atoms = 12g

(ii) 1 mole of carbon is burnt in 16 g of dioxygen.

Thus, 6.022 × 1023 atoms of 12C have mass = 12g

(iii) 2 moles of carbon are burnt in 16 g of dioxygen.

12 g ? 1 atom of 12C will have mass = 6.022 u10 23

Sol. The balanced equation for the combustion of carbon in dioxygen/air is

= 1.9927 × 10–23 g

C (s)  O2 (g)  o CO2 (g)

Example - 15

1 mole

1 mole (32 g )

1 mole (44 g)

Calculate the molecular mass of : (i) H2O

(i) In air, combustion is complete. Therefore, CO2 produced from the combustion of 1 mole of carbon = 44 g.

(ii) CO2 (iiii) CH4

Sol. (i) Molecular mass of H2O = 2 (1.008 amu) + 16.00 amu

(ii) As only 16 g of dioxygen is available, it can combine only with 0.5 mole of carbon, i.e., dioxygen is the limiting reactant. Hence, CO2 produced = 22 g.

= 18.016 amu (ii) Molecular mass of CO2 = 12.01 amu + 2 × 16.00 amu = 44.01 amu

(iii) Here again, dioxygen is the limiting reactant. 16 g of dioxygen can combine only with 0.5 mole of carbon. CO2 produced again is equal to 22 g.

(iii) Molecular mass of CH4 = 12.01 amu + 4 (1.008 amu) = 16.042 amu

32

SOME BASIC CONCEPTS OF CHEMISTRY

Example - 18

Example - 21 Boron has two isotopes boron-10 and boron-11 whose percentage abundances are 19.6% and 80.4% respectively. What is the average atomic mass of boron ?

Hydrogen chloride (HCl) on oxidation gives water and chlorine. How many litres of chlorine at STP can be obtained starting with 36.50 g HCl ? Sol. Oxidation of HCl takes place according to the following equation :

Sol. Average atomic mass of B =

4HCl  O 2  o 2Cl2  2H 2O 4 mol

(10 u19.6)  (11u 80.4) = 10.804 amu 100

2 mol

Mass Moles of HCl = Molecular mass

36.5 1 mole 36.5

Example - 22 Carbon occurs in nature as a mixture of carbon-12 and carbon-13. The average atomic mass of carbon is 12.011. What is the percentage abundance of carbon-12 in nature ?

' 4 moles HCl give 2 moles Cl2 ? 1 mole will give

2 moles Cl2 = 0.5 moles Cl2 4

Sol. Let x be the percentage abundance of carbon-12; then (100 – x) will be the percentage abundance of carbon-13.

Volume of Cl2 at STP = 22.4 × 0.5 = 11.2 litre Example - 19 Why is air sometimes considered as a heterogeneous mixture ?

Therefore,

12x  13 (100  x) 12.011 100

or

12x + 1300 – 13x = 1201.1 x = 98.9

Sol. This is due to the presence of dust particles which form a separate phase.

Abundance of carbon-12 is 98.9%

Example - 20

Example - 23

Calculate the mass of sodium acetate (CH3COONa) required to make 500 mL of 0.375 molar aqueous solution. Molar mass of sodium acetate is 82.0245 g mol–1.

Calculate the mass of 2.5 gram atoms of oxygen. Sol. We know that

Sol. 0.375 M aqueous solution means that 1000 mL of the solution contain sodium acetate = 0.375 mole

Mass of an element in grams Number of gram atoms = Atomic mass of the element in grams

? 500 mL of the solution should contain sodium acetate =

So, Mass of oxygen = 2.5 × 16 = 40.0 g

0.375 mole 2

Example - 24 Calculate the gram atoms in 2.3 g of sodium.

Molar mass of sodium acetate = 82.0245 g mol–1 ?Mass of sodium acetate acquired =

Sol. Number of gram atoms =

0.375 mole × 82.0245g mol–1 2

2.3 23

0.1

[Atomic mass of sodium = 23 g]

= 15.380 g.

33

SOME BASIC CONCEPTS OF CHEMISTRY

Example - 28

Example - 25

What is the concentration of sugar (C12H22O11) in mol L–1 if its 20 g are dissolved in enough water to make a final volume up to 2 L ?

Calculate the mass of 1.5 gram molecule of sulphuric acid. Sol. Molecular mass of H2SO4

Sol.

= 2 × 1 + 32 + 4 × 16 = 98.0 amu

Molar mass of sugar (C12H22O11) = 12 × 12 + 22 × 1 + 11 × 16 = 342 g mol–1

Gram-molecular mass of H2SO4 = 98.0 g

20g No. of moles in 20 g of sugar = 342g mol 1 = 0.0585 mole

Mass of 1.5 gram molecule of H2SO4 = 98.0 × 1.5 = 147.0g Example - 26

Volume of solution = 2 L

Calculate the actual mass of one molecule of carbon dioxide (CO2).

(Given)

Moles of solute Molar concentration = Volume of sol in L

Sol. Molecular mass of CO2 = 44 amu

0.0582 mol 2L

= 0.0293 mol L–1 = 0.0293 M

1 amu = 1.66 × 10–24 g

Example - 29

So, the actual mass of CO2 = 44 × 1.66 × 10–24

How many molecules of water and oxygen atoms are present in 0.9 g of water ?

= 7.304 × 10–23g Sol. Given :

Example - 27

Mass of water = 0.9 g

Calculate the mass of a single atom of sulphur and a single molecule of carbon dioxide.

Molar mass of water = 18 g mol–1 Number of molecules of water and number of oxygen atoms present in water are to be calculated.

Sol. Gram-atomic mass of sulphur = 32 g Mass of one sulphur atom =

Gram  atomic mass 6.02 u10 23

To find :

Mass Number of moles, n = Molar mass

32 = 5.31 × 10–23 g 6.02 u 1023

Number of molecules = n × 6.02 × 1023

Formula of carbon dioxide = CO2

Solution :

Molecular mass of CO2 = 12 + 2 × 16 = 44

n

Gram-molecular mass of CO2 = 44 g

0.9 18

0.05

Number of molecules of water = 0.05 × 6.02 × 1023

Gram  molecular mass Mass of one molecule of CO2= 6.02 u 1023

= 3.01 × 1022 As one molecule of water contains one oxygen atom,

44 = 7.308 × 10–23 g 6.02 u 1023

So, number of oxygen atoms in 3.01 × 1022 molecules of water = 3.01 × 1022

34

SOME BASIC CONCEPTS OF CHEMISTRY One molecule of methane has = 6 + 4 = 10 electrons

Example - 30

So, 6.02 × 1022 molecules of methane have

What is the mass of 3.01 × 1022 molecules of ammonia ?

= 10 × 6.02 × 1022 electrons Sol. Gram-molecular mass of ammonia = 17 g = 6.02 × 1023 electrons

Number of molecules in 17 g (one mole) of NH3 = 6.02 × 1023 Example - 33 Let the mass of 3.01 × 1022 molecules of NH3 be = x g 3.01 u 10 22 6.02 u 10 23

So,

Calculate the number of moles in 25 g of calcium carbonate and number of oxygen atoms.

x 17

Sol. Formula mass of calcium carbonate or

x

17 u 3.01 u 10 6.02 u10 23

22

(CaCO3) = 100

= 0.85 g

Mass in grams No. of moles of CaCO3 = Formula mass

Example - 31 How many molecules and atoms of oxygen are present in 5.6 litres of oxygen (O2) at NTP ?

25 100

= 0.25 mole No. of oxygen atoms in one mole of CaCO3

Sol. We know that 22.4 litres of oxygen at NTP contain 6.02 × 1023 molecules of oxygen.

= 3 × 6.02 × 1023

So, 5.6 litres of oxygen at NTP contain

No. of oxygen atoms in 0.25 mole of CaCO3

5.6 ×6.02 × 1023 molecules 22.4

= 0.25 × 3 × 6.02 × 1023 = 4.515 × 1023

= 1.505 × 1023 molecules Example - 34

1 molecule of oxygen contains

One atom of an element weighs 6.644 × 10 –23 g. Calculate the number of gram atoms in 40 kg of it.

= 2 atoms of oxygen So, 1.505 × 1023 molecules of oxygen contain = 2 × 1.505 × 1023 atoms

Sol. Atomic mass of the element

23

= 3.01 × 10 atoms

= Mass of one atom × 6.02 × 1023

Example - 32

= 6.644 × 10–23 × 6.02 × 1023

How many electrons are present in 1.6 g of methane ? = 40 g Sol. Gram-molecular mass of methane

40 kg = 40,000 g

(CH4) = 12 + 4 = 16 g

1.6 Number of moles in 1.6 g of methane = 16

Number of grams atoms =

0.1

Number of molecules of methane in 0.1 mole = 0.1 × 6.02 × 10

0ass of the element in grams Atomic mass in grams

40000 = 1000 40

23

= 6.02 × 1022

35

SOME BASIC CONCEPTS OF CHEMISTRY

Example - 37

Example - 35

How much copper can be obtained from 100 g of copper sulphate (CuSO4) ? (Atomic mass of cu = 63.5 amu)

250 cm3 of sulphuric acid solution contain 24.5 g of H2SO4. If the density of the solution is 1.98 g cm–3, determine (i) molarity and (ii) molality.

Sol. 1 mole of CuSO4 contains 1 mole (1 g atom) of Cu Molar mass of CuSO4 = 63.5 + 32 + 4 × 16 = 159.5 g mol–1

Sol. (i) Molecular mass of H2SO4 = 2 + 32 + 64 = 98 No. of moles of H2SO4 in solution

Thus, Cu that can be obtained from 159.5 g of CuSO4

24.5 = 0.25 98

= 63.5 g ? Cu that can be obtained from 100 g of CuSO4

Volume of solution = 250 cm3 = 0.250 L

=

0.25 =1M 0.250

Molarity

63.5 u 100g 159.5

= 39.81 g Example - 38

(ii) Mass of solution = 250 × 1.98 = 495.0 g

If the density of methanol is 0.793 kg L–1, what is the volume needed for making 2.5 L of its 0.25 M solution ?

Mass of solvent = Mass of solution – Mass of solute = 495.0 – 24.5 = 470.5g = 0.4705 kg

Sol. Molar mass of methanol (CH3OH) = 32 g mol–1 Moality =

0.25 = 0.53 m 0.4705

= 0.032 kg mol–1 Molarity of the given solution

Example - 36 Calculate the concentration of nitric acid in moles per

0.793 kg L1 0.032 kg mol 1

= 24.78 mol L–1

–1

litre in a sample which has a density, 1.41 g mL and mass per cent of nitric acid in it being 69%.

Applying

M V

2 2 (Solution to be prepared)

24.78 × V1 = 0.25 × 2.5 L or V1 = 0.02522 L = 25.22 mL

Sol. Mass percent of 69% means that 100 g of nitric acid solution contain 69 g of nitric acid by mass.

Example - 39 Pressure is determined as force per unit area of the surface. The S.I. unit of pressure, pascal, is

Molar mass of nitric acid (HNO3) = 1 + 14 + 48 = 63 g mol–1

1 Pa = 1 N m–2

69 g ? Moles of 68 g HNO3 = 63 g mol 1 = 1.095 mole

If mass of air at sea level is 1034g cm–2, calculate the pressure in pascal.

100 g Volume of 100 g nitric acid solution = 1.41g mL1

Sol. Pressure is the force (i.e., weight) acting per unit area But weight = mg

= 70.92 mL = 0.07092 L ? Conc. of HNO3 in moles per litre =

M uV

1 1 (Given solution )

? Pressure = Weight per unit area =

1.095 mole 0.07092L

1034g u 9.8 ms 2 cm 2

1034g u 9.8 ms 2 1kg 100 cm 100cm 1N 1Pa u u u u u 2 2 cm 1000g 1m 1m kg ms 1N m 2

= 15.44 M = 1.01332 × 105 Pa

36

SOME BASIC CONCEPTS OF CHEMISTRY Example - 42

Example - 40

Calculate the number of atoms in each of the following samples:

Calculate the empirical formula of a compound that contains 26.6% potassium, 35.4% chromium and 38.1% oxygen [Given K = 39.1; Cr = 52; O = 16] Sol. Element

Percentage

Atomic mass

Potassium

26.6

39.1

Chromium

35.4

52.0

Oxygen

38.1

16.0

Simplest ratio

Simplest whole

Relative no. of atoms

26.6 39.1

(a) 800 amu of Ca (b) 800 grams of Ca Sol : (a) Atomic Mass of Ca = 40 amu

Ÿ 40 amu is the mass of 1 Ca atom Thus, 800 amu is the mass of 800/40 Ca atoms = 20 Ca atoms Ans. (b) Atomic mass of Ca = 40 g/mole

Ÿ 40g is the mass of 1 mole Ca atoms

no. ratio

= 6.022 × 1023 Ca atoms

0.68 1 0.68

0.68

1×2=2

Thus, 800g is the mass of (800 × 6.022 × 1023)/40 Ca atoms = 20 mole Ca atoms

35.4 52

0.68 1 0.68

0.68

38.1 2.38 16

2.38 0.68

= 1.2044 × 1025 Ca atoms Ans.

1×2=2

Example - 43

3.5

Calculate the mass of carbon in 1kg of sugar (C12H22O11)

3.5 × 2 = 7

Sol : Molecular mass of sugar = 12 × 12 + 22 × 1 + 11 × 16

Therefore, empirical formula is K2Cr2O7.

= 342 g/mol 342g sugar contains = 144g carbon

Example - 41

1000g sugar contains = 421g carbon

(a) Calculate the mass of KClO3 necessary to produce 1.23 g O2.

Example - 44

(b) What mass of KCl is produced along with this quantity of oxygen ?

Find the amount of weight of NH3 being produced when 1kg of N2 reacts with 1kg of H2. Which reactant is in excess and how much?

Sol. (a) The reaction involved is :

2KClO3  o 2KCl  3O2 2 mol 2 u 122.5 g

2 mol 2 u 74.5 g

Sol : N2 + 3H2

3 mol 3 u 32 g

1 mole of N2 reacts with 3 moles of H2 to produce 2 moles of NH3. Thus, 28g N2 reacts with 6g of H2 to produce 34g of NH3.

' 3 × 32 g O2 is produced by 2 × 122.5 g KClO3 ? 1.23 g O2 will be produced by

245 × 1.23 = 3.139 g 96

Since the weight of N2 and H2 taken are equal, so N2- will be consumed before H2. So, N2 is th LR and

(b) ' 2 × 122.5 g KClO3 give 2 × 74.5 g KCl ? 3.139 g KClO3 will give

Æ 2NH3

H2 is the ER.

2 u 74.5 u 3.139 = 1.909 g 2 u122.5

Since, 28g N2- reacts with = 6g H2;

37

SOME BASIC CONCEPTS OF CHEMISTRY Moles of NaOH in 2 nd solution = 3 × 5 = 15.

1000g N2 reacts = with 1000 × 6/28 = 214.3g H2

Thus on mixing the total moles of NaOH = 24.

So, H2 is the ER and the amount of H2 in excess = 1000-214.3 = 785.7g Ans.

Final Molarity = 1M

Also, 28g N2 produces = 34g NH3;

Final moles = 24

so, 1000g N2 produces = 1000 × 34/28

Total Volume of solutions = 8L.

= 1.214kg NH3 Ans.

§ Ÿ V = 24L ¨ As M ©

Example - 45

n· ¸ Ans. v¹

The mixture needs to be diluted 3 folds

Calculate the Molarity and molality of a 98% Example - 47

by mass of H2SO4 solution having a density of 1.25g/cc.

An organic containing C,H and N gave the following analysis: C: 40% H:13.3%, N:46.67%. If its molecular formula weight is three times its empirical formula weight then find out its empirical and molecular formula of the compound.

Sol : H 2SO 4 taken = 98% Ÿ 100g of solution contains 98g H2SO4. mass of solution = 100g mass of solute, H2SO4 = 98g

Sol: Relative no. of atoms of C = 40/12 = 3.33

mass of solvent = 100 – 98 = 2g = 0.002 kg moles of solute, H2SO4 =

volume of solution =

100 1.25

Relative no. of atoms of H = 13.3/1 = 13.3 and that for N = 46.67/14 = 3.33

98 1 98

Thus, simplest atomic ratio C:H:N

mass of solution density

= 3.33:13.33:3.33 = 1:4:1 Therefore the empirical formula of the compound is CH4N

80mL 0.08L

Molarity, M

moles of solute volume of solution (L)

Ans. Molecular Formula Mass

1 0.08

Also, given: Empirical Formula Mass = 3 = n-factor Therefore, molecular formula is (CH4N)3 i.e. C3H12N3

= 12.5 M Ans.

molality, m

moles of solute mass of solvent (kg)

Example - 48

1 0.02

Calculate the number of equivalents in the following samples:

= 500 m Ans. (a) 490g H2SO4 Example - 46

(c) 730g HCl

(b) 1600g NaOH (d) 0.37g Ca(OH)2

A 3M 3L solution of NaOH is mixed with another 3M 5L solution of NaOH. How much should the mixture be diluted so that the final Molarity of the solution become 1M ?

Sol : Eq. wt. of H2SO4 = 98/2 = 49; NaOH = 40/1 = 40; HCl = 36.5/1 = 36.5; Ca(OH)2 = 74/2 = 37 (a) No. of eq. of H2SO4 = 490/49 = 10 Ans.

Sol : Moles of NaOH in 1 st solution = MV = 3 × 3 = 9.

(b) No. of eq. of NaOH = 1600/40 = 40 Ans.

38

SOME BASIC CONCEPTS OF CHEMISTRY (c) No. of eq. of HCl = 730/36.5 = 20 Ans.

?

(d) No. of eq. of Ca(OH)2 = 0.37/37 = 0.01

N Na 2CO3 = 3.948 Now Na2CO3 fresh solution reacts with H2SO4

= 10 milli-eq. Ans

Wt. of Na2CO3 solution = 125 g

Example - 49 A mixture of three acids 3.65 g of HCl, 4.9 g H2SO4 and 9 g H2C2O4 is made to react with a mixture of two bases x g NaOH and 7.4 g Ca(OH)2. Calculate w for complete neutralisation.

125 = 100 mL 1.25

?

Volume of Na2CO3 solution

?

Meq. of H2SO4 = Meq. of Na2CO3 0.84 × V = 100 × 3.948

Sol : We know that total equivalents of acids must be equal to total equivalents of bases.’ 6 (w/E)ACIDS

?

Example - 52

= 6 (w/E)BASES

5 mL of 8N HNO3, 4.8 mL of 5N HCl and a certain volume of 17M H2SO4 are mixed together and made upto 2 litre 30 mL of this acid mixture exactly neutralizes 42.9 mL of Na2CO3 solution containing 1g of Na 2CO3. 10H2O in 100 mL of water. Calculate the amount of sulphate ions in g present in solution.

3.65/36.5 + 4.9/49 + 9/45 = x/40 + 7.4/37 Ÿx

Volume of H2SO4 required = 470 mL

= 8g

Example - 50 Calculate the Equivalent mass of Al2 (SO4)3 ?

Sol. Meq. of HNO3 = 5 × 8 = 40 Sol : 1 equivalent of Al2(SO4)3 = 1 equivalent of Al3+ + 1 equivalent of SO42-

Meq. of HCl = 4.8 × 5 = 24 Meq. of H2SO4 = V × 17 × 2 = 34 V (Let V mL of H2SO4)

E (Al2(SO4)3) = E (Al3+) + E (SO42-)

?

§ 27 · § 96 · ¨ ¸  ¨ ¸ 9  48 57g © 3 ¹ © 2¹

Total Meq. of acid in 2 litre solution = 40 + 24 + 34V

= 64 + 34V Now Meq. of acid in 30 mL solution = Meq. of Na2CO3 used for it

This can be tallied by the method for the salt. For this salt z = 6 and M = 342 g therefore E = 342/6 = 57 g.

Meq. of Na2CO3

Example - 51

42.9 u

25 mL of a solution of Na2CO3 having a specific gravity of 1.25g mL–1 required 32.9 mL of a solution of HCl containing 109.5 g of the acid per litre for complete neutralization. Calculate the volume of 0.84 N H2SO4 that will be completely neutralized by 125g of Na2CO3 solution.

109.5 Sol : equivalents of HCl = 36.5 N HCl

§ 3 ¨ N Na 2CO3 ©

3

3 3 1

Meq. of Na2CO3 = Meq. of HCl N × 25 = 32.9 × 3

39

1 1000 · u ¸ 286 / 2 100 ¹

?

Meq. of acid in 2 litre solution =

?

64 + 34V = 200

?

3 u 2000 30

200

34V = 200 – 64 = 136

Now Meq. of H2SO4 = Meq. of

Since Na2CO3 is completely neutralized by HCl ?

1u1000 286 / 2 u100

?

Meq. of SO 4 2 = 136

?

Weight of SO 4 2 = 6.528g

SO42

?

= 34V = 136

w u1000 136 96/ 2

CHEMICAL EQUILIBRIUM

EXERCISE - 1 : BASIC OBJECTIVE QUESTIONS Equilibrium : Definition 1.

(a)

In chemical reaction A

B, the system will be known 7.

(a) A completely changes to B

(d) 4

On a given condition, the equilibrium concentration of HI,

(a) 64 (c) 8

(d) Only 10% of A changes to B

(b) 12 (d) 0.8

(a) Concentration of reactants

C(g),the equilibrium For the system A(g) + 2B(g) concentrations are (A) 0.06 mole/litre (B) 0.12 mole/litre (C) 0.216 mole/litre. The Keq for the reaction is

(b) Molar concentration of reactants

(a) 250

8.

According to law of mass action rate of a chemical reaction is proportional to

(c) 4 × 10

(c) Concentration of products 9.

(d) Molar concentration of products The active mass of 64 gm of HI in a two litre flask would be (a) 2

(a)

(d) 0.25

The rate constant for forward and backward reactions of

CH3COOH + C2H5OH

(a) 4.33 (c) 6.33

(b) 5.33 (d) 7.33

[A ] [B]2 [C ]

10.

(a) NO(g)

equilibrium constant is

6.

(a)

[3A][2B] C

[C] (b) [3A][2B]

(c)

[A]3[B]2 [C]

[C] (d) [A]3[B]2

(b)

[A] [B] [C]

[C] (d) 2 [B] [ A ]

1 1 N 2 (g)  O2 (g) 2 2

(b) Zn (s)  Cu 2 (aq)

C, the expression for

C, the expression for

For which of the following reactions does the equilibrium constant depend on the units of concentration

Equilibrium Constant : Expression For the system 3A + 2B

(d) 125

[C ] (c) [A ] [B]2

hydrolysis of ester are 1.1 × 10–2 and 1.5 × 10–3 per minute respectively. Equilibrium constant for the reaction is CH3COOC2H5 + H2O

(b) 416 –3

For the reaction A + 2B equilibrium constant is

(b) 1

(c) 5

5.

2HI will be

constant for the reaction H2 + I2

(c) The rate of change of A to B and B to A on both the sides are same

4.

1 2

H2 and I2 are 0.80, 0.10 and 0.10 mole/litre. The equilibrium

(b) 50% of A changes to B

3.

(b)

(c) 1

in equilibrium when

2.

1 4

Cu (s)  Zn 2 (aq)

CH3COOC2H5(l)+ H2O (l) (c) C2H5OH (l) + CH3COOH (Reaction carried in an inert solvent) (d) COCl2(g) 11.

CO(g) + Cl2(g)

Unit of equilibrium constant for the reversible reaction 2HI is H2 + I 2

4 moles of A are mixed with 4 moles of B. At equilibrium for

(a) mol–1 litre

(b) mol–2 litre

C + D, 2 moles of C and D are

(c) mol litre–1

(d) None of these

the reaction A + B

formed. The equilibrium constant for the reaction will be

A-1

CHEMICAL EQUILIBRIUM 12.

(a) 0.1 (c) 10 13.

19.

In a reaction A + B C + D, the concentrations of A, B, C and D (in moles/litre) are 0.5, 0.8, 0.4 and 1.0 respectively. The equilibrium constant is

had 0.5 mole H2S, 0.10 mole H2 and 0.4 mole S2 in one litre vessel. The value of equilibrium constant (K) in mole litre-1 is (a) 0.004 (c) 0.016

The suitable expression for the equilibrium constant of 2NOCl (g) is 20. (a) K c

(c) K c

14.

15.

[2 NOCl] [2 NO] [Cl 2 ] [ NOCl]2 [ NO] [Cl 2 ]2

2H2 (g) + S2(g)

2H2S(g)

(b) 1.0 (d) f

the reaction 2NO (g) + Cl2(g)

An equilibrium mixture of the reaction

(b) K c

(d) K c

[ NOCl]

2

For the equilibrium N2 + 3H2 2.37 × 10

[ NO]2 [Cl 2 ]

(b) 0.008 (d) 0.160 –3

2NH3, Kc at 1000K is

If at equilibrium [N2] = 2M, [H2] = 3M, the

concentration of NH3 is (a) 0.00358 M (c) 0.358 M

[ NOCl]2 [ NO]2 [Cl 2 ]2

21.

(b) 0.0358 M (d) 3.58 M

Two moles of NH3 when put into a previously evacuated

C + D. If finally the concentration of A and B A+ B are both equal but at equilibrium concentration of D will be twice of that of A then what will be the equilibrium constant of reaction.

vessel (one litre), partially dissociate into N2 and H2. If at

(a) 3/4 mol2 litre–2

(b) 27/64 mol2 litre–2

(a) 4 / 9 (c) 1 / 9

(c) 27/32 mol2 litre–2

(d) 27/16 mol2 litre–2

equilibrium one mole of NH3 is present, the equilibrium constant is

(b) 9 / 4 (d) 4

For the reaction 2SO2 + O2 (a) litre/mole

22.

2SO3, the units of Kc are (b) mol/litre

–1 2

(c) (mol litre )

In Haber process 30 litres of dihydrogen and 30 litres of dinitrogen were taken for reaction which yielded only 50% of the expected product. What will be the composition of gaseous mixture under the aforesaid condition in the end (a) 20 litres ammonia, 25 litres nitrogen, 15 litres hydrogen

(d) (litre mole–1)2

(b) 20 litres ammonia, 20 litres nitrogen, 20 litres hydrogen 16.

The equilibrium concentration of X, Y and YX2 are 4, 2 and 2 moles respectively for the equilibrium 2X + Y

(d) 20 litres ammonia, 10 litres nitrogen, 30 litres hydrogen

YX2. The value of Kc is

(a) 0.625 (c) 6.25 17.

(c) 10 litres ammonia, 25 litres nitrogen, 15 litres hydrogen 23.

(b) 0.0625 (d) 0.00625

4.8 × 10–2 and 1.2 × 10–2 mol litre–1 respectively. The value of Kc for the reaction is

COCl2. At equilibrium, it contains 0.2 moles of COCl2 and 0.1 mole of each of CO and CO2. The equilibrium constant

(a) 5 (c) 15 18.

24.

COCl2 is

(b) 3 × 101 mol litre–1

(c) 3 × 10–3 mol litre–1

(d) 3 × 103 mol litre–1

The equilibrium constant for the reaction 2NO (g) at temperature T is 4 × 10–4.

The value of Kc for the reaction

4.5 moles each of hydrogen and iodine heated in a sealed ten litre vessel. At equilibrium, 3 moles of HI were found.

(a) 1 (c) 5

(a) 3.3 × 102 mole litre–1

N2(g) + O2(g)

(b) 10 (d) 20

The equilibrium constant for H2(g) + I2(g)

2NO2(g), the

concentrations of N 2 O4 and NO2 at equilibrium are

In a 500 ml capacity vessel CO and Cl2 are mixed to form

Kc for the reaction CO + Cl2

For the reaction equilibrium N2O4

1 1 N 2 (g )  O 2 (g ) at the same temperature 2 2

NO (g)

2 HI(g) is

is (a) 4 × 10–4

(b) 10 (d) 0.33

2

(c) 2.5 × 10

A-2

(b) 50 (d) 0.02

CHEMICAL EQUILIBRIUM 25.

In the reaction, H2 + I2

Position of Equilibrium : Extent of a Reaction

2HI. In a 2 litre flask 0.4 moles of

each H2 and I2 are taken. At equilibrium 0.5 moles of HI are

26.

formed. What will be the value of equilibrium constant, Kc

In which of the following, the reaction proceeds towards completion

(a) 20.2 (c) 0.284

(a) K = 103 (c) K = 10

30.

(b) 25.4 (d) 11.1 31.

2 mol of N2 is mixed with 6 mol of H2 in a closed vessel of

equilibrium constant is 6 × 1022

2NH3(g) is (b) 27/4 (d) 24

32.

Equilibrium constants K1 and K 2 for the following

(a) K 2

1 K1

(c) K 2

K1 2

NO2(g)

(d) 1.6 × 10–4 M

In which of the following reaction, the value of K p will be

(c) 2NH3

2NO(g) + O2(g) are related as (b) K 2 (d) K 2

33.

1 O 2 (g ) 2

34.

K12

(c) N2 + O2 (d) CO + H2O 35.

2SO2(g) + O2(g) have equilibrium constants

(c) K2 = 1/K12

(d) K2 = 1/K1

(b) Kc = Kp (RT) (d) Kp = Kc (RT)–1

2HI 2 NH3 2NO CO2 + H2

For the reaction H2(g) + I2(g)

2HI(g) at 721K the

equilibrium concentration of both is 0.5M, the value of Kp under the same conditions will be

relationships between K1 and K2 is correct (b) K2 = K12

2SO3

value of equilibrium constant (K c) is 50. When the

K1 and K2 respectively at 298 K. Which of the following (a) K1 = K2

(d) 2SO2 + O2

In which of the following equilibria, the value of Kp is less than Kc (a) H2 + I2

SO3(g)

PCl3 + Cl2

2NH3 + heat

(c) Kp = Kc (RT)

1

1 O 2 (g ) 2

N2 + 3H2

For N2 + 3H2

(b) N2 + 3H2 SO 2 (g ) 

(b) PCl5

–2

and 2SO3(g)

2HI

(a) Kp = Kc (RT)

K12

Two gaseous equilibria SO 2 (g ) 

29.

(c) 3.2 × 10–4 M

(a) H2 + I2

1 O 2 2

and 2NO2(g)

28.

(b) 3.2 × 10–6 M

equal to K c

equilibria NO(g) +

(a) 1.6 × 10–8 M

Kp and Kc Relationship

Equilibrium Constant : Effect of Stoichiometry 27.

C6H12O6

What is the concentration of HCHO at equilibrium if

at equilibrium, the value of K c for the reaction

(a) 4/27 (c) 1/27

A 1 M solution of glucose reaches dissociation equilibrium according to equation given below 6HCHO

one litre capacity. If 50% of N2 is converted into NH3 N2(g) + 3H2(g)

(b) K = 10–2 (d) K = 1

(a) 0.002 (c) 50.0

(b) 0.2 (d) 50 / RT

At a given temperature, the equilibrium constant for 36.

PCl3(g) + Cl2(g) is 2.4 × 10–3. At the reaction PCl5(g) same temperature, the equilibrium constant for reaction PCl3(g) + Cl2(g)

The reaction between N2 and H2 to form ammonia has Kc = 6 × 10–2 at the temperature 500°C. The numerical value of Kp for this reaction is

PCl5(g) is

(a) 2.4 × 10–3

(b) – 2.4 × 10–3

(a) 1.5 × 10–5

(b) 1.5 × 105

(c) 4.2 × 102

(d) 4.8 × 10–2

(c) 1.5 × 10–6

(d) 1.5 × 106

A-3

CHEMICAL EQUILIBRIUM 37.

For the reversible reaction, N2(g) + 3H2(g)

42.

2NH3(g)

N 2 (g)  3H 2 (g) U 2NH 3 (g)

at 500°C, the value of Kp is 1.44 × 10–5 when partial pressure is measured in atmospheres. The corresponding value of Kc with concentration in mole litre-1, is

at constant temperature and pressure. Then the ratio of the final volume to the initial volume of gases are

(a) 1.44 × 10–5/(0.082 × 500)–2

(a) 4 : 5

(b) 5 : 4

(c) 7 : 10

(d) 8 : 5

–5

–2

(b) 1.44 × 10 /(8.314 × 773)

43.

(c) 1.44 × 10–5/(0.082 × 773)2 –5

–2

Equilibrium : Analysis

Initial pressure of XY2 is 600 mm Hg. The total pressure at equilibrium is 800 mm Hg. Assuming volume of system to remain constant, the value of Kp is

A mixture of 0.3 mole of H2 and 0.3 mole of I2 is allowed to react in a 10 litre evacuated flask at 500ºC. The reaction is 2HI, the K is found to be 64. The amount of H2 + I 2 unreacted I2 at equilibrium is (a) 0.15 mole (c) 0.03 mole

39.

(b) 0.06 mole (d) 0.2 mole

(b) 100

(c) 200

(d) 400

0.6 mole of NH3 in a reaction vessel of 2dm3 capacity was brought to equilibrium. The vessel was then found to contain 0.15 mole of H2 formed by the reaction

(a) 50

(b) 25

(c) 200

(d) 15

44.

N2(g) + 3H2(g)

Which of the following statements is true

45.

For the equilibrium AB (g)

U A(g) + B(g). Kp is equal to

(a) 0.15 mole of the original NH3 had dissociated at equilibrium

four times the total pressure. Calculate the number moles of A formed if one mol of AB is taken initially.

(b) 0.55 mole of ammonia is left in the vessel

(a) 0.45

(b) 0.30

(c) 0.60

(d) 0.90

(c) At equilibrium the vessel contained 0.45 mole of N2 46.

(d) The concentration of NH3 at equilibrium is 0.25 mole per dm3

I 2  I  U I 3 This reaction is set-up in aqueous medium. We start with – 1 mol of I2 and 0.5 mol of I in 1L flask. After equilibrium is

5 moles of SO2 and 5 moles of O2 are allowed to react to

reached, excess of AgNO 3 gave 0.25 mol of yellow precipitate. Equilibrium constant is

form SO3 in a closed vessel. At the equilibrium stage 60% of SO2 is used up. The total number of moles of SO2, O2

41.

(a) 50

15 mol of H2 and 5.2 moles of I2 are mixed and allowed to attain equilibrium at 500ºC. At equilibrium, the concentration of HI is found to be 10 mol. The equilibrium constant for the formation of HI is.

2NH3(g)

40.

XY2 dissociates as :

XY2 (g) U XY(g)  Y(g)

(d) 1.44 × 10 /(0.082 × 773)

38.

40% of a mixture of 0.2 mol of N2 and 0.6 mol of H2 reacts to give NH3 according to the equation :

and SO3 in the vessel now is

(a) 1.33

(b) 2.66

(a) 10.0 (c) 10.5

(c) 2.00

(d) 3.00

(b) 8.5 (d) 3.9

The equilibrium P4 (s) + 6 Cl2 (g)

47.

ZZX 4 PCl3(g) is attained YZZ

by mixing equal moles of P4 and Cl2 in an evacuated vessel. Then at equilibrium (a) [Cl2] > [PCl3]

(b) [Cl2] > [P4]

(c) [P4] > [Cl2]

(d) [PCl3] > [P4]

A-4

One mole of N2O4(g) at 100 K is kept in a closed container at 1.0 atm pressure. It is heated to 400 K, where 30% by mass of N2O4(g) decomposes of NO2(g). The resultant pressure will be (a) 4.2

(b) 5.2

(c) 3.2

(d) 6.2

CHEMICAL EQUILIBRIUM

Equilibrium Constant and 'G 48.

at 25ºC for 2HI(g)

49.

to yield NH3 and H2 S gases in the flask. When the decomposition reaction reaches equilibrium, the total pressure in the flask rises to 0.84 atm. The equilibrium constant for NH4HS decomposition at this temperature is

'Gº(HI; g) #  1.7 kJ. What is the equilibrium constant H2(g) + I2(g)

(a) 24.0

(b) 3.9

(c) 2.0

(d) 0.5

(a) 0.30 (c) 0.17 55.

Calculate 'Gº for conversion of oxygen to ozone 3/2 O2(g) o O3(g) at 298 K, if Kp for this conversion is

–1

(c)1.63 kJ mol

–1

(a) 0.2 (c) 1.6

(d) 2.38 × 10 kJ mol

Homogeneneous and Heterogeneous Equilibria, Degree of Dissociation 50.

56.

Pure ammonia is placed in a vessel at temperature where its dissociation ( D ) is appreciable. At equilibrium

PCl3 (g) + Cl2(g)

(c) N2(g) + 3H2(g)

(c) Concentration of NH does not change with pressure

(d) SO2Cl2(g) 57.

2NH3(g) SO2(g) + Cl2(g)

The equilibrium constant (K p ) for the reaction

H2 + I2 Only 50% HI is At a certain temp. 2HI dissociated at equilibrium. The equilibrium constant is

PCl5(g) o PCl3(g) + Cl2(g) is 16. If the volume of the container is reduced to one half its original volume, the value of Kp for the reaction at the same temperature will be

(a) 0.25 (c) 3.0

(a) 32 (c) 16

(b) 1.0 (d) 0.50 58.

3.2 moles of hydrogen iodide were heated in a sealed bulb at 444°C till the equilibrium state was reached. Its degree of dissociation at this temperature was found to be 22%. The number of moles of hydrogen iodide present at equilibrium are

(b) 64 (d) 4

Consider the following equilibrium in a closed container : N2O4 (g)

U 2 NO2(g)

At a fixed temperature, the volume of the reaction container is halved. For this change, which of the following statements, holds true regarding the equilibrium constant (Kp) and degree of dissociation (D) ?

(b) 1.87 (d) 4

(a) neither Kp nor D changes (b) both Vp and D change

The vapour density of completely dissociated NH4Cl would be

(c) Kp changes but D does not change (d) Kp does not change but D changes

(a) Slight less than half that of NH4Cl 59.

(b) Half that of NH4Cl (c) Double that of NH4Cl

Calculate the partial pressure of carbon monoxide from the following data : ' o CaO (s)  CO 2 n, Kp = 8 × 10–2 CaCO3 (s) 

(d) Determined by the amount of solid NH4Cl in the experiment 54.

2NO(g)

(b) D does not change with pressure

(a) 2.496 (c) 2 53.

Change in volume of the system does not alter the number of moles in which of the following equilibrium

(b) PCl5 (g)

(d) Concentration of H2 is less than that of N2

52.

(b) 0.4 (d) 4

(a) N2(g)+ O2(g)

(a) Kp does not change significantly with pressure

51.

CaO(s) + CO2 n; Kp = 8 × 10–2

CO2(g) + C (s) o 2CO (g) ; Kp = 2

(b) 2.4 × 102 kJ mol–1 6

Calculate the partial pressure of carbon monoxide from the following CaCO3 (s)

2.47 × 10–29

(a) 163 kJ mol–1

(b) 0.18 (d) 0.11

CO2(g) + C (s)  o 2 CO (g), Kp = 2

An amount of solid NH4HS is placed in a flask already containing ammonia gas at a certain temperature and 0.50 atm. pressure. Ammonium hydrogen sulphide decomposes

A-5

(a) 0.2

(b) 0.4

(c) 1.6

(d) 4

CHEMICAL EQUILIBRIUM 60.

61.

64.

A small amount of NH4HS is placed in a flask already containing ammonia gas at a certain temperature and 0.50 atm pressure. Ammonium hydrogen sulphide decomposes to yield NH3 and H2S gases in the flask. When the decomposition reaction reaches equilibrium, the total pressure in the flask rises to 0.84 atm. The equilibrium constant for NH4HS decomposition at this temperature is (a) 0.30

(b) 0.18

(c) 0.17

(d) 0.11

N2O4 (g) U 2 NO2 (g), the observed molecular weight of N2O4 is 80 g mol–1 at 350 K. The percentage of dissociation of N2O4 (g) at 350 K is

The vapour density of completely dissociated NH4Cl is

(b) 15%

(c) 20%

(d) 18%

(a) double than that of NH4Cl

What is the effect of halving the pressure by doubling the volume on the following system at 500°C

(b) half than that of NH4Cl

H2(g) + I2(g)

(c) same as that of NH4Cl

(a) Shift to product side

65.

(c) Liquifaction of HI (d) No effect 66.

COCl2 (g) YZZ ZZX CO (g) + Cl2 (g)

67.

(d) Cl2 will increase The equilibrium constant (Kp) for the decomposition of gaseous H2O 1 O (g) 2 2

(b) CH COOH is added

(c) Catalyst is added

(d) Mixture is heated

(c) The same state of equilibrium is reached whether one starts with the reactants or the products (d) The forward reaction is favoured by the addition of a catalyst

D p1/ 2 (1  D) (2  D)1/ 2 D p3 / 2 (1  D ) (2  D)1/ 2

(c) K p

D3 / 2 p2 (1  D ) (2  D )1/ 2

(d) K p

D 3 / 2 p1/ 2 (1  D ) (2  D)1/ 2

Which of the following statements regarding a chemical equilibrium is wrong

(b) An equilibrium is dynamic

68. (b) K p

(a) CH COO– are added

(a) An equilibrium can be shifted by altering the temperature or pressure

is related to degree of dissociation (D) at a total pressure p, given by (a) K p

CH COO– + H+

(b) PCl5 wil decrease

(c) PCl5 will remain unaffected

H2O (g) U H2 (g) +

In equilibrium CH3COOH + H2O

The equilibrium constant may change when

If some CO is added into the vessel, then after the equilibrium is attained again, concentration of (a) PCl5 will increase

2HI(g)

(b) Shift to product formation

The following two equilibria exist simultaneously in a closed vessel : ZZX PCl3(g) + Cl2 (g) PCl5(g) YZZ

63.

(a) 10%

Le Chatelier’s Principle

(d) determined by the amount of solid NH4Cl taken 62.

At the equilibrium of the reaction,

In the reaction, A2(g) + 4 B2(g)

2AB4(g)

'H  0 the formation of AB4 is will be favoured at

(a) Low temperature, high pressure (b) High temperature, low pressure (c) Low temperature, low pressure (d) High temperature, high pressure

A-6

CHEMICAL EQUILIBRIUM 69.

The formation of SO takes place according to the following reaction, 2SO2 + O2

2SO3 ;

75.

In the reaction A(g) + 2B(g)

C(g) + Q kJ, greater

product will be obtained or the forward reaction is favoured by

'H = – 45.2 kcal

(a) At high temperature and high pressure The formation of SO is favoured by

(b) At high temperature and low pressure

(a) Increasing in temperature

(c) At low temperature and high pressure

(b) Removal of oxygen

(d) At low temperature and low pressure

(c) Increase of volume

76.

(d) Increasing of pressure 70.

N2 + O 2

Which will increase the equilibrium concentration of C2H6

In the above reaction which is the essential condition for the higher production of NO

71.

(a) Increase of temperature (b) By reducing temperature

(b) High pressure (d) Low pressure

(c) By removing some hydrogen

Which of the following reactions proceed at low pressure 2NH

(a) N2 + 3H2 (c) PCl5 72.

PCl3 + Cl2

(b) H2 + I2 (d) N2 + O2

(d) By adding some C2H6

2HI 77. 2NO

The effect of increasing the pressure on the equilibrium 2A + 3B

In the following reversible reaction 2SO2 + O2

3A + 2B is

(a) Forward reaction is favoured

2SO + Q Cal

(b) Backward reaction is favoured

Most suitable condition for the higher production of SO is

(c) No effect (d) None of the above

(a) High temperature and high pressure (c) Low temperature and high pressure

The exothermic formation of ClF is represented by the equation

(d) Low temperature and low pressure

Cl2(g) + 3F2(g)

C + D +heat has reached The reaction A + B equilibrium. The reaction may be made to proceed forward by

Which of the following will increase the quantity of ClF

(a) Adding more C

(a) Increasing the temperature

(b) Adding more D

(b) Removing Cl2

78.

(b) High temperature and low pressure

73.

C2H6; 'H = – 32.7 Kcal

C2H4 + H2

2NO - Q cals

(a) High temperature (c) Low temperature

Following gaseous reaction is undergoing in a vessel

2ClF3(g); 'H = – 329 kJ

in an equilibrium mixture of Cl2, F2 and ClF

(c) Decreasing the temperature

(c) Increasing the volume of the container

(d) Increasing the temperature 74.

(d) Adding F2

According to Le-chatelier principle, if heat is given to solidliquid system, then (a) Quantity of solid will reduce

In which of the following system, doubling the volume of the container cause a shift to the right

(b) Quantity of liquid will reduce

(a) H2(g) + Cl2(g)

(c) Increase in temperature

(b) 2CO(g) + O2(g)

79.

2HCl(g) 2CO2(g)

(d) Decrease in temperature (c) N2(g) + 3H2(g) (d) PCl5(g)

A-7

2NH3(g)

PCl3(g) + Cl2(g)

CHEMICAL EQUILIBRIUM 80.

81.

Which of the following information can be obtained on the basis of Le-chatelier’s principle

83.

N2 + O 2

2NO. 'H = 43.200 kcal is favoured by

(a) Entropy change in a reaction

(a) Low temperature and low pressure

(b) Dissociation constant of a weak acid

(b) Low temperature and high pressure

(c) Equilibrium constant of a chemical reaction

(c) High temperature and high pressure

(d) Shift in equilibrium position on changing value of a constant

(d) High temperature and excess reactants concentration

The equilibrium SO2Cl2 (g)

84.

SO2(g) + Cl2(g) is attained

The yield of product in the reaction A2(g)+ 2B(g)

at 25°C in a closed container and an inert gas helium is introduced which of the following statement is correct

C(g) +

Q.kJ. would be high at

(a) High temperature and high pressure

(a) More chlorine is formed

(b) High temperature and low pressure

(b) Concentration of SO2 is reduced

(c) Low temperature and high pressure (d) Low temperature and low pressure

(c) More SO2Cl2 is formed (d) Concentration of SO2Cl2(SO2) and Cl2 does not change 82.

The formation of nitric oxide by contact process

H2(g) + I2(g)

2HI(g) 'H = + q cal, then formation of HI

(a) Is favoured by lowering the temperature (b) Is favoured by increasing the pressure (c) Is unaffected by change in pressure (d) Is unaffected by change in temperature

A-8

CHEMICAL EQUILIBRIUM

ANSWER KEY

EXERCISE ­ 1 : (Basic Objective Questions) 1. (c)

2. (b)

3. (d)

4. (d)

5. (d)

6. (c)

7. (a)

8. (a)

9. (c)

10. (d)

11. (d)

12. (b)

13. (b)

14. (d)

15. (a)

16. (b)

17. (b)

18. (a)

19. (c)

20. (c)

21. (d)

22. (c)

23. (c)

24. (b)

25. (d)

26. (a)

27. (d)

28. (c)

29. (c)

30. (a)

31. (d)

32. (a)

33. (c)

34. (b)

35. (c)

36. (a)

37. (d)

38. (b)

39. (d)

40. (b)

41. (c)

42. (a)

43. (b)

44. (a)

45. (d)

46. (a)

47. (b)

48. (d)

49. (a)

50. (a)

51. (a)

52. (a)

53. (b)

54. (d)

55. (b)

56. (a)

57. (c)

58. (d)

59. (b)

60. (d)

61. (b)

62. (b)

63. (d)

64. (b)

65. (d)

66. (d)

67. (d)

68. (a)

69. (d)

70. (a)

71. (c)

72. (c)

73. (c)

74. (a)

75. (c)

76. (b)

77. (c)

78. (d)

79. (d)

80. (d)

81. (d)

82. (c)

83. (d)

84. (c)

Dream on !!

[\]^[\]^

A-9

IONIC EQUILIBRIUM

EXERCISE - 1 : BASIC OBJECTIVE QUESTIONS Theory of Electrolytes, Acid­Base Strength Comparison 1.

2.

3.

Which one of the following substance has the highest proton affinity (a) H2O

(b) H2S

(c) NH3

(d) PH3

(a) BF3

(b) FeCl3

(c) SiF4

(d) C2H4

+

8.

(c) NH24

(d) N2

According to Bronsted principle, an aqueous solution of HNO3 will contain

2–

–

14.

(b) NO3

–

+

(d) NO

+

Ammonium ion is (a) Neither an acid nor base (b) Both an acid and a base (c) A conjugate acid (d) A conjugate base Which shows weak ionisation in water (a) H2SO4

(b) NaCl

(c) HBr

(d) HI

(c) HNO3

(d) NH3

Among the following, the weakest Lewis base is

15.

2–

The conjugate acid of HPO4 is –

–

(b) OH

–

–

(d) HCO3

–

(b) PO4

(c) H3PO4

(d) H3PO3

Weak Acids and Bases : Analysis

(a) HClO4

(b) HCl

16.

(c) HOCl

(d) HClO3

–

Which one is the weakest acid (a) HNO3

(b) HClO4

(c) H2SO4

(d) HBr

17.

Which one is lewis acid (a) Cl

(b) Ag

(c) C2H5OH

(d) S

3–

(a) H2PO4

Cl is the conjugate base of

A monoprotic acid in 1.00 M solution is 0.01% ionised. The dissociation constant of this acid is (a) 1 × 10

–8

(b) 1 × 10

–4

(c) 1 × 10

–6

(d) 10

–5

If D is the degree of ionization, C the concentration of a weak electrolyte and Ka the acid ionization constant, then the correct relationship between D, C and Ka is

+

2 (a) D

2–

Ka C

2 (b) D

C Ka

The correct order of acid strength is (c) D

(a) HClO < HClO2 < HClO3 < HClO4 (b) HClO4 < HClO < HClO2 < HClO3 (c) HClO2 < HClO3 < HClO4 < HClO

18.

(a) K a

Which of the following is the weakest base (a) NaOH

(b) Ca(OH)2

(c) NH4OH

(d) KOH

(c) D

A-10

Ka C

(d) D

C Ka

For a weak acid HA, Ostwald’s dilution law is represented by the equation

(d) HClO4 < HClO3 < HClO2 < HClO 10.

–

(b) HCl

–

9.

(b) NH2†

(a) HF

(c) Cl

7.

13.

Which of the following is the weakest acid

(a) H

6.

(d) N3

(a) NH†4

(c) NO2

–

(b) NH

Conjugate base of NH3 is

(a) NO2

The conjugate base of NH2 is (c) NH4

5.

12.

Which of the following is not a Lewis acid

(a) NH3 4.

11.

Dc 1 D

Kac 1 c

2

(b) K a

(d) K a

D 2c 1 D D 2c 1 D2

IONIC EQUILIBRIUM 19.

20.

In which of the following dissociation of NH4OH will be minimum

27.

+

(a) NaOH

(b) H2O

(a) HClO4 is the conjugate acid of H2O

(c) NH4Cl

(d) NaCl

(b) H3O is the conjugate base of H2O

+

The following equilibrium exists inaqueous solution, –

(c) H2O is the conjugate acid of H3O

+

–

(a) Concentration of CH3COO will increase

(b) A weak acid

(c) The equilibrium constant will increase

(c) Both a weak acid and a weak base

(d) The equilibrium constant will decrease

(d) Neither an acid nor a base

The degree of dissociation of 0.1 M HCN solution is 0.01%. Its ionisation constant would be

The unit of ionic product of water Kw are –1

–1

(b) Mol L

–2

–1

(d) Mol L

–3

(b) 10

–5

(a) Mol L

(c) 10

–7

(d) 10

–9

(c) Mol L

The hydrogen ion concentration in weak acid of dissociation constant Ka and concentration c is nearly equal to

(d)

30.

–

(c) OH is added

Kac

Degree of dissociation of 0.1 N CH3COOH is 32.

–5

(a) 10

–5

(b) 10

–4

(c) 10

–3

(d) 10

–2

0.2 molar solution of formic acid is ionized 3.2%. Its ionization constant is (a) 9.6 × 10

–3

–6

(c) 1.25 × 10

(b) 2.1 × 10

–4

(d) 4.8 × 10

–5

–2

(b) 1.6 × 10

–4

–10

(d) 4.3 × 10

–7

(c) 4.4 × 10

–

34.

(b) 4.5 × 10 M

–6

(d) 9.2 × 10 M

(c) 6.3 × 10 M

+

(d) Temperature increases

(b) 10–12

(c) 10–14

(d) 10–8

HCOOH and CH3COOH solution have equal pH. If K1/K2 (ratio of acid dissociation constants) is 4, ratio of their molar concentration will be : (a) 2

(b) 0.5

(c) 4

(d) 0.25 H3O+ + OH–, Kw = 1 × 10–14 at 25ºC hence Ka is:

2H2O (a) 1 × 10

(b) 5.55 × 10–3

(c) 18 × 10–17

(d) 1.00 × 10–7

For a “c molar” concentrated solution of a weak electrolyte AxBy, the degree of dissociation is given as (a) D

K eq / c ( x  y )

(b) D

K eq c /( xy )

–10

–6

(c) D

–6

K eq / c x  y 1x x y y

–6

(d)

A-11

–2

(b) H is added

(a) 10–6

Concentration CN in 0.1 M HCN is [Ka = 4 × 10 ] (a) 2.5 × 10 M

2

–2

At 90ºC, pure water has [H3O+] = 10–6 mole litre–1. What is the value of Kw at 90º C ?

–14

The values of dissociation constants of some acids (at 25°C) are as follows. Indicate which is the strongest acid in water (a) 1.4 × 10

33.

–2

Ionic product of water increases, if (a) Pressure is reduced

31.

(b) c/Ka

Ka / c

(Dissociation constant = 1 × 10 )

26.

29.

(a) 10

(c) Kac

25.

H3O+ + OH–, water is

(b) Concentration of CH3COO will decrease

(a)

24.

In the reaction 2H2O (a) A weak base

–

23.

+

(d) ClO4 is the conjugate base of HClO4 28.

–

22.

–

ZZZX HClO4 + H2O YZZ Z H3O + ClO4

CH3COOH YZZ ZZZX Z CH3COO + H if dil HCl is added, without change in temperature, the

21.

Review the equilibrium and choose the correct statement

K eq / xyc

1 /( x  y )

IONIC EQUILIBRIUM 35.

36.

The pH of a 0.1 M aqueous solution of a weak acid (HA) is 3. What is its degree of dissociation ? (a) 1%

(b) 10%

(c) 50%

(d) 25%

(b) pH decreases and pOH increases with increase in temperature

(a) [NH 4 ] [NH 3 ]

(b) [NH 2 ] [NH 3 ]

 4

 4

 2

(c) both pH and pOH increase with increase in temperature (d) both pH and pOH decrease with increase in temperature

 2

(d) [NH ] /[NH ] 45.

pH of Solutions What is the pH value of N/1000 KOH solution (a) 10

–11

(c) 2 38.

39.

40.

(a) less than 1

(b) Between 1 and 2

(d) 11

(c) 3

(d) Between 4 and 5

The pH of a 0.001 M NaOH will be (a) 3

(b) 2

(c) 11

(d) 12

46.

–4

–9

The pH of a 10 M solution of HCl in water is (a) 8

(b) – 8

(c) Between 7 and 8

(d) Between 6 and 7

A is an aqueous acid; B is an aqueous base. They are diluted separately, then

47.

(a) A < B < C < D

(b) A > B > C > D

(c) A = B = C = D

(d) None of these

The pH of a solution is increased from 3 to 6; its H+ ion concentration will be :

(a) pH of A increases and pH of B decreases

(a) reduced to half

(b) pH of A increases and pH of B decreases till pH in each case is 7

(c) reduced by 1000 times (d) increased by 1000 times 48.

Pure water is kept in a vessel and it remains exposed to atmospheric CO2 which is absorbed, then its pH will be (a) Greater than 7

(b) Less than 7

(c) 7

49.

(d) Depends on ionic product of water

43.

The dissociation constants of monobasic acids A, B, C and D are 6 × 10 , 5 × 10–5, 3.6 × 10–6 and 7 × 10–10 respectively. The pH values of their 0.1 molar aqueous solutions are in the order :

(d) pH of B and A decrease

42.

Equal volumes of two solutions of hydrochloric acid are mixed. One solution has a pH 1 while the other has a pH5. The pH of the resulting solution is :

(b) 3

(c) pH of A and B increase

41.

For a pure water, (a) pH increases and pOH decreases with increase in temperature

Autoprotolysis constant of NH3 is

(c) [NH ] [NH ]

37.

44.

(b) 6

(c) 4

(d) 8

Aspirin (acetyl salicyclic acid, molar mass = 180 g mol–1) used as analgesic has pKa value of 2. Two tablets of aspirin each weighing 90 mg are dissolved in 100 mL of water. The pH of the solution is (a) 0.5

(b) 1.0

(c) 2.0

(d) 4.0

A patient is said to suffer from acidosis when the pH of his blood (a) falls below 7.35

An acid solution of pH = 6 is diluted hundred times. The pH the solution becomes : (a) 6.95

(b) doubled

(b) rises above 7.35 (c) Shows sudden fall and rise (d) has strong basic character

The number of H+ ions present in 1 mL of a solution having pH = 13 is :

50.

What is the pH of a 0.015 M Ba(OH) 2 solution ?

(a) 1013

(b) 6.023 × 1013

(a) 1.82

(b) 1.52

(c) 6.023 × 107

(d) 6.023 × 1010

(c) 12.48

(d) 12.18

A-12

IONIC EQUILIBRIUM 51.

52.

Equal volumes of two solutions of HCl are mixed. One solution has a pH = 1, while the other has a pH = 5. The pH of the resulting solution is (a) < 1

(b) Between 1 and 2

(c) 3

(d) Between 4 and 5

At a certain temperature the value of pK w is 13.4 and the measured pH of soln is 7. The solution is (a) Acidic

(b) Basic

(c) Neutral

(d) Unpredictable

59.

(Given : Ka for HCOOH = 2 × 10–4)

60.

Buffer Solutions 53.

54.

55.

56.

For preparing a buffer solution of pH 6 by mixing sodium acetate and acetic acid, the ratio of the concentration of –5 salt and acid should be (Ka = 10 ) (a) 1 : 10

(b) 10 : 1

(c) 100 : 1

(d) 1 : 100

In a mixture of a weak acid and its salt, the ratio of concentration of acid to salt is increased ten-fold. The pH of the solution (a) Decreases by one

(b) Increases by one-tenth

(c) Increases by one

(d) Increases ten-fold

(a) 7.302

(b) 9.302

(c) 8.302

(d) 10.302

63.

(b) 0.05

(c) 0.1

(d) 0.2

What is the pH of a buffer solution which is 0.250 M in benzoic acid, C 6 H5 COOH, and 0.150 M in sodium benzoate, C 6 H 5 COONa, if Ka for benzoic acid is 6.5 × 10 –5 ? (a) 3.40

(b) 3.97

(c) 4.19

(d) 4.41

Calculate the pH of a buffer prepared by mixing 0.10 mol of sodium formate and 0.05 mole of formic acid in 1.0 L of solution. [HCO2H : Ka =1.8 × 10 –4] (a) 1.8 × 10 –4

(b) 3.44

(c) 4.05

(d) 5.31

100 mL of a buffer solution contains 0.1 M each of weak acid HA and salt NaA. How many gram of NaOH should be added to the buffer so that its pH will be 6 ? (Ka of HA = 10–5). (a) 0.328

(b) 0458

(c) 4.19

(d) None

Two buffer solutions, A and B, each made with acetic acid and sodium acetate differ in their pH by one unit, A has salt : acid = x : y, B has salt : acid = y : x. If x > y, then the value of x : y is (a) 10,000

(b) 3.17

(a) NaH2PO4 + Na2HPO4

(c) 6.61

(d) 2.10

64.

Buffer solutions can be prepared from mixtures of

(c) CH3COOH + CH3COONa

(a) HCl and NaCl

(b) NaH2PO4 and Na2HPO4

(d) H2CO3 + HCO 3

(c) CH3COOH + NaCl

(d) NH4OH + NH3

0.1 mole of CH3NH2 (Kb = 5 × 10–4) is mixed with 0.08 mole of HCl and the solution diluted to one litre. The H+ ion concentration in the solution will be (a) 1.6 × 10–11 (c) 5 × 10

58.

62.

(a) 0.01

The principal buffer present in human blood is

(b) H3PO4 + NaH2PO4

57.

61.

–10

The dissociation constant of HCN is 5 × 10 . The pH of the solution prepared by mixing 1.5 mole of HCN and 0.15 moles of KCN in water and making up the total volume 3 to 0.5 dm is

How many moles of HCOONa must be added to 1L of 0.1 M HCOOH to prepare a buffer solution with a pH of 3.4 ?

–5

65.

4 of X4 and HX. The K b for X is 10 –10. The pH of the buffer is

(b) 8 × 10–11 (d) 8 × 10

–2

The pKa of weak acid (HA) is 4.5. The pOH of an aqueous buffered solution of HA in which 50% of the acid is ionized is

A certain buffer solution contains equal concentration

66.

(a) 4

(b) 7

(c) 10

(d) 14

The pKb of CN 4 is 4.7. The pH of solution prepared by mixing 2.5 mol of KCN of 2.5 mol of HCN in water and making the total volume upto 500 mL is

(a) 7.0

(b) 4.5

(a) 10.3

(b) 9.3

(c) 2.5

(d) 9.5

(c) 8.3

(d) 4.7

A-13

IONIC EQUILIBRIUM 67.

68.

69.

70.

The pH of a dilute solution of acetic acid was found to be 4.3. The addition of a small crystal of sodium acetate will cause pH to

74.

(a) Become less than 4.3

(b) Become more than 4.3

(a) [H2SO4] > [H+]

(b) [H+] = [SO 24 ]

(c) Remain equal to 4.3

(d) Unpredictable

(c) [SO 24 ] > [HSO 4 ]

(d) [H+] > [HSO 4 ]

To 1.0 L solution containing 0.1 mol each of NH3 and NH 4Cl, 0.05 mol NaOH is added. The change in pH will be (pKa for CH3COOH = 4.74) (a) 0.30

(b) –0.30

(c) 0.48

(d) –0.48

The pH of blood is maintained by the balance between H2CO3 and NaHCO 3. If the amount of CO 2 in the blood is increased, how will it effect the pH of blood ?

75.

73.

4.5 × 10–3 and K a2

1.7 × 10–10 at 298 K

(a) 3.0

(b) 10.0

(c) 6.1

(d) 7.2

Salt Hydrolysis 76.

Aq. solution of sodium cyanide is

(a) pH will remain same

(b) pH will be 7

(a) Acidic

(b) Amphoteric

(c) pH will increase

(d) pH will decrease

(c) Basic

(d) Netural

Fixed volume of 0.1 M benzoic acid (pKa = 4.2) solution is added into 0.2 M sodium benzoate solution and formed a 300 mL, resulting acidic buffer solution. If pH of the resulting solution is 3.9, then added volume of benzoic acid is (a) 240 mL

(b) 150 mL

(c) 100 mL

(d) None

77.

78.

In a 0.010 M solution of oxalic acid, H 2 C 2 O 4 ,

79.

The solution of strong acid and weak base (FeCl3) is (a) Acidic

(b) Basic

(c) Neutral

(d) none of these

Which one of the following salts gives an acidic solution in water (a) CH3COONa

(b) NH4Cl

(c) NaCl

(d) CH3COONH4

An aqueous solution of aluminium sulphate would show (a) An acidic reaction

Ka1 = 5.9 × 10 –2, Ka2 = 6.4 × 10 –5, the species present in the lowest concentration is

72.

What is the pH of 0.01 M glycine solution ? For glycine

K a1

Polyprotic Acids and Bases 71.

Which one of the following statements is true with regard to a 0.10 M H2 SO4 solution ?

(b) A neutral reaction

(a) H2C2O4

(b) H3O+

(c) A basic reaction

(c) HC2 O 4

(d) C2 O42

(d) Both acidic and basic reaction 80.

Calculate the carbonate ion concentration in a 0.10 M solution of the weak acid, carbonic acid are (H 2CO3). The dissociation constants of carbonic acid are Ka1 = 4.5 × 10 –7 and Ka2 = 4.7 × 10 –11. (a) 4.7 × 10 –11 M

(b) 1.0 × 10 –7 M

(c) 4.5 × 10 –7 M

(d) 2.1 × 10 –4 M

(a) Weak acid and weak base (b) Strong acid and weak base (c) Weak acid and strong base (d) Strong acid and strong base

In a saturated solution of H2S, DECREASING the pH of the solution will cause

81.

(a) the S2– concentration to decrease (b) the H 2S concentration to decrease

An aqueous solution of sodium carbonate is alkaline because sodium carbonate is a salt of

82.

The aqueous solution of ammonium chloride is (a) Neutral

(b) Basic

(c) Acidic

(d) Amphoteric

The aqueous solution of FeCl3 is acidic due to

(c) the S 2– concentration to increase

(a) Acidic impurities

(b) Ionisation

(d) no change in either the H2S or S 2– concentration

(c) Hydrolysis

(d) Dissociation

A-14

IONIC EQUILIBRIUM 83.

Which is the correct alternate for hydrolysis constant of

Solubility Equilibria

NH4CN

91.

(a)

(c) 84.

Kw Ka

Kw (b) K u K a b

Kb c

Ka (d) K b

92.

93.

94.

HCN is a weak acid (Ka = 6.2 × 10–10). NH4OH is a weak base (Kb = 1.8 × 10–5). A 1M solutionk of NH4CN would be : (a) strongly acidic (b) weakly acidic (c) neutral (d) weakly basic

–1

(a) 4 × 10

–15

(b) 4 × 10

–10

(c) 1 × 10

–15

(d) 1 × 10

–10

The solubility of CaF2 is a moles/litre. Then its solubility product is........ 2

(c) 3s 95.

(b) 4s 2

(d) s

+ 2

–2

+

96.

3

+

–2

+ 2

A precipitate of AgCl is formed when equal volumes of –10

(b) 3.6 × 10

the following are mixed. [Ksp for AgCl = 10 ]

–7

–4

–7

–5

–6

–5

–4

–6

–6

(a) 10 M AgNO3 and 10 M HCl

(d) 2.8 × 10 –6

(b) 10 M AgNO3 and 10 M HCl

F (aq.)  H 2 O(l ) U HF(aq.)  OH  (aq.)

(d) 10 M AgNO3 and 10 M HCl

(a) 6.9 × 10–11

(b) 1.4 × 10 –11

(c) 2.6 × 10 –9

(d) 8.3 × 10 –6

(c) 10 M AgNO3 and 10 M HCl

97.

Ka for HF is 3.5 × 10 –4. Calculate Kb for the fluoride ion. –4

(b) 1.0 × 10

–11

(d) 1.0 × 10 –14

(c) 2.9 × 10

–6

(a) 2 × 10 (c) 5 × 10–9

4+

2– 2

2+

2–

(a) [Sn ] [S ]

(c) [Sn ] [2S ] 98.

[H † ] in an aqueous so lution in whic h t he concentration of ASc

The correct representation for solubility product of SnS2 is 2+

2– 2

(b) [Sn ] [S ]

–7

Ka for ascorbic acid (HASc) is 5 × 10–5. Calculate the

4

–2

(d) [2Ag ] [CrO4 ]

Ka for hydrofluoric acid is 6.9 × 10–4 . What is the equilibrium constant K for the following reaction ?

(a) 3.5 × 10

–2

(b) [Ag ] [CrO4 ]

(c) [2Ag ] [CrO4 ]



3

Which is the correct representation of the solubility product constant of Ag2CrO4 (a) [Ag ] [CrO4 ]

What is Ka for HOCl ?

90.

The solubility in water of a sparingly soluble salt AB2 is

(a) s

The equilibrium constant for this reaction is 3.6×10 –7.

OCl (aq.)  H 2 O(l ) U HOCl(aq.)  OH (aq.)

89.

(d) 9x –5



88.

5

1.0 × 10 mol l . Its solubility product number will be

(b) 7.4 (d) 3.7

(c) 6 × 10

M

(b) 108x 4

respectively. The isoelectric point of amino acid is :

–4

–10

Let the solubility of an aqueous solution of Mg(OH)2 be 3

The pK a1 and pK a 2 of an amino acid are 2.3 and 9.7

(a) 2.8 × 10

(d) 4.0 × 10

(a) 4x

(d) HCl < NaCl < NaCN < NH4Cl.

–8

–4

x then its ksp is

(c) NaCN < NH4Cl < NaCl < HCl

87.

–4

(c) 27x

(b) HCl < NH4Cl < NaCl < NaCN

86.

(b) 1.0 × 10 M

(c) 1.6 × 10 M

The pH 0.1 M solution of the following salts increases in the order :

(a) 12 (c) 6.0

–6

(a) 2.0 × 10 M

(a) NaCl < NH4Cl < NaCN < HCl

85.

The solubility product of a salt having general formula –12 2+ MX2, in water is : 4 × 10 . The concentration of M ions in the aqueous solution of the salt is

–9

–3

(b) 1 × 10

–27

(d) 1 × 10

(c) 4 × 10

(b) 2 × 10 –7 (d) 5 × 10 –10

A-15

2– 2

Solubility product of BaCl2 is 4 × 10 . Its solubility in moles/litre would be (a) 1 × 10

ions is 0.02 M.

4+

(d) [Sn ] [2S ]

–9

–27

IONIC EQUILIBRIUM 99.

100.

107. Ksp of Mg(OH)2 is 4.0 × 10 –6. At what minimum pH, Mg2+ ions starts precipitating 0.01 MgCl

Solubility of AgCl will be minimum in (a) 0.001 M AgNO3

(b) Pure water

(c) 0.01 M CaCl2

(d) 0.01 M NaCl –6

At 298 K, the solubility product of PbCl2 is 1.0 × 10 . What will be the solubility of PbCl2 in moles/litre

101.

102.

103.

(a) 6.3 × 10

–3

(b) 1.0 × 10

–3

(c) 3.0 × 10

–3

(d) 4.6 × 10

–14

(a) 2 + log 2

(b) 2 – log 2

(c) 12 + log 2

(d) 12 – log 2

108. The pH of an aqueous solution of Ba(OH)2 is 10. If the Ksp of Ba(OH)2 is 1 × 10–9, then the concentration of Ba2+ ions in the solution in mol L–1 is (a) 1 × 10–2

(b) 1 × 10–4

(c) 1 × 10–1

(d) 1 × 10–5

–6

Ksp for Ca(OH)2 is 5.5 × 10 . What is the maximum pH that can be attained in a sewage tank treated with slaked lime ? (a) 9.35

(b) 10.35

(c) 11.35

(d) 12.35

For a sparingly soluble salt Ap Bq, the relationship of its solubility product (LS) with its solubility (S) is : (a) LS = Sp+q . pp. qq

(b) LS = Sp+q . pq . qp

(c) LS = Spq . pp . qq

(d) LS = Spq (pq)p+q

109. CaCO 3 and BaCO 3 have solubility product values 1 × 10–8 and 5 × 10–9, respectively. If water is shaken up with both solids till equilibrium is reached, the concentration of CO32  ion is (b) 1.225 × 10–4

(c) 2.25 × 10–9

(d) None of these

Indicators and Titrations 110.

When equal volumes of the following solutions are mixed, precipitation of AgCl (Ksp = 1.8 × 10 with :

(a) 1.5 × 10–8

–10

) will occur only 111.

(a) 10–4 M (Ag+) and 10–4M (Cl–) (b) 10–5 M (Ag+) and 10–5 M (Cl–)

What will be the pH of a solution formed by mixing 40 ml of 0.10 M HCl with 10 ml of 0.45 M NaOH (a) 12

(b) 10

(c) 8

(d) 6

The rapid change of pH near the stoichiometric point of an acid-base titration is the basis of indicator detection. pH of the solution is related to ratio of the concentrations –

of the conjugate acid (HIn) and base (In ) forms of the indicator by the expression

(c) 10–6 M (Ag+) and 10–6 M (Cl–) (d) 10–10 M (Ag+) and 10–10 M (Cl–) 104.

The solubility product of different sparingly soluble salts are : 1. XY = 4 × 10 –20

2. X2Y = 3.2 × 10–11

3. XY3 = 2.7 × 10–31

(a) log

[HIn] [In  ]

pH  pK In

(b) log

[In  ] [HIn ]

pH  pK In

(c) log

[In  ] [HIn]

pK In  pH

(d) log

[HIn] [In  ]

pK In  pH

The increasing order of solubility is :

105.

(a) 1, 3, 2

(b) 2, 1, 3

(c) 1, 2, 3

(d) 3, 1, 2

Solubility of AgCN is maximum in : (a) acidic buffer solution (b) basic buffer solution (c) in pure water

(d) equal in all solution

106. The solubility of CH3COOAg in a buffer solution with

112.

104 is pH = 4, whose K sp = 10 –12 and Ka = 3 (a) 10 –6

(b) 0.5 × 10 –6

(c) 5 × 10 –6

(d) 2 × 10 –6

The pH indicators are : (a) salts of strong acids and strong bases (b) salts of weak acids and weak bases (c) either weak acids or weak bases (d) either strong acids or strong bases

A-16

IONIC EQUILIBRIUM 113.

In which of the following acid-base titration, pH is greater than 8 at the equivalence point ?

(a) (i) o A, (ii) o B, (iii) o C

(a) Acetic acid versus ammonia

(c) (i) o B, (ii) o C, (iii) o A

(b) Acetic acid versus sodium hydroxide

(d) (i) o C, (ii) o A, (iii) o B

(c) Hydrochloric acid versus ammonia

119. In which of the following acid-base titration, pH is greater than 8 at the equivalence point ?

(d) Hydrochloric and versus sodium hydroxide 114.

Why are strong acids generally used as standard solutions in acid-base titrations ?

(a) Aceitic acid vs ammonia

(a) The pH at the equivalent point will always be 7

(c) Hydrochloric acid vs ammonia

(b) They can be used to titrate both strong and weak bases

(d) Hydrochloric acid vs sodium hydroxide

(c) Strong acids form more stable solutions than weak acids 115.

(b) Acetic acid vs sodium ammonia

120. 20 cm3 of x M solution of HCl is exactly neutralised by 40 cm3 of 0.05 M NaOH solution, the pH of HCl solution is

(d) The salts of strong acid do not hydrolyze

(a) 1.0

(b) 2

The best indicator for detection of end point in titration of a weak acid and a strong base is :

(c) 1.5

(d) 2.5

(a) methyl orange (3 to 4)

121. Phenolphthalein does not act as an indicator for the titration between

(b) methyl red (5 to 6)

(a) HCl and NH4OH

(b) Ca(OH)2 and HCl

(c) bromothymol blue (6 to 7.5)

(c) NaOH and H 2SO4

(d) KOH and CH3COOH

122. Methyl orange gives red colour in

(d) phenolphthalein (8 to 9.6) 116.

(b) (i) o A, (ii) o C, (iii) o B

Which of the following mixture will have the pH close to 1 ? (a) 100 ml of M/10 HCl + 100 ml of M/10 NaOH (b) 55 ml of M/10 HCl + 45ml of M/10 NaOH

123.

(c) 10 ml of M/10 HCl + 90 ml of M/10 NaOH (d) 75 ml of M/5 HCl + 25 ml of M/5 NaOH 117.

Consider the titratiokns listed below. In which is the pH at the equivalence point described incorrectly ? Acid

(a) KOH solution

(b) HCl solution

(c) Na 2CO3 solution

(d) NaCl solution

Which acid-base indicator should be used in an aqueous solution titration which is complete at about 0.001 M H+ (aq) ? The transition range in pH is given in parenthesis. (a) Methyl violet (0.5 – 1.5)

Base

pH at equiv.pt.

(b) Methyl red (4.2 – 6.3)

(a) CH3COOH

NaOH

> 7.00

(c) Methyl yellow (2.9 – 4.0)

(b) HNO3

Ca(OH)2

= 7.00

(d) Phenol red (6.4 – 8.0)

(c) HCl

NH 3

< 7.00

(d) HF

NaOH

< 7.00

124.

118. Which indicator is suitable for the titrations : Titration (i) HCOOH/NaOH

Indicator (A) Bromothymol blue or phenolphthalein or methyl orange or thymolphthalein

(ii) HBr/KOH

(B) Methyl orange or methyl red or bromocresol green

(iii) NH4OH.HNO3

(C) Phenolphthalein or thymolphthalein

A-17

A sample of 50.0 mL of 0.10 M NH3 (Kb = 1.8 × 10–5) is titrated with 0.10 M HCl. Calculate the pH at the equivalence point. (a) 6.98

(b) 2.87

(c) 7.78

(d) 5.28

IONIC EQUILIBRIUM

Answer Key Exercise­1 (Basic Objective Questions) 1. (c)

2. (d)

3. (b)

4. (a)

5. (c)

6. (b)

7. (a)

8. (b)

9. (a)

10. (c)

11. (c)

12. (b)

13. (c)

14. (d)

15. (a)

16. (a)

17. (c)

18. (b)

19. (a,c)

20. (b)

21. (d)

22. (d)

23. (d)

24. (b)

25. (a)

26. (c)

27. (d)

28. (c)

29. (d)

30. (d)

31. (b)

32. (d)

33. (c)

34. (c)

35. (a)

36. (c)

37. (d)

38. (c)

39. (d)

40. (a)

41. (b)

42. (a)

43. (c)

44. (d)

45. (b)

46. (a)

47. (c)

48. (c)

49. (a)

50. (c)

51. (b)

52. (b)

53. (b)

54. (a)

55. (c)

56. (d)

57. (b)

58. (d)

59. (b)

60. (b)

61. (c)

62. (a)

63. (b)

64. (b)

65. (a)

66. (b)

67. (b)

68. (c)

69. (d)

70. (a)

71. (d)

72. (a)

73. (a)

74. (d)

75. (c)

76. (c)

77. (a)

78. (b)

79. (a)

80. (c)

81. (c)

82. (c)

83. (b)

84. (b)

85. (c)

86. (d)

87. (a)

88. (b)

89. (c)

90. (c)

91. (b)

92. (a)

93. (a)

94. (b)

95. (a)

96. (c)

97. (a)

98. (a)

99. (c)

100. (a)

101. (d)

102. (a)

103. (a)

104. (a)

105. (a)

106. (d)

107. (c)

108. (c)

109. (b)

110. (a)

111. (b)

112. (c)

113. (b)

114. (c)

115. (d)

116. (d)

117. (d)

118. (d)

119. (b)

120. (a)

121. (a)

122. (b)

123. (c)

124. (d)

Dream on !!

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A-18

SOME BASIC CONCEPTS OF CHEMISTRY

EXERCISE - 1 : BASIC OBJECTIVE QUESTIONS Molecules

Atoms 1.

2.

3.

4.

5.

Which of the following contains atoms equal to those in 12 g Mg ? (At. wt. Mg = 24) (a) 12 gm C

(b) 7 gm N2

(c) 32 gm O2

(d) None of These

8.

9.

1 moles of oxygen combine with Al to form Al2O3, the 2 weight of Al used in the reaction is (Al = 27)

If 1

(a) 27 g

(b) 54 g

(c) 40.5 g

(d) 81 g

11.

Which has the highest mass ? (a) 50 g of iron

(b) 5 moles of N2

(c) 0.1 mol atom of Ag

(d) 1023 atoms of carbon

11.

(b) 64 g of S

(c) 8 g of O

(d) 48 g of Mg

(a) 1.0 × 1023

(b) 1.5 × 1023

(c) 2.0 × 1023

(d) 2.5 × 1023

If 20% nitrogen is present in a compound, its minimum molecular weight can be (a) 144

(b) 28

(c) 100

(d) 70

The weight of molecule of the compound C60H122 is (a) 1.4 × 10–21 g

(b) 1.09 × 10–21 g

(c) 5.025 × 1023 g

(d) 16.023 × 1023 g

Choose the wrong statement : (a) 1 mole means 6.02 × 1023 particles

The number of atoms present in 0.5 mole of nitrogen is same as the atoms in (a) 12 g of C

The number of molecules in 4.25 g of ammonia is about

(b) Molar mass is mass of one molecule (c) Molar mass is mass of one mole of a substance (d) Molar mass is molecular mass expressed in grams

Which of the following weighs the least ? 12.

Which among the following is the heaviest ?

(a) 2 g atom of N (at. wt. of N = 14) (a) One mole of oxygen

(b) 3 × 1023 atoms of C (at. wt. of C = 12)

(b) One molecule of sulphur trioxide

(c) 1 mole of S (at. wt. of S = 32)

6.

7.

(d) 7 g silver (at. wt. of Ag = 108)

(c) 100 amu of uranium

If A is Avogadro’s number then number of valence electrons in 4.2 g of nitride ions (N3–) is

(d) 44g of carbon dioxide

(a) 2.4 NA

(b) 4.2 NA

(c) 1.6 NA

(d) 3.2 NA

13.

Haemoglobin contains 0.33% of iron by weight. The molecular weight of haemoglobin is approximately 67200. The number of iron atoms (at. wt. of Fe = 56) present in one molecule of haemoglobin is (a) 6

(b) 1

(c) 4

(d) 2

A-19

Rearrange the following I to IV in order of increasing masses and choose the correct answer [At. wt. of N = 14 u, O = 16 u, Cu = 63 u] I

1 molecule of oxygen

II 1 atom of nitrogen III 1 × 10–10 mol molecule of oxygen IV 1 × 10–10 mol atom of copper (a) II < I < III < IV

(b) IV < III < II < I

(c) II > I > III > IV

(d) I < II < IV < III

SOME BASIC CONCEPTS OF CHEMISTRY 14.

15.

The number of moles of SO2Cl2 in 13.5 g is : (a) 0.1

(b) 0.2

(c) 0.3

(d) 0.4

21.

The largest number of molecules is in

(a) 20

(b) 40

(a) 36 g of water

(c) 16

(d) 35.5

(b) 28 g of carbon monoxide

16.

(c) 46 g of ethyl alcohol

30g of magnesium and 30g of oxygen are reacted, then the residual mixture contains

(d) 54 g of nitrogen pentoxide.

(a) 60g of Magnesium oxide only

Which of the following contains maximum number of atoms ?

(b) 40g of Magnesium oxide and 20 g of oxygen

22.

(a) 6.023 × 1021 molecules of CO2

(c) 45 g of Magnesium oxide and 15g of oxygen

(b) 22.4 L of CO2 at STP

(d) 50 g of Magnesium oxide and 10g of oxygen 23.

(c) 0.44 g of CO2 (d) None of these

18.

19.

20.

How many grams of SiC could be formed by reacting 2.00 g of SiO2 and 2.0 g of C ?

If 0.5 mol of BaCl2 is mixed with 0.2 mol of Na3PO4, the maximum number of mole of Ba3(PO4)2 that can be formed is (a) 0.7

(b) 0.5

(c) 0.30

(d) 0.10

24.

One mole of a mixture of CO and CO2 requires exactly 20 gram of NaOH in solution for complete conversion of all the CO2 into Na2 CO3. How many moles more of NaOH would it require for conversion into Na2CO3 if the mixture (one mole) is completely oxidised to CO2. (a) 0.2

(b) 0.5

(c) 0.4

(d) 1.5

Silicon carbide, is produced by heating SiO2 and C to high temperatures according to the equation : SiO2 (s) + 3C (s) o SiC (s) + 2CO (g)

Stoichiometric Calculations 17.

A sample of pure calcium weighing 1.35 g was quantitatively converted to 1.88 g of pure calcium oxide. Atomic mass of calcium would be :

The number of water molecules present in a drop of water (volume = 0.0018 ml) at room temperature is (density of H2O = 1 g/mL)

(a) 1.33

(b) 2.56

(c) 3.59

(d) 4.0

Given the reaction Pb(NO3)2 (aq) + 2KI o PbI2 (s) + 2KNO3 (aq) What is the mass of PbI2 that will precipitate if 10.2 g of Pb(NO3)2 is mixed with 5.73 g of KI in a sufficient quantity of H2O ?

25.

(a) 2.06 g

(b) 4.13 g

(c) 7.96 g

(d) 15.9 g

If 9 moles of O2 and 14 moles of N2 are placed in a container and allowed to react according to the equation :

(a) 6.023 × 1019

(b) 1.084 × 1018

3O2 + 2N2 o 2N2O3

(c) 4.84 × 1017

(d) 6.023 × 1023

The reaction proceeds until 3 moles of O2 remain, how

What is the weight of oxygen required for the complete combustion of 2.8 kg of ethylene ? (a) 2.8 kg

(b) 6.4 kg

(c) 9.6 kg

(d) 96 kg

A-20

many moles of N2O3 are present at that instant ? (a) 6

(b) 3

(c) 4

(d) 12

SOME BASIC CONCEPTS OF CHEMISTRY 26.

Iron (III) oxide can be reduced with CO to foum metalic iron as described by unbalanced chemical reaction

32.

Fe2O3 + CO o Fe + CO2 The number of moles of CO required to form one mole of Fe from its oxide is

28.

(d) 54

(c) 2

(d) 3

33.

A molal solution is one that contains one mole of a solute in (a) 1000 g of the solvent

The mass of CaO that shall be obtained by heating 20 kg of 90% pure lime-stone (CaCO3) is

(b) one litre of the solvent

(a) 11.2 kg

(b) 8.4 kg

(c) one litre of the solution

(c) 10.08 kg

(d) 16.8 kg

(d) 22.4 litres of the solution

If potassium chlorate is 80% pure, then 48 g of oxygen would be produced from (atomic mass of K = 39)

34.

An aqueous solution of ethanol has density 1.025 g/mL and it is 2 M. What is the molality of this solution ?

(a) 153.12g of KClO3

(b) 122.5 g of KClO3

(a) 1.79

(b) 2.143

(c) 245 g of KClO3

(d) 98.0 g of KClO3

(c) 1.951

(d) None of these

35.

Antimony reacts with sulphur according to the equation

What is the percentrage yield for a reaction in which 1.40 g of Sb2S3 is obtained from 1.73 g of antimony and a slight excess of sulphur ?

What volume of 0.4 M FeCl3 . 6H2O will contain 600 mg of Fe3+ ?

The molar mass of Sb2S3 is 340 g mol–1.

(a) 49.85 mL

(b) 26.78 mL

(c) 147.55 mL

(d) 87.65 mL

(a) 80.9 %

(b) 58.0 %

A sample of H2SO4 (density 1.8 g/ml) is 90% by weight. What is the volume of the acid that has to be used to make 1 litre of 0.2 M H2SO4?

(c) 40.5 %

(d) 29.0 %

(a) 16 mL

(b) 10 mL

(c) 12 mL

(d) 18 mL

36.

NH3 is produced according to the following reaction : (g) + 3H2(g) o 2NH3(g) 2 In an experiment 0.25 mol of NH3 is formed when 0.5 mol of is reacted with 0.5 mol of H2. What is % yield ? 2 (a) 75%

(b) 50%

(c) 33%

(d) 25%

37.

What is the weight % sulhuric acid in an aqueous solution which is 0.502 M in sulphuric acid ? The specify gravity of the solution is 1.07

The density (in g mL–1) of a 3.60 M sulphuric acid solution that is 29% H2SO4 (molar mass = 98 g mol–1) by mass will be

Strength : Mass Percent 31.

(c) 46

Strength : Molality

2Sb(s) + 3S(s) o Sb2S3(s)

30.

(b) 75

(b) 1.5

Percentage Yield 29.

(a) 25

(a) 1

Percentage Purity 27.

Mole fraction of ethanol in ethanol - water mixture is 0.25. Hence, percentage concentration of ethanol (C2H6O) by weight of mixture is

(a) 1.45

(b) 1.64

(c) 1.88

(d) 1.22

(a) 4.77 %

(b) 5.67 %

An antifreeze mixture contains 40% ethylene glycol (C2H6O2) by weight in the aqueous solution. If the density of this solution is 1.05 g mL, what is the molar concentration?

(c) 9.53 %

(d) 22.0 %

(a) 6.77 M

(b) 6.45 M

(c) 0.0017 M

(d) 16.9 M

38.

A-21

SOME BASIC CONCEPTS OF CHEMISTRY 39.

What is the molarity of SO 24  ion in aqueous solution that

46.

contain 34.2 ppm of Al2(SO4)3 ? (Assume complete dissociation and density of solution 1 g/mL) (a) 3 × 10–4 M –4

(c) 10 M

(b) 2 × 10–4 M (d) None of these

42.

(b) 18 mL

(c) 12 mL

(d) 10 mL

The mole fraction of a given sample of I2 in C 6 H 6 is 0.2.

What is the concentration of nitrate ions if equal volumes of 0.1 M AgNO3 and 0.1 M NaCl are mixed together ?

The molality of I2 in C6H6 is

(a) 0.1 M

(b) 0.2 M

(a) 0.32

(b) 3.2

(c) 0.05 M

(d) 0.25 M

(c) 0.032

(d) 0.48

47.

48.

Strength : Variation 41.

(a) 16 mL

Strength : Stoichiometric Calculations

Strength : Mole Fraction 40.

A sample of H2SO4 (density 1.8 g mL–1) is 90% by weight. What is the volume of the acid that has to be used to make 1 L of 0.2 M H2SO4 ?

How many grams of NaBr could be formed if 14.2 g of NaI are reacted with 40.0 mL of a 0.800 M Br2 ? 2NaI + Br2 o 2NaBr + I2

In which mode of expression, the concentration of a solution remains independent of temperature ?

(a) 3.30

(b) 4.80

(a) Molarity

(b) Normality

(c) 6.59

(d) 9.75

(c) Formality

(d) Molality

49.

With increase of temperature, which of these changes ?

If AgBr is assumed to be completely insoluble, What mass of AgBr precipitates when 30.0 mL of a 0.500 mol/L solution of AgNO3 is added to 50.0 mL of an 0.400 mol/L solution of NaBr ?

(a) molality (b) weight fraction of solute (c) fraction of solute present in unit volume of water

50.

(d) mole fraction. 43.

Molarity and Normality changes with temperature because they involve : (a) Moles

(b) equivalents

(c) weights

(d) volumes

51.

44.

45.

When 500.0 mL of 1.0 M LaCl3 and 3.0 M NaCl are mixed.

(a) 3.76 g

(b) 1.28 g

(c) 2.82 g

(d) 3.76 kg

In a titration, 15.0 cm3 of 0.100 M HCl neutralizes 30.0 cm3 of Ca(OH)2. What is the molarity of Ca(OH)2 solution ? (a) 0.0125

(b) 0.0250

(c) 0.0500

(d) 0.200

10 mL of 1 M BaCl2 solution and 5 mL 0.5 M K2SO4 are mixed together to precipitate out BaSO4. The amount of BaSO4 precipated will be

What is molarity of Cl– ion ?

(a) 0.005 mol

(b) 0.00025 mol

(a) 4.0 M

(b) 3.0 M

(c) 0.025 mol

(d) 0.0025 mol

(c) 2.0 M

(d) 1.5 M

Molar Volume of Gas based Calculations

When 50 mL of 2.00 M HCl, 100 mL of 1.00 M HCl and 100 mL of 0.500 M HCl are mixed together, the resulting HCl concentration of the solution is (a) 0.25 M

(b) 1.00 M

(c) 3.50 M

(d) 6.25 M

52.

A-22

M g of a substance when vaporised occupy a volume of 5.6 litre at NTP. The molecular mass of the substance will be : (a) M

(b) 2M

(c) 3M

(d) 4M

SOME BASIC CONCEPTS OF CHEMISTRY 53.

Number of molecules in 1 litre of oxygen at NTP is :

(a)

6.02 u10 23 32

(b)

(c) 32 × 22.4 54.

55.

(d)

61.

6.02 u10 23 22.4 32 22.4

62.

57.

58.

59.

(a) 13

(b) 26

(c) 52

(d) 78

A hydrate of Na2SO3 losses 22.2% of H2O by mass on

The number of molecules in 89.6 litre of a gas at NTP are :

strong heating. The hydrate is

(a) 6.02×1023

(b) 2×6.02×1023

(a) Na2SO3 ˜ 4H2O

(b) Na2SO3 ˜ 6H2O

(c) 3×6.02×1023

(d) 4×6.02×1023

(c) Na2SO3 ˜ H2O

(d) Na2SO3 ˜ 2H2O

3

The mass of 112 cm of CH4 gas at STP is

Laws of Chemical Combination

(a) 0.16 g

(b) 0.8 g

63.

(c) 0.08 g

(d) 1.6 g

Empirical Formula 56.

The Ew of an element is 13. It forms an acidic oxide which with KOH forms a salt isomorphous with K2SO4. The atomic weight of element is

An oxide of metal (M) has 40% by mass of oxygen. Metal M has atomic mass of 24. The empirical formula of the oxide is (a) M2O

(b) M2O3

(c) MO

(d) M3O4

64.

One of the following combinations illustrate law of reciprocal proportions (a) N 2 O 3 , N 2 O 4 , N 2 O 5

(b) NaCl, NaBr, Nal

(c) CS 2 , CO 2 , SO 2

(d) PH 3 , P2 O 3 , P2 O 5

If water samples are taken from sea, river, clouds, lake or snow, they will be found to contain H 2 and O2 in the approximate ratio of 1 : 8. This indicates the law of

What is the empirical formula of a compound composed of O and Mn in equal weight ratio ? (a) MnO

(b) MnO2

(c) Mn2O3

(d) Mn2O7

(a) Multiple proportion

(c) Reciprocal proportions (d) none of these 65.

(b) C7H5

(c) C7H9NO

(d) C7H5NO

The law of multiple proportion is illustrated by (a) Carbon monoxide and carbon dioxide

Determine the empirical formula of Kelvar, used in making bullet proof vests, is 70.6% C, 4.2% H, 11.8% N and 13.4% O: (a) C7H5NO2

(b) Potassium bromide and potassium chloride (c) Water and heavy water (d) Calcium hydroxide and barium hydroxide

O 2

66.

A compound contains atoms of three elements A, B and C. If the oxidation number of A is +2, B is +5 and C is –2, the possible formula of the compound is :

The percentage of copper and oxygen in samples of CuO obtained by different methods were found to be the same. This illustrates the law of (a) constant proportions (b) conservation of mass (c) multiple proportions

60.

(a) A(BC3)2

(b) A3(BC4)2

(c) A3(B4C)2

(d) ABC2

(b) Definite proportion

67.

The carbonate of a metal is isomorphous (similar formula) with magnesium carbonate and contains 6.091 percent of carbon. The atomic weight of metal is

(d) reciprocal proportions

Two samples of lead oxide were separately reduced to metallic lead by heating in a current of hydrogen. The weight of lead from one oxide was half the weight of lead obtained from the other oxide. The data illustrates. (a) law of reciprocal proportions

(a) 24

(b) 56

(b) law of constant proportions

(c) 137

(d) 260

(c) law of multiple proportions (d) law of equivalent proportions

A-23

SOME BASIC CONCEPTS OF CHEMISTRY 68.

One part of an element A combines with two parts of another element B. Six parts of the element C combine with four parts of the element B. If A and C combine together the ratio of their weights will be governed by

73.

hydrogen while SO2 contains 50% sulphur. These figures illustrate the law of :

(a) law of definite proportions

(a) conservation of mass

(b) constant proportions

(b) law of multiple proportions

(c) multiple proportions

(d) reciprocal proportions

(c) law of reciprocal proportions

74.

(d) law of conservation of mass. 69.

n g of substance X reacts with m g of substance Y to form p g of substance R and q g of substance S. This reaction can be represented as follows : X+Y=R+S

70.

(a) 23 : 35.5

(b) 35.5 : 23

(c) 1 : 1

(d) 23 : 1

Principle of Atom Conservation

(a) n – m = p – q

(b) n + m = p + q

75.

(c) n = m

(d) p = q

Which one is the best example of law of conservation of mass ?

76.

(b) 6 g of carbon combines with 16 g of oxygen to form 22 g of CO2 (c) 6 g water is completely converted into steam (d) A sample of air is heated at constant pressure when its volume increases but there in no change in mass.

77.

SO2 gas was prepared by (i) burning sulphur in oxygen,

x g of Ag was dissolved in HNO3 and the solution was treated with excess of NaCl when 2.87 g of AgCl was precipitated. The value of x is (a) 1.08 g

(b) 2.16 g

(c) 2.70 g

(d) 1.62 g

A 1.50 g sample of an ore containing silver was dossolved, and all the Ag+ was converted to 0.125 g Ag2S. What was the percentage of silver in the ore ? (a) 14.23%

(b) 10.8%

(c) 8.27%

(d) 720%

NaOH is formed according to the reaction

1 2Na  O 2 o Na 2O 2

(ii) reacting sodium sulphite with dilute H 2SO 4 and (iii) heating copper with conc. H2SO4. It was found that in

72.

Hydrogen combines with chlorine to form HCl. It also combines with sodium to form NaH. If sodium and chlorine also combine with each other, they will do so in the ratio of their masses as :

The relation which can be established in the amounts of the reactants and the products will be

(a) 6 g of carbon is heated in vacuum, there is no change in mass

71.

H2S contains 5.88% hydrogen, H2O contains 11.11%

Na 2O  H 2 O o 2NaOH

each case sulphur and oxygen combined in the ratio of 1 : 1. The data illustrates the law of :

To make 4g of NaOH, Na required is

(a) conservation of mass

(b) multiple proportions

(a) 4.6g

(b) 4.0g

(c) constant proportions

(d) reciprocal proportions

(c) 2.3g

(d) 0.23g

A sample of CaCO3 has Ca = 40%, C = 12% and O = 48%. If the law of constant proportions is true, then the mass of Ca in 5 g of CaCO3 from another source will be :

Equivalent Concept 78.

2H3PO4 + 3 Ca(OH)2 o Ca3(PO4)2 + 6 H2O

(a) 2.0g

(b) 0.2g

Equivalent weight of H3PO4 in this reaction is

(c) 0.02g

(d) 20.0g

(a) 98

(b) 49

(c) 32.66

(d) 24.5

A-24

SOME BASIC CONCEPTS OF CHEMISTRY 79.

80.

81.

82.

The Ew of H3PO4 in the reaction is

87.

Normality of 0.74 g Ca(OH)2 in 5 mL solution is

Ca(OH)2 + H3PO4 o CaHPO4 + 2H2O

(a) 8 N

(b) 4 N

(Ca = 40, P = 31, O = 16)

(c) 0.4 N

(d) 2 N

(a) 49

(b) 98

(c) 32.66

(d) 147

88.

What weight of a metal of equivalent weight 12 will give 0.475 g of its chloride ?

Normality of a 2 M sulphuric acid is (a) 2 N

(b) 4 N

(c) N / 2

(d) N / 4

(a) 0.12 g

(b) 0.24 g

1 L of a normal solution is diluted to 2000 ml. The resulting normality is :

(c) 0.36 g

(d) 0.48 g

(a) N / 2

(b) N/ 4

(c) N

(d) 2 N

How many grams of phosphoric acid would be needed to neutralise 100 g of magnesium hydroxide ? (The molecular weights are : H3PO4 = 98 and Mg (OH)2 = 58.3) (a) 66.7 g

(b) 252 g

(c) 112 g

(d) 168 g

89.

90.

milliequivalent of solute ?

0.116 g of C4H4O4 (A) is neutralised by 0.074 g of Ca(OH)2. 91.

Hence, protonic hydrogen (H † ) in (A) will be

What volume of 0.232 N solution contains 3.17

(a) 137 mL

(b) 13.7 mL

(c) 27.3 mL

(d) 12.7 mL

1L solution of NaOH contains 4.0 g of it. What shall be the difference between molarity and the normality ?

83.

84.

85.

(a) 1

(b) 2

(c) 3

(d) 4

4.2 g of metallic carbonate MCO3 was heated in a hard glass tube and CO2 evolved was found to have 1120 mL of volume at STP. The EW of the metal is (a) 12

(b) 24

(c) 18

(d) 15

1.0 g of a monobasic acid when completely aceted upon Mg gave 1.301 g of anhydrous Mg salt. Equivalent weight of acid is (a) 35.54

(b) 36.54

(c) 17.77

(d) 18.27

0.1 g of metal combines with 46.6 mL of oxygen at STP. The equivalent weight of metal is (a) 12

(b) 24

(c) 6

(d) 36

92.

93.

(b) acidic

(c) strongly acidic

(d) neutral

(c) 0.05

(d) 0.20

100 ml of 0.3 N HCl is mixed with 200 ml of 0.6 N H2SO4. The (a) 0.1 N

(b) 0.2 N

(c) 0.3 N

(d) 0.5 N

Normality of a mixture of 30 mL of 1N H2SO4 and 20 mL of 4N H 2SO4 is

94.

(a) 1.0 N

(b) 1.1 N

(c) 2.0 N

(d) 2.2 N

Normality of solution obtained by mixing 10 mL of 1N HCl, 20 mL of 2N H2SO4 and 30 mL of 3N HNO3 is

When 100 ml of 1 M NaOH solution and 10 ml of 10 N H2SO4 solution are mixed together, the resulting solution will be : (a) alkaline

(b) zero

final normality of the resulting solution will be

Normality 86.

(a) 0.10

A-25

(a) 1.11 N

(b) 2.22 N

(c) 2.33 N

(d) 3.33 N

(Use the Final volume as sum of all volumes).

SOME BASIC CONCEPTS OF CHEMISTRY

ANSWER KEY Exercise­1 : (Basic Objective Questions) 1. (b)

2. (b)

3. (b)

4. (a)

5. (b)

6. (a)

7. (c)

8. (b)

9. (d)

10. (a)

11. (b)

12. (d)

13. (a)

14. (a)

15. (a)

16. (b)

17. (d)

18. (d)

19. (a)

20. (c)

21. (b)

22. (d)

23. (a)

24. (c)

25. (c)

26. (b)

27. (c)

28. (a)

29. (b)

30. (a)

31. (a)

32. (c)

33. (a)

34. (b)

35. (b)

36. (c)

37. (d)

38. (a)

39. (a)

40. (b)

41. (d)

42. (c)

43. (d)

44. (b)

45. (b)

46. (c)

47. (c)

48. (c)

49. (c)

50. (b)

51. (d)

52. (d)

53. (b)

54. (d)

55. (c)

56. (c)

57. (d)

58. (d)

59. (a,b)

60. (c)

61. (d)

62. (d)

63. (c)

64. (b)

65. (a)

66. (a)

67. (c)

68. (c)

69. (b)

70. (b)

71. (c)

72. (a)

73. (d)

74. (a)

75. (b)

76. (d)

77. (c)

78. (c)

79. (a)

80. (a)

81. (c)

82. (b)

83. (a)

84. (b)

85. (a)

86. (d)

87. (b)

88. (b)

89. (a)

90. (b)

91. (b)

92. (d)

93. (d)

94. (c)

Dream on !!

[\]^[\]^

A-26

CHEMICAL EQUILIBRIUM

EXERCISE - 2 : PREVIOUS YEAR AIEEE/JEE MAINS QUESTIONS 1.

(b) PCl5 (g)

PCl3 (g) + Cl2 (g)

SO2 (g) + Cl2 (g)

For the reaction

7. CO2 (g), Kp/Kc is

(c) (RT)

–1/2

(d) (RT)

(a) (c)

CO(g)

(c) H2(g) + 1/2O2(g)

8.

H2O(g)

Kc

–2

2

(b) 1.0 (d) RT

RT

The equilibrium constant for the reaction 2NO(g)

(b) 3.3 × 10 mol L

(c) 3 × 10–1 mol L–1

(d) 3 × 10–3 mol L–1

9.

The conditions favourable for the reaction 2SO3(g);

are

1 1 N2(g) + O2(g) 2 2 (2004)

(a) 2.5 × 102

(b) 0.02

(c) 4 × 10–4

(d) 50

–1

(a) 3 × 10 mol L

2SO2(g) + O2(g)

COCl2 (g),

at the same temperature is

–1

4.8 × 10 and 1.2 × 10 mol L respectively. The value of Kc for the reaction is (2003) –1

[P4 O10 ] 5 [P4 ] [O 2 ] (2004)

1 RT

NO2(g)

the concentrations of N2O4 and NO2 at equilibrium are

3

[O 2 ]5

at temperature T is 4 × 10–4. The value of Kc for the reaction

2NO2 (g)

–2

(d) K c

1

is equal to

For the reaction equilibrium N2O4 (g)

5.

Kp

N2(g) + O2(g)

2HI(g)

(d) H2(g) + I2(g) 4.

the

1/2

N2(g) + 3H2(g)

(b) C(g) + 1/2O2(g)

[P4 ] [O 2 ]

(b) K c

5

CO(g) + Cl2 (g)

In which of the following reactions, increase in the pressure at constant temperature does not affect the moles at equilibrium (2002) (a) 2NH3(g)

[P4 O10 ]

For the reaction

(b) (RT)–1

(a) RT

P4O10 (s) ?

(c) Kc = [O2]5

(2002)

1 O (g) 2 2

CO (g) +

3.

(a) K c

2NH3 (g)

(c) N2 (g) + 3H2 (g) (d) SO2Cl2

What is the equilibrium expression for the reaction (2004) P4 (s) + 5O2 (g)

2NO (g)

(a) N2(g) + O2 (g)

2.

6.

In which of the following equilibrium change in the volume of the system does not alter the number of moles (2002)

'Hº = – 198 kJ (2003)

(a) Low temperature, high pressure (b) Any value of T and P (c) Low temperature and low pressure (d) High temperature and high pressure

B-1

A small amount of NH4HS is placed in a flask already containing ammonia gas at a certain temperature and 0.50 atm pressure. Ammonium hydrogen sulphide decomposes to yield NH 3 and H2S gases in the flask. When the decomposition reaction reaches equilibrium, the total pressure in the flask rises to 0.84 atm. The equilibrium constant for NH4HS decomposition at this temperature is (a) 0.30

(b) 0.18

(c) 0.17

(d) 0.11

CHEMICAL EQUILIBRIUM 10.

For the reaction 2NO2(g)

13.

2NO (g) + O2(g)

The equilibrium constants K p and K p for the reactions 2 1 X

–6

2Y and Z

P + Q, respectively are in the ratio

of 1 : 9. If the degree of dissociation of X and Z be equal then the ratio of total pressure at these equilibria is

(Kc = 1.8 × 10 at 184ºC) (R = 0.0831 kJ/(mol.K)

(2008)

When Kp and Kc are compared at 184ºC it is found that (2005) (a) Whether Kp is greater than, less than or equal of Kc depends upon the total gas pressure

14.

(b) Kp = Kc

(c) 1 : 3

(d) 1 : 9

For the following three reactions 1, 2 and 3 equilibrium constants are given CO2(g) + H2(g);

K1

(d) Kp is greater than Kc

(2) CH4(g) + H2O(g)

CO(g) + 3H2(g);

K2

Phosphourus pentachloride dissociates as follows, in a closed reaction vessel,

(3) CH4(g) + 2H2O(g)

PCl3(g) + Cl2(g)

PCl5 (g)

§ x · ¸P (a) ¨ © x 1¹

§ 2x · ¸P (b) ¨ ©1 x ¹

§ x · ¸P (c) ¨ © x 1 ¹

§ x · ¸P (d) ¨ ©1 x ¹

SO2(g) +

(a) K 1 K 2 (c) K3 = K1K2 15.

(a) 416

(b) 2.40 × 10–3

(c) 9.8 × 10–2

(d) 4.9 × 10–2

K

(b) K 2 K

K1

(d) K3 . K 2 K12

A vessel at 1000 K contains CO2 with a pressure of 0.5 atm.

value of Kp is

1 O (g) 2 2

2SO3(g) will be

(2008)

Some of the CO2 is converted into CO on the addition of graphite. If the total pressure at equilibrium is 0.8 atm, the

is KC = 4.9 × 10–2. The value of KC for the reaction 2SO2 + O2(g)

K

(2006)

The equilibrium constant for the reaction SO3(g)

CO2(g) + 4H2(g);

Which of the following relations is correct ?

If total pressure at equilibrium of the reaction mixture is P and degree of dissociation of PCl5 is x, the partial pressure of PCl3 will be

12.

(b) 1 : 1

(1) CO(g) + H2O(g)

(c) Kp is greater than Kc 11.

(a) 1 : 36

(2006)

B-2

(2011)

(a) 1.8 atm

(b) 3 atm

(c) 0.3 atm

(d) 0.18 atm

CHEMICAL EQUILIBRIUM

ANSWER KEY EXERCISE ­ 2 : (Previous year AIEEE/JEE Mains Questions) 1. (a)

2. (c)

3. (d)

4. (d)

5. (a)

11. (a)

12. (a)

13. (a)

14. (c)

15. (a)

6. (b)

Dream on !!

[\]^[\]^

B-3

7. (a)

8. (d)

9. (d)

10. (d)

IONIC EQUILIBRIUM

EXERCISE - 2 : PREVIOUS YEAR AIEEE/JEE MAINS QUESTIONS 1.

The molar solubility (in mol L–1) of a sparingly soluble salt

7.

are 1.0 × 10–5 and 5.0 × 10–10 respectively. The overall dissociation constant of the acid will be (2007)

MX4 is ‘s’. The corresponding solubility product is Ksp. ‘s’ is given in terms of Ksp by the relation (2004) 1/ 4

§ K sp · ¨ ¸ ¨ 128 ¸ © ¹

(a) s

(c) s = (256 Ksp)1/5 2.

3.

1/ 5

§ K sp · ¨ ¸ ¨ 256 ¸ © ¹

(b) s

8.

(d) s = (128 Ksp)1/4

The conjugate base of H 2 PO 4

(2004)

(a) PO34

(b) HPO 24

(c) H3PO4

(d) P2O5

The solubility product of a salt having general formula

(c) 5.0 × 10–15

(d) 0.2 × 1015

Four species are listed below (i) HCO3

(ii) H 3O 

(iii) HSO 4

(iv) HSO 3 F

(a) iv < ii < iii < i

(b) ii < iii < i < iv

(c) i < iii < ii < iv

(d) iii < i < iv < ii

(a) 4.0 × 10–10 M

(b) 1.6 × 10–4 M

(c) 1.0 × 10–4 M

(d) 2.0 × 10–6 M

(a) 8.58

(b) 4.79

(c) 7.01

(d) 9.22

What is the conjugate base of OH– ? 2–

(c) H2O

6.

(b) 5.0 × 1015

The pKa of a weak acid, HA, is 4.80. The pKb of a weak base, BOH, is 4.78. The pH of an aqueous solution of the corresponding salt, BA, will be (2008)

(a) O

5.

(a) 5.0 × 10–5

Which of the following is the correct sequence of their acid strength ? (2008)

MX2; in water is 4 × 10–12. The concentration of M2+ ions in the aqueous solution of the salt is (2005)

4.

The first and second dissociation constant of an acid H2A

9.

(2005) 10.

–

(b) O

Na2CO3 solution. At what concentration of Ba+2 will a

(d) O2

precipitate being to form ? (Ksp for BaCO3 = 5.1 × 10–9);

The pKa of a weak acid (HA) is 4.5. The pOH of an aqueous buffered solution of HA in which 50% of the acid is ionized is (2007) (a) 4.5

(b) 2.5

(c) 9.5

(d) 7.0

(2009)

11.

(a) 4.1 × 10–5 M

(b) 5.1 × 10–5 M

(c) 8.1 × 10–8 M

(d) 8.1 × 10–7 M

At 25ºC, the solubility product of Mg(OH)2 is 1.0 × 10–11. At which pH, will Mg2+ ions start precipitating in the form

In a saturated solution of the sparingly soluble strong electrolyte AgIO3 (molecular mass = 283) the equilibrium which sets in is

AgIO3(s )

Solid Ba (NO3)2 is gradually dissolved in a 1.0 × 10–4 M

of Mg(OH)2 from a solution of 0.001 M Mg2+ ions ? (2010)

Ag (aq .) + IO 3( aq .)

(a) 9

(b) 10

(c) 11

(d) 8

If the solubility product constant Ksp of AgIO3 at a given

Solubility product of silver bromie is 5.0 × 10–13. The quantity of potassium bromide (molar mass taken as

temperature is 1.0 × 10–8, what is the mass of AgIO3 contained in 100 mL of its saturated solution ? (2007)

120 g mol–1) to be added to 1L of 0.05 M solution of silver nitrate to start the precipitation of AgBr is (2010)

(a) 28.3 × 10–2g

(b) 2.83 × 10–3g

(a) 1.2 × 10–10 g

(b) 1.2 × 10–9 g

(c) 1.0 × 10–7g

(d) 1.0 × 10–4g

(c) 6.2 × 10–5 g

(d) 5.0 × 10–8 g

12.

B-4

IONIC EQUILIBRIUM 13.

In aqueous solution, the ionisation constants for carbonic acid are (2010)

15.

K1 = 4.2 × 10–7 and K2 = 4.8 × 10–11 Select the correct statement for a saturated 0.034 M solution of the carbonic acid. 16.

(a) The concentration of CO32 is 0.034 M

The Ksp for Cr(OH)3 is 1.6 × 10–30. The molar solubility of this compound in water is (2011) (a) 2 1.6 u1030

(b) 4 1.6 u1030

(c) 4 1.6 u1030 / 27

(d) 1.6 u10 30 / 27

An acid HA ionises as ZZX H   A  HA YZZ

(b) The concentration of CO32 is greater than that of

The pH of 1.0 M solution is 5. Its dissociation constant would be (2011)

HCO3 (c) The concentration of H+ and HCO3 are approximately equal (d) The concentration of H+ is double that of CO32 14.

Three reactions involving H 2 PO 4 are given below : (2010) I. H3PO4 + H2O o H3O+ + H 2 PO 4 II. H 2 PO 4  H 2O o HPO24  H3O  III. H 2 PO4  OH o H 3PO4  O2 In which of the above does H 2 PO 4 act as an acid ? (a) II only

(b) I and II

(c) III only

(d) I only

B-5

(a) 1 × 10–10

(b) 5

(c) 5 × 10–8

(d) 1 × 10–5

IONIC EQUILIBRIUM

Answer Key Exercise­2 (Previous Year AIEEE/JEE Mains Questions) 1. (b)

2. (b)

3. (c)

4. (a)

5. (c)

6. (b)

11. (b)

12. (b)

13. (c)

14. (a)

15. (c)

16. (a)

Dream on !!

[\]^[\]^

B-6

7. (c)

8. (c)

9. (c)

10. (b)

SOME BASIC CONCEPTS OF CHEMISTRY

EXERCISE - 2 : PREVIOUS YEAR AIEEE QUESTIONS 1.

2.

3.

4.

The molarity of a solution obtained by mixing 750 mL of 0.5 (M) HCl with 250 mL of 2 (M) HCl will be (2013)

7.

Density of a 2.05 M solution of acetic acid in water is 1.02 g/mL. The molality of the solution is (2006)

(a) 0.875 M

(b) 1.00 M

(a) 0.44 mol Kg–1

(b) 1.14 mol kg–1

(c) 1.75 M

(d) 0.0975 M

(c) 3.28 mol kg–1

(d) 2.28 mol kg–1

The density of a solution prepared by dissolving 120g of urea (mol. mass = 60 u) in 1000 g of water is 1.15g/mL. The molarity of this solution is (2012) (a) 0.50 M

(b) 1.78 M

(c) 1.02 M

(d) 2.05 M

8.

(a) be a function of the molecular mass of the substance (b) remain unchanged

The mass of potassium dichromate crystals required to oxidise 750 cm3 of 0.6 M Mohr’s salt solution is (Given molar mass = 392) (2011) (a) 0.49 g

(b) 0.45 g

(c) 29.4 g

(d) 2.2 g

If we consider that 1/6, in place of 1/12, mass of carbon atom is taken to be the relative atomic mass unit, the mass of one mole of a substance will (2005)

(c) increase two fold (d) decrease twice 9.

Amount of oxalic acid present in a solution can be determined by its titration with KMnO4 solution in the presence of H2SO4. The titration gives unsatisfactory result when carried out in the presence of HCl because HCl (2008)

10.

(a) gets oxidised by oxalic acid to chlorine

25 mL of a solution of Ba(OH)2 on titration with a 0.1 M solution of HCl gave a titre value of 35 mL. The molarity of barium hydroxide solution was (2003) (a) 0.07

(b) 0.14

(c) 0.28

(d) 0.35

What volume of H 2 gas at 273 K and 1 atm pressure will be consumed in obtaining 21.6 g of boron (At. mass 10.8 u) from reduction of boron trichloride by H 2 (2003)

(b) furnishes H+ ions in addition to those from oxalic acid (c) reduces permanganate to Mn2+ (d) oxidises oxalic acid to carbon dioxide and water 5.

11.

The density (in g mL–1) of a 3.60 M sulphuric acid solution that is 29% H 2SO4 (Molar mass = 98g mol–1) by mass will be (2007) (a) 1.64

(b) 1.88

(c) 1.22

(d) 1.45 12.

6.

How many moles of magnesium phosphate, Mg3 (PO4)2 will contain 0.25 mole of oxygen atoms ? (2006)

(a) 89.6 L

(b) 67.2 L

(c) 44.8 L

(d) 22.4 L

6.023 × 10 20 molecules of urea are present in 100 mL of its solution. The concentration of urea solution is (2004) (a) 0.001 M

(b) 0.1 M

(c) 0.02 M

(d) 0.01 M

To neutralize completely 20 mL of 0.1 M aqueous solution of phosphorus (H3PO 3) acid, the volume of 0.1 M aqueous KOH solution required is (2004)

(a) 0.02

(b) 3.125×10–2

(a) 60 mL

(b) 20 mL

(c) 1.25×10–2

(d) 2.5×10–2

(c) 40 mL

(d) 10 mL

B-7

SOME BASIC CONCEPTS OF CHEMISTRY 13.

14.

15.

Number of atoms in 560g of Fe (atomic mass 56 g mol–1) is (2002) (a) twice that of 70 g N

(b) half that of 20 g H

(c) Both (a) and (b)

(d) None of the above

In an organic compound of molar mass 108 g mol–1C, H and N atoms are present in 9 : 1 : 3.5 by weight. Molecular formula can be (2002) (a) C6H8

2

(b) C7H10

(c) C5H6

3

(d) C4H18

3

Number of atoms in 558.5 Fe (at. wt. 55.85) is (2002) (a) Twice that in 60 g carbon (b) 6.023 × 10 22 (c) Half in 8 g He

(d) 558.5 × 6.023 × 1023

B-8

SOME BASIC CONCEPTS OF CHEMISTRY

ANSWER KEY Exercise ­ 2 : (Previous Year AIEEE Questions) 1. (a)

2. (d)

3. (c)

4. (a)

5. (c)

6. (b)

7. (d)

9. (a)

10. (b)

11. (d)

12. (c)

13. (c)

14. (a)

15. (a)

Dream on !!

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B-9

8. (b)

CHEMICAL EQUILIBRIUM

CHEMICAL EQUILIBRIUM EXERCISE - 3: ADVANCED OBJECTIVE QUESTIONS 1. 2. 3. 4.

All questions marked “S” are single choice questions All questions marked “M” are multiple choice questions All questions marked “C” are comprehension based questions All questions marked “A” are assertion–reason type questions (A) If both assertion and reason are correct and reason is the correct explanation of assertion. (B) If both assertion and reason are true but reason is not the correct explanation of assertion. (C) If assertion is true but reason is false. (D) If reason is true but assertion is false.

5. 6.

All questions marked “X” are matrix–match type questions All questions marked “I” are integer type questions

4. (S)

Equilibrium : Definition and Expression 1. (A)

Assertion : For a reaction at equilibrium, the free energy

Reason : The free energy for both reactants and

where x is the amount of ‘A’ dissociated. The value of equilibrium constant (Keq) is :

products decreases and become equal.

2. (A)

(b) B

(c) C

(d) D

(a) 10 (c) 20

Assertion : Chemical equilibrium represents a state of a

5. (A)

reversible reaction in which measurable properties of the system (pressure, concentration, colour etc.) become constant under the given set of conditions.

of rest in which two opposing reactions are proceeding at the same rate.

3. (S)

(b) B

(c) C

(d) D

6. (A)

Which of the following statements about the reaction quotient, Q are correct ? (a) the reaction quotient, Q and the equilibrium constant always have the same numerical value

(b) 0.05 (d) Can’t be calculated

Assertion : The reaction quotient, Q has the same form as the equilibrium constant Keq and is evaluated using any given concentrations of the species involved in the reaction, and not necessarily equilibrium concentrations. Reason : If the numerical value of Q is not the same as the value of equilibrium constant, a reaction will proceed.

Reason : The chemical equilibrium is an apparent state

(a) A

(a) A

(b) B

(c) C

(d) D

Assertion : The active mass of pure solids and pure liquids is taken unity. Reason : The active mass of pure solids and liquids depends on density and molecular mass. The density and molecular mass of pure liquids and solids are constant.

(b) Q may be lesser than, equal to or greater than Keq

(a) A

(b) B

(c) Q (numerical value) varies as reaction proceeds

(c) C

(d) D

(d) Q = 1 at equilibrium

C-1

C

§ dx · ¨ ¸ = 2.0 × 103 L mol–1 s–1 [A] [B] – 1.0 × 102 s–1 [C] © dt ¹

for the reaction is minimum.

(a) A

For a reversible reaction : A + B

CHEMICAL EQUILIBRIUM 7. (A)

Assertion : If some PCl 5 (g) containing labelled phosphorus p31 is added to a system with following equilibrium, after sometime the system was found to contain radioactive PCl3.

11. (S)

1 NH 3 (g) U N 2 (g)  H 2 (g) 2 2

PCl3(g) + Cl2(g)

PCl5(g)

The value of kp for the following reaction will be :

Reason : Chemical equilibrium is dynamic in nature.

8. (A)

(a) A

(b) B

(c) C

(d) D

N 2 (g)  3H 2 (g) U 2NH 3 (g)

§ 800R · (a) ¨ ¸ © 4 ¹

Assertion : As a reversible system approaches equilibrium, entropy of the system increases.

(b) B

(c) C

(d) D

(b) 16 × (800R)2

2

(d) (800R)1/2 4

Equilibrium : Analysis 12. (S)

9. (M)

2

1 ª º (c) « » ¬ 4 u 800R ¼

Reason : The state of equilibrium is the most disordered state of a reversible system. (a) A

Equilibrium constant Kc for the following reaction at 800K is, 4.

§1· §1· N 2O2 U 2KO K1 ; ¨ ¸ N 2  ¨ ¸ O 2 U NO K 2 ©2¹ ©2¹

N2O4

2NO2, Kc = 4. This reversible reaction is

studied graphically as shown in the given figure. Select the correct statement out of I, II and III. I : Reaction quotient has maximum value at point A II : Reaction proceeds left to right at a point when [N2O4] = [NO2] = 0.1 M

§1· §1· 2NO U N 2  O 2 K ; NO U ¨ ¸ N 2  ¨ ¸ O 2 K 4 ©2¹ ©2¹

III : Kc = Q when point D or F is reached :

Correct relation(s) between K1, K2, K3 and K4 is/are : (a) K2 × K4 = 1 (c)

p

K u K2

1

(b)

K1 u K 4

1

(d)

K u K2

1

and KC

10. (X)

Match the following systems with their K p /K c relationship Equilibrium types

Kp/Kc

(a) I, II (c) II

relationship (a) NO(g)  NO 3 (g) U 2NO 2 (g)

(p)Kp > Kc

(b) N 2 (g)  3H 2 (g) U 2NH 3 (g)

(q) Kp= Kc (RT)2

(c) 2O 3 (g) U 3O 2 (g)

(r) Kp= Kc

(d) BrF5 (g) U

1 5 Br2 (g)  F2 (g) 2 2

13. (S)

I2 + I –

(b) II, III (d) I, II, III

I

This reaction is set-up in aqueous medium. We start with 1 mol of I2 and 0.5 mol of I– in 1L flask. After equilibrium is reached, excess of AgNO3 gave 0.25 mol of yellow ppt. Equilibrium constant is : [Given : AgNO gives yellow ppt with I–]

(s) Kp< Kc

(a) 1.33 (c) 2.00

C-2

(b) 2.66 (d) 3.00

CHEMICAL EQUILIBRIUM 14. (I)

Given the hypothetical reaction :

respectively, what is the molarity of HBr ?

2A(s) + nB(g) U 3C(g) Kp = 0.0105

(a) 1.8 × 10 M

–2

0

(c) 1.5 × 10 M

and Kc = 0.45 at 250ºC. What is the value of coefficient ‘n’ ? 15. (S)

21. (S)

In the reaction : A + B U 2C + D. The initial concentration of A and B are 1M each. The value of Kc 8 is 10 . What is the equilibrium concentration of A ? –4

16. (S)

4

(a) 2 × 10 M

(b) 2 × 10 M

(c) 0.005 M

(d) 0.0025 M 22. (S)

Equilibrium constant for two complexes are : 37

A : K4 [Fe(CN)6]

2.6 × 10 (for dissociation)

B : K3 [Fe(CN)6]

1.9 × 10 (for dissociation)

–1

(b) 6.7 × 10 M 1

(d) 3.6 × 10 M

For the all gas reaction at 1000 K : 2HI U H2 + I2, K = 0.0344. If you were to begin with 8.0 M of HI in a fixed volume container, what would be the equilibrium concentration of H2 ? (a) 1.10 M

(b) 0.48 M

(c) 5.8 M

(d) 2.92 M k

1 ZZZ X In reversible reaction A YZZ Z B , the initial k 2

concentration of A and B are a and b in moles per litre and the equilibrium concentration are (a – x) and (b – x) respectively ; express x in terms of k1, k2, a and b.

17

(a) A and B are equally stable (b) A is more stable than A

(a)

k1a  k 2 b k1  k 2

(b)

k1a  k 2 b k1  k 2

(c)

k1a  k 2 b k1k 2

(d)

k1a  k 2 b k1  k 2

(c) B is more stable than A (d) the predictable stability 17. (S)

+

–

–

For the reaction, [Ag(CN)2] U Ag + 2CN , the –19

+

equilibrium constant Kc at 25ºC is 4 × 10 ; then Ag concentration in a solution which has 0.1 M KCN and 0.03M AgNO3 is : 18

(b) 7.5 × 10

19

(d) 7.5 × 10

(a) 7.5 × 10 (c) 7.5 × 10 18. (S)

19. (S)

Equilibrium Constant and 'G 23. (S)

–18

CaCO3(s) U CaO(s) + CO2 (g) ; K = 1 at 1 atm,

–19

the temperature is given by :

A vessel at 1000 K contains CO2 with a pressure of 0.5 atm. Some of the CO2 is converted to CO on addition of graphite. If the total pressure at equilibrium is 0.8 atm, the value of K is : (a) 0.18 atm

(b) 1.8 atm

(c) 3 atm

(d) 0.3 atm

24. (X)

The equilibrium reaction : 2AB(aq) U 2A(aq) + B2(g) has an equilibrium constant equal to 1.28. If the concentrations of AB and A are 5 mol/L and 4 mol/L respectively, what is [B2] ? (a) 0.50 M

(b) 1.60 M

(c) 2.00 M

(d) 2.84 M

If for the heterogeneous equilibrium

(a) T

'Sº 'Hº

(b) T

'Hº 'Sº

(c) T

'Gº R

(d) T

'Gº 'Hº

Match the Column-I with Column-II : Column-I

Column-II

(Equations)

(Type of process)

(a) Kp > Q

(p) Non spontaneous

(b) 'Gº < RT loge Q

(q) Equilibrium

(c) Kp = Q

(r) Spontaneous and endothermic

20. (S)

Consider the reaction NaBr(aq) + H 2 SO 4(aq) U NaHSO4(aq) + HBr(aq). The equilibrium constant is –2 8.3 × 10 . If the equilibrium concentrations of NaBr, H2SO4 and NaHSO 4 are 6.0 M, 9.0 M and 3.0 M

(d) T !

C-3

'H 'S

(s) Spontaneous

CHEMICAL EQUILIBRIUM Passage (For Questions 25 to 27)

26. (C)

Effect of temperature on the equilibrium process is analyzed by using the thermodynamics. From the thermodynamics relation. 'Gº = – 2.303 RT logK

For exothermic reaction is 'Sº < 0, then the sketch of log K vs 1/T may be

...(i)

'Gº : Standard free energy change 'Gº = 'Hº – T'Sº

...(ii)

(a)

'Hº : Standard heat of the reaction From (i) and (ii) –2.303 RT log K = 'Hº – 7'Sº 'Sº : Standard entropy change

Ÿ logK 

'Hº 'Sº  ...(iii) 2.303RT 2.303R

If a plot of log K vs 1/T is made then it is a straight line having slope

'Hº and Y intercept 2.303R

(b)

'Sº 2.303R

If at temperature T1, equilibrium constant be K1 and at temperature T2, equilibrium constant be K2 then the above equation reduces to :

'Hº 'Sº  2.303RT1 2.303R

Ÿ log K1



Ÿ log K 2

'Hº 'Sº   2.303RT2 2.303R

(c)

Substracting (iv) from (v) we get.

K2 Ÿ log K 1

'Hº § 1 1 · ¨  ¸ 2.303R © T1 T2 ¹

From the above relation we can conclude that the value of equilibrium constant increases with increase in temperature for an endothermic reaction and the same decreases with the increase in temperature for an exothermic reaction. Answer the following three questions based on the above information. 25. (C)

(d)

27. (C)

If standard heat of dissociation of PCl5 is 230 cal, then slope of the graph of log k vs

If for a particular reversible reaction : Kc = 57 at 355ºC and Kc = 69 at 450º, then (a) 'H < 0

1 is T

(b) 'H > 0

(a) +50

(b) –50

(c) 'H = 0

(c) 10

(d) None of these

(d) Sign of 'H can’t be determined

C-4

CHEMICAL EQUILIBRIUM 28. (M) Variation of equilibrium constant K for the reaction

­ d log e K ½ ¾ ¯ dT ¿P



(b) ®

2A(s)  B(g) U C(g)  2D(g)

'H RT

is plotted against absolute temperature T in figure as : ln K Vs (1/T) (c) loge K = Constant 

RT 'H

(d) loge K = Constant 

'H RT

Homogeneneous and Heterogeneous Equilibria, Degree of Dissociation 32. (S) ln K vs 1/T (a) the forward reaction is exothermic (b) the forward reaction is endothermic (c) the slope of line is proportional to 'H

(a) N 2  3H 2

(d) adding ‘A’ favours forward reaction

(g )

29. (M) Which is/are correct ? (a) 2.303 log K

(d) 2.303 log K 30. (S)



(g )

(g )

PCl  Cl 2 , endothermic

33. (S)

(g )

In each of the following, total pressure set-up at equilibrium is assumed to be equal and is one atm with equilibrium constants KP given : I : CaCO3 (s)

–1

II : NH4HS (s)

CaO (s) + CO2 (g), K1 NH3 (g) + H2S (g), K2

III : NH2COONH4 (s)

1 1 N 2 (g)  O2 (g) U NO(g) 2 2

2NH3 (g) + CO2 (g), K

In the increasing order :

–5

(b) 7.1 × 10

–10

(d) 1.7 × 10

(a) K1 = K2 = K (c) K3 < K2 < K1

–9 –19

34. (S)

31. (M) The variation of equilibrium constant K with temperature is represented by :



(g )

(d) none of these

The free energy of formation of NO is 78 kJ mol at the temperature of an automobile engine (1000 K). What is the equilibrium constant for this reaction at this reaction at 1000 K ?

(a) log e K 2  log e K1

2 NO, endothermic

(g)

1 ('Hº 'Sº ) RT

(c) 4.2 × 10

(g)

(g )

(c) PCl 5

'Hº 'Sº  RT 2 R

(a) 8.4 × 10

2 NH 3 , exothermic

(g)

(b) N 2  O 2

'Hº 'Sº   RT R

(b) 'Gº = – 2.303 RT log K (c) –2.303 log K

Volume of the flask in which the following equilibria are separately established are transferred to a flask that is double the size of the earlier flask. In which of the following cases, equilibrium constants are affected ?

'H T2 § 1 · ¨ ¸ dT R ³T1 © T ¹

During thermal dissociation, the observed vapour density of N2O4 (g) is 26.0. The extent of dissociation of N2O4 (g) is : (a) 50% (c) 77%

C-5

(b) K1 < K2 < K (d) None of these

(b) 87% (d) 23%

CHEMICAL EQUILIBRIUM 35. (S)

For the reaction N2O4 (g)

2NO2 (g), the value of Kp

3

38. (I)

is 1.7 × 10 at 500 K and 1.78 × 10 at 600 K. Which of the following is correct ? I.

temperature

The proportion of NO2 in the equilibrium mixture is increased by decreasing the pressure 39. (I)

III. units of Kp are atm IV. at 500 K, the degree of dissociation N2O4 decreases by 50% by increasing the pressure by 100%.

(b) I (d) III

What weight of solid ammonium carbamate (NH2COONH4) when vaporised at 200ºC will have a volume of 8.96 litre at 1 atm ? Assume that solid completely decomposes into CO2 and NH3 at 200ºC and 1 atm.

41. (A)

Assertion : The dissociation of CaCO 3 can be represented as, CaCO3(s) U CaO(s) + CO2(g). Some

The equilibrium constant K is given by (when D n(A2) at

5

equilibrium the total pressure 3

(c)

equilibrium

is 2.0 atm.

2

(d) 2A(g) U A2(g)Kp = 2. A mixture, initially containing

(s) n(A) < n(A2) at equilibrium

both A and A2 at 1.0 atm partial pressure is prepared. (d)

69. (X)

Match the Column–I with Column–II

Column–I 67. (S)

Column–II

(A) N2(g) + 3H2(g) U 2NH3(g)

For the following reaction through stages I, II and III

(p) Forward shift by increase in pressure

III I II A B o C o D o

(B) N2(g) + O2(g) U 2NO(g)

quantity of the product formed (x) varies with temperature (T) as given. Select correct statement :

(p) No effect of pressure change

(C) CaCO3(s) U CaO(s)+CO2(g)

(r) No effect of inert gas addition at constant volume

(D) NH2COONH4(s) U 2NH3(g) + CO2(g) (s) Addition of inert gas will favour backward direction (Not at constant volume)

(a) Stages I and III are endothermic but II is exothermic (b) Stages I and III are exothermic but II is endothermic (c) Stages II and III are exothermic but I is endothermic

Passage : For Questions (70 to 72)

(d) Stage I is exothermic but stages II and III are endothermic 68. (X)

Match the conditions on the left column with states on the right column.

Equilibrium types

Kp/Kc

Change of Pressure :

relationship (a) A2(g) U 2A(g); Kc = 1. Mixture of 1.0 mole of each

If a system in equilibrium consists of gases, then the concentrations of all components can be altered by changing the pressure. When the pressure on the system is increased, the volume decreases proportionately. The total number of moles per unit volume will now be more and the equilibrium will shift in that direction in which there is decrease in number of moles i.e, towards the direction in which there is decrease in volume.

(p) System is at equilibrium

is prepared in a 1.0 L flask. (b) A(g) + 2B(g) U AB2(g)

Le-Chatelier’s principle : If a system at equilbrium is subjected to a change of any one of the factors such as concentration, pressure or temperature, the system adjusts itself in such a way as to “Nullify” the effect of that change.

(q) Goes to right

Kc= 2. A mixture of 2.0 moles of each A, B and AB2 is prepared in a 1.0 L flask.

C-10

CHEMICAL EQUILIBRIUM Effect of Pressure on Melting Point : 72. (C)

There are two types of solids : (a) Solids whose volume decreases on melting e.g., ice, diamond, carborundum, magnesium nitride and quartz.

1 1 N 2 (g)  O2 (g) U NO(g) 2 2

If pressure is increased by reducing the volume of the container, then

Solid (higher volume) U liquid (lower volume)

(a) total pressure at equilibrium will change

The process of melting is facilitated at higher pressure, thus the melting point is lowered.

(b) concentration of all the components at equilibrium will change

(b) Solids whose volume increase on melting e.g. Fe, Cu, Ag, Au, etc.

(c) concentration of all the components at equilibrium will remain same (d) equilibrium will shift in the forward direction

Solid (lower volume) U liquid (higher volume) In this case, the process of melting becomes difficult at high pressure ; thus melting point becomes high.

73. (M) For the reaction : PCl5(g) U PCl3(g) + Cl2(g), The forward reaction at constant temperature is favoured by

(c) Solubility of substances : When solid substances are dissolved in water, either heat is evolved (exothermic) or heat is absorbed (endothermic) :

(a) introducing an inert gas at constant volume (b) introducing chlorine gas at constant volume (c) introducing an inert gas at constant pressure

KCl + aq U KCl(aq) + heat In such cases, solubility increases with increase in temperature. Consider the case of KOH; when this is dissolved, heat is evolved;

(d) increasing the volume of the container 74. (M) Which of the following factors will favour the backward reaction ? Cl2(g) + 3F2(g) U 2ClF3(g) ; 'H = –ve

KOH + aq U KOH(aq) – heat

70. (C)

For the reaction :

In such cases, solubility decrease with increse in temperature.

(a) Addition of inert gas at constant pressure

(d) Solubility of gases in liquids : When a gas dissolves in liquid, there is decrease in volume. Thus, increase of pressure will favour the dissolution of gas in liquid. Answer the following three questions based on the above information.

(c) Increase in the temperature of reaction

(b) Addition of Cl2 gas (d) Increasing the volume of the container 75. (M) For the gas phase reaction, C2H4(g) + H2(g) U C2H6(g) ; 'Hº = –136.8 kJ mol

–1

When a gas “X” is dissolved in water, the heat is evolved. Then solubility of ‘X’ will increase at

carried out in a vessel, the equilibrium concentration of C2H4 can be increased by :

(a) low pressure, high temperature

(a) increase in temperature(b) decrease in pressure

(b) low pressure, low temperature

(c) removing some H2

(c) high pressure, high temperature

(d) adding some C2H6

Physical Equilibria

(d) high pressure, low temperature Au(s) U Au(l)

Densities of diamond and graphite are 3.5 and 2.3 g/mL respectively. Increase of pressure on the equilibrium

Above equilibrium is favoured at

C (diamond) U C (graphite)

(a) high pressure low temperature

(a) favours backward reaction

(b) high pressure high temperature

(b) favours forward reaction

(c) low pressure, high temperature

(c) has no effect

(d) low pressure, low temperature

(d) increase the reaction rate

76. (S) 71. (C)

C-11

CHEMICAL EQUILIBRIUM 77. (S)

Some quantity of water is contained in a container as shown in figure below. As neon is added to this system at constant pressure, the amount of liquid water in the vessel :

If initially only NO and NO2 are present in a 3 : 5 mole ratio and the total pressure at equilibrium is 5.5 atm with the pressure of NO2 is 0.5 atm, calculate K p (in 10–1). 1

82. (I)

Two solids A and C dissociate into gas products as follows. A(s) U B(g) + D(g); K p = 400 1

C(s) U E(g) + D(g); K p = 900 2

At 25ºC, the pressure over excess solid A only is 40 atm, and that over solid C only is 60 atm. Find the pressure over the solid mixture. 83. (S) On heating a mixture of SO2Cl2 and CO, two equilibria are simultaneously established :

78. (S)

79. (S)

(a) increases

(b) decrease

(c) remains same

(d) change unpredictably

SO 2 Cl 2 (g) U SO 2 (g)  Cl 2 (g)

CO(g)  Cl 2 (g) U COCl 2 (g)

Solubility of a substance which dissolves with a decrease in volume and absorption of heat will be favoured by (a) high p and high T

(b) low p and low T

(c) high p and Low T

(d) low p and high T

A solution is transferred to a 1 L flask, some pure ice is added and a stopper is inserted to close the flask. After the system has reached constant temperature, there are still a few cubes of ice left in the flask, but no other solid. Which of the following statements is false relative to this system at equilibrium ?

On adding more SO2 at equilibrium what will happen ? (a) Amount of CO will decrese (b) Amount of SO2Cl2 and COCl2 will increase (c) Amount of CO will remain unaffected 84. (S)

(a) The system is a heterogeneous one. (b) There are four phases present in the system, not including the container and stopper. (c) The vapour pressure is constant. (d) The temperature of the system is below 0ºC. 80. (S)

When a liquid in equilibrium with its vapour is heated (a) the rate of the reaction, vapour o liquid, is increased

85. (S)

(d) Amount of CO2Cl2 and CO will increase Two solid compounds X(s) and Y(s) undergo dissociation as follows : X(s) U A(g)  2B(g) ;

K

Y(s) U 2B(g)  C(g) ;

K

2

9 u 10 3 bar 3

4.5 u 10 3 bar 3

The total pressure of gaseous mixture at equilibrium will be : (a) 4.5 bar (b) 0.45 bar (c) 0.6 bar (d) 1.2 bar Given the following reactions and associated equilibrium constants, select the correct expression for the third equilibrium constant in terms of the first two. (A) 2CO(g)  O 2 (g) U 2CO 2 (g)K c 1

(b) the same equilibrium mixture is re-established (c) a new equilibrium mixture with a higher vapour pressure is established (d) a new equilibrium mixture is established in which the rate of evaporation is greater than the rate of condensation

1 (B) H 2 (g)  O 2 (g) U H 2 O(g)K c2 2 (C) CO(g)  H 2 O(g) U CO 2 (g)  H2 (g)K c3

Simultaneous Equilibria 81. (I)

1

Two equilibria are simultaneously existing in a vessel at 25ºC NO(g) + NO2(g) U N2O3(g); K p (say)

(a) K c3

(c) K c3

1

K c1

K c1 u K c2

2NO2(g) U N2O4(g); K p = 8 atm–1 2

(d) K c3

C-12

(b) K c3

K c2

Kc1

1/ 2

1/ 2

u K c2

1/ 2

K c1 K c2

1/ 2

CHEMICAL EQUILIBRIUM

ANSWER KEY Advanced Objective Questions 1. (a)

2. (a)

3. (b,c)

4. (c)

5. (b)

6. (a)

7. (a)

8. (a)

9. (a,c)

11. (c)

12. (b)

13. (a)

14. (0004)

15. (a)

16. (c)

17. (b)

18. (b)

19. (c)

20. (c)

21. (a)

22. (a)

23. (b)

24. (A–s), (B–p), (C–q), (D–r)

25. (b)

26. (b)

27. (b)

28. (a,c,d)

29. (a,b)

30. (a)

31. (a,d)

32. (d)

33. (c)

34. (c)

35. (a)

36. (a)

37. (A–p,q,s), (B-q,r), (C–q), (D–p,s)

38. (0008)

39. (0008)

40. (0006)

41. (b)

42. (c)

43. (b)

44. (d)

45. (d)

46. (b)

47. (d)

48. (a,c)

49. (a)

50. (b)

51. (b)

52. (b)

53. (a)

54. (a)

55. (d)

56. (a,c)

57. (a,b,d)

58. (c,d)

59. (b,d)

60. (b,c,d)

61. (a,b,c)

62. (a,c)

63. (a,c)

64. (c)

65. (c)

66. (d)

67. (a)

69. (A-p,r,s), (B-q,r), (C-r), (D-r)

70. (d)

71. (c)

72. (b)

73. (c,d)

74. (a,c,d)

75. (a,b,c,d) 76. (a)

79. (b)

80. (c)

81. (0004)

82. (0072)

83. (d)

84. (b)

77. (b)

78. (a)

85. (b)

Dream on !!

[\]^[\]^

C-13

IONIC EQUILIBRIUM

IONIC EQUILIBRIUM EXERCISE - 3: ADVANCED OBJECTIVE QUESTIONS 1. 2. 3. 4.

All questions marked “S” are single choice questions All questions marked “M” are multiple choice questions All questions marked “C” are comprehension based questions All questions marked “A” are assertion–reason type questions (A) If both assertion and reason are correct and reason is the correct explanation of assertion. (B) If both assertion and reason are true but reason is not the correct explanation of assertion. (C) If assertion is true but reason is false. (D) If reason is true but assertion is false.

5.

All questions marked “X” are matrix–match type questions

6.

All questions marked “I” are integer type questions

Theory of Electrolytes, Acid­Base Strength Comparison 1. (A)

5. (S)

Assertion : [Al(H 2O) 6] 3+ is a stronger acid than [Mg(H2O)6]2+.

I. HClO4 III. H3PO4

2+

Reason : Size of [Al(H2O)6]3+ smaller than [Mg(H2O)6] and possesses more effective nuclear charge.

2. (M)

(a) A

(b) B

(c) C

(d) D

6. (S)

Which among the following represent the conjugate acid/base pairs ? (a) H3O+/H2O

(b) H2SO4/ SO 24 

(a) III < II < I

(b) I < II < III

(c) II < III < I

(d) III < I < II

Consider the following reaction which proceeds predominantly from

From the information given it is clear that (a) HSO4 is a stronger acid than NH 4 .

(d) All are conjugate acid/base paires

(b) NH3 is a weaker base than SO 24 

The equilibrium constant for this reaction is approximately 10–3.

(c) HSO4

HPO42 (aq.)  HCO3 (aq.) U H2 PO4 (aq.)  CO32 (aq.)

(d) NH3 is a weaker base than NH 4

Which is the strongest conjugate base in this reaction ?

4. (S)

II. H2SO4

HSO 4  NH3 U NH4  SO42

(c) HCO3 / CO32 3. (S)

Arrange the following acids in increasing order of their acid strength :

(a) HPO 24  (aq.)

(b) HCO3 (aq.)

(c) H2 PO4 (aq.)

(d) CO32  (aq.)

In the following reaction :  3

2 3

HCO  H 2 O U CO  H3O

7. (A)

t onge

e th n NH 4

Assertion : H2SO4 acts as a base in the presence of HClO4. Reason : Perchloric acid is stronger acid than H2SO4. (a) A

(b) B

(c) C

(d) D

Weak Acids and Bases : Analysis 

8. (S)

Which two substances are Bronsted bases ? (a) CO32 and H3O 

(b) HCO3 and H3O+

(c) HCO3 and CO32

(d) CO32 and H2O

A mixture of weak acid is 0.1 M in HCOOH –4 –4 (Ka = 1.8 × 10 ) and 0.1 M in HOCN (Ka = 3.1 × 10 ). Hence, [H3O† ] is –3

C-14

–4

(a) 7.0 × 10 M

(b) 4.1 × 10 M

(c) 0.20 M

(d) 4.1 × 10 M

–3

IONIC EQUILIBRIUM 9. (S)

Like water, ammonia (NH3) is an amphoteric substance that can be used as a solvent for acid base reactions. The phase label (am) means solvated by ammonia (ammoniated). Which statement is correct, extrapolating from your knowledge of acid base reactions in water ?

[H  ]Total

mixed) [H+] of polyprotic acid (weak, H3A) (has dissociation constants K a1 !! K a 2 !! K a 3 ); will be contributed by

(a) Auto-ionization of ammonia is described by  4

first dissociation at the most.

 2

2NH3 (l ) U NH (am)  NH (am)

12. (C)

(b) The strongest acid that can exist in ammonia solution is H3O+ (am). solution, is NH 4 (am).

+

–6

–7

(c) 9.5 × 10 M 13. (C)

(d) The addition of an acid to an ammonia solution will increase the concentration of NH2– (am). –8

The concentration of NaOH solution is 10 M. Find

1u 107 ; K a 2

out the (OH ) concentration

11. (M)

(a) 10

–8

(c) 10

–6

–11

(b) Greater than 10

(a) 1.25 × 10

–6

–9

–6

(d) Lies between 10 and 10

14. (C)

The percentage ionization of a weak base is given by

§ Ka · ¨ ¸ (a) ¨ c ¸ u100 © ¹

1 § · ¸ u100 (b) ¨ pK b  pOH © 1  10 ¹

§ Kb · ¨ ¸ (c) ¨ c ¸ u 100 © ¹ § Kw ¨ (d) ¨ c u K of conjugate acid a ©

(c) 1.45 × 10

–7

(b) 1.3 × 10

–13

(d) 1.30 × 10

–15 –4

The pH of 0.01 M HCOOH (Ka = 1.4 × 10 ) will be : (a) 2.926

(b) 3.296

(c) 4.962

(d) 5.926 –12

Ionic product of water is 10 . pH of water is _____.

16. (M)

Which of the following statements is/are correct about the ionic product of water ?

· ¸ u 100 ¸ ¹

(b) pK > pKw (c) At 100ºC, Kw of water becomes 10

–12

(d) Ionic product of water at 25ºC is 10 17. (A)

For mono basic acids :

Ka C 18. (A)

where, Ka = Dissociation constant of acid C = molarity of acid Ka , [H  ] D

–14

Assertion : Water acts as levelling solvent for various acids. Reason : Levelling effect of water is due to high dielectric constant and strong proton accepting tendency.

CH 3COOH U CH3COO   H 

CD,[H  ]

1.3 u 10 13

(a) Ka (equilibrium constant of water) < Kw (ionic product of water)

The concentration of hydrogen ion is a measure of acidity or alkalinity of a solution.

[H  ]

–8

(d) 10 M

15. (I)

Comprehension

D

–6

(b) 10 M

What will be sulphide ion concentration of a dilute solution that has been saturated with 0.1 M H2S if the pH of the solution is 3 ? K a1

–

–6

What will be the value of [H ] of 10 M CH3COOH ? –5 (Ka = 1.8 × 10 ) (a) 4.24 × 10 M

(c) The strongest base, which can exist in ammonia

10. (S)

C1K a1  C 2 K a 2 (When two weak acids are

(a) A

(b) B

(c) C

(d) D

Assertion : Addition of HCl(aq.) to HCOOH(aq.) decreases the ionization of HCOOH. +

Reason : Common ion effect of H ion, reduces the ionization of HCOOH.

CK a +

pH of a weak acid can be calculated using [H ] by any of above methods.

C-15

(a) A

(b) B

(c) C

(d) D

IONIC EQUILIBRIUM 19. (S)

The charge balance equation of species in 0.1 M CH3COOH solution is given by : +

25. (A)

–

Reason : In equimolar solutions, the number of titrable protons present in HCl in less than that present in acetic acid.

(a) [H ] = [OH ] +

–

(b) [H ] = [CH3COO ] +

–

–

(c) [H ] = [OH ] + [CH3COO ] +

–

A solution of 2M formic acid (HCOOH) is 0.95% ionized. What is the Ka of formic acid ? –2

21. (S)

(b) 1.8 × 10

–5

(d) 4.5 × 10

What is the Kb of a weak base that produces one OH per molecule if a 0.05 M solution is 2.5% ionized ? –5

27. (A)

A solution is prepared by mixing 100 mL 0.50 M –4 hydrazoic acid (HN3), whose Ka = 3.6 × 10 , with 400 –4 mL of 0.10M cyanic acid (HOCN), whose Ka = 8 × 10 . Which of the following is (are) true regarding

28. (S)

(b) [N3 ] = 3.6 × 10 –3

(c) [OCN ] = 4.57 × 10

–

(d) [OH ] = 7.14 × 10

(d) D –8

Assertion : pH of 10 M HCl lies between 6 and 7.

(a) A

(b) B

(c) C

(d) D

In our body, carbon dioxide (CO2) combines with water (H2O) to form carbonic acid. o H2CO3 H2O + CO2 

–13

At 25ºC, pH of a 0.01 M solution of a monobasic acid (HA) is 4. The correct statement(s) regarding HA and its given solution is (are)

Carbonic acid undergoes dissociation as,

(a) HA is a weak acid

During the physical and mental stress, the rate of respiration increses, which results in the decrease in concentration of CO2 in the blood. What will be the effect on pH of human blood during the stress ?

H 2 CO3 U H   HCO3

(b) The ionization constant (Ka) of acid is approximately –6 10 at 25ºC (c) Increasing the temperature of solution would cause the pH to decrease (d) Addition of 0.1 M HCl solution would lower pH by increasing degree of ionization.

29. (S)

pH of Solutions 24. (S)

(b) B

Reason : For very dilute solutions of acids, H ion contribution from water is also taken into consideration.

–3

–2

Assertion : pH of water increases with an increase in temperature.

+

–3

(d) 1.2 × 10

(a) [H ] = 10 M

23. (M)

(d) D

(c) C

(b) 1.6 × 10

(c) 3.2 × 10

–

(c) C

(a) A

–

–6

(a) 7.8 × 10

+

(b) B

Reason : Kw of water increases with increase in temperature.

–5

–8

22. (M)

26. (A)

–4

(a) 1.9 × 10 (c) 9 × 10

(a) A

–

(d) 2[H ] = [OH ] + [CH3COO ] 20. (S)

Assertion : pH of HCl solution is less than that of acetic acid of the same concentration.

The pKa of acetylsalicylic acid (aspirin) is 3.5. The pH of gastric juice in human stomach is about 2-3 and the pH in the small intestine is about 8. Aspirin will be :

(b) Remains same

(c) Increases

(d) Cannot be predicted –3

–4

Assertion : If Ka of HA is 10 and Ka of HB is 10 at 25ºC, pH of an aqueous solution of HB will be one unit greater than pH of equimolar solution of HA. Reason : For weak acids, both concentration and ionization constant affects the pH.

(a) unionised in the small intestine and in the stomach (b) completely ionized in the small intestine and in the stomach

(a) Decreases

30. (S)

(c) ionised in the stomach and almost unionised in the small intestine (d) ionised in the small intestine and almost unionised in the stomach

C-16

(a) A

(b) B

(c) C

(d) D

Assertion : Increasing the temperature of an aqueous acetic acid solution decreases pH. Reason : Ionization of acetic acid is endothermic in nature. (a) A

(b) B

(c) C

(d) D

IONIC EQUILIBRIUM 31. (S)

32. (S)

33. (S)

What is the pH of a solution at 25ºC that is 0.010 M in HCl and 0.025 M in HNO3 ? (a) 1.49

(b) 1.60

(c) 1.82

(d) 3.60

38. (M)

(a) Acidic buffer will be effective within in the pH range (pK a r 1)

If 112 mL of HCl(g) at S.T.P. conditions become the solute in 500 mL of water solution, what will be the pH of this mixture ? (a) 0.5

(b) 1.0

(c) 1.5

(d) 2.0

(b) Basic buffer will be effective within the pH range (pK w  pK b r 1) (c) H3PO4 + NaH2PO4 is not a buffer solution

Why is it necessary to take the acid-base properties of water into account when computing the hydronium ion concentration of very dilute solutions of strong acids ?

(d) Buffer behaves most effectively when the [Salt]/ [Acid] ratio equal to one 39. (A)

(a) The hydroxide ion produced from the dissociation + of water reacts with most of the H ion produced from the acid.

+

(c) The amount of H ion produced by the dissociation of water is significant compared to that produced by the acid.

35. (S)

(b) 1.0 (d) 3.7 –5

The pH of 10 M HCl solution if 1 ml of it is diluted to 1000 ml is : (a) 5

(b) 8

(c) 7.02

(d) 6.98

37. (M)

On diluting a buffer solution, its pH : (b) decreases

(c) remains same

(d) can’t say

(d) D

kidney cells to produce the base ion ammonia (NH4 )

40. (C)

(a) increases

(c) C

which combines with acids brought to the kidney and is then excreted as ammonium salts.

Buffer Solutions 36. (S)

(b) B

Hydrogen carbonate and phosphate buffers in the + blood prevent excess hydrogen ions [H ] produced by metabolic activity, from decreasing the pH of the blood. Carbon dioxide released into the blood during respiration is regulated by this system and prevented from causing changes in plasma pH prior to its excretion from the lungs. Excessive change in blood chemistry which would change the plasma pH from its normal level 7.4. This excretes hydrogen ions and retains hydrogen carbonate ions and retains hydrogen ions if the pH rises. This may produce change in the pH of the urine from 4.5 to 8.5. A fall in pH also stimulates the

When one drop of a concentrated HCl is added to 1 L of pure water at 25ºC, the pH drops suddenly from 7 to 4. When the second drop of the same acid is added, the pH of the solution further drops to about (c) 2.0

(a) A

Comprehension

(d) The conjugate base of the strong acid reacts with the hydroxide ion produced from the dissociation of water.

(a) 0

Assertion : pH value of HCN solution decreases when NaCN is added to it Reason : NaCN provides a common ion CN– to HCN.

(b) The dissociation constant for water is larger in dilute rather than in concentrated solutions of acids.

34. (S)

Which of the following is (are) correct for buffer solution ?

41. (C)

Choose the correct statement : (a) pH of acidic buffer solution decreases if more salt is added

The normal pH of blood is (a) 4.5

(b) 8.5

(c) 7.32

(d) 7.4

Which of the following buffer present in the blood ? (a) HCO 3  H 2 CO3 and PO34  HPO 24  (b) HCO3  CO 32  and HPO 24   H 2 PO 4

(b) pH of acidic buffer solution increases if more salt is added

(c) HCO 3  H 2 CO 3 and PO 34  H 3 PO 4

(c) pH of basic buffer decreases if more salt is added

(d) None of these

(d) pH of basic buffer increases if more salt is added

C-17

IONIC EQUILIBRIUM 42. (C)

49. (S)

Assuming that the buffer in blood is CO 2  HCO3 .

Which of these aqueous solutions are buffers ? 1.

A solution that is 0.1 M HNO3 and 0.1 M HCl.

2.

A solution that is 0.1 M NaH2PO4 and 0.1 M Na2HPO4.

3.

A solution made by mixing 10 mL of 0.1 M HF and 5.0 mL of 0.1 M NaOH.

Assertion : An aqueous solution of ammonium acetate can act as a buffer.

4.

A solution made by mixing 10 mL of 0.1 M KOH and 20 mL of 0.1 M HCl.

Reason : Acetic acid is a weak acid and NH4OH is a weak base.

(a) 1 and 2

(b) 2 and 3

(c) 3 and 4

(d) 1 and 4

Calculate the ratio of conjugate base to acid necessary to maintain blood at its proper pH. –7

(K1 of H2CO3 = 4.2 × 10 ) (a) 15 (c) 14 43. (A)

44. (M)

(b) 16 (d) 11

(a) A

(b) B

(c) C

(d) D

Which of the following will function as buffer ?

A solution was prepared by mixing 10.0 mL of 0.50 M NaOH with 10.0 mL of 1.00 M acetic acid, Ka=1.8 ×10–5. Find the pH of solution.

(a) NaCl + NaOH

(a) 2.45

(b) 1.67

(c) 2.37

(d) 4.74

50. (S)

(b) Borax + Boric acid

(c) NaH2PO4 + NaHPO4 (d) NH4Cl + NH4OH 45. (M)

(a) 10 mL 0.1 M NH4Cl + 10 mL 0.08 M NaOH

How many milliliters (mL) of a 0.0500 M NaOH (a strong base) solution should be added to 1.00 L of 0.100 M H3PO4, solution to produce a buffer of pH = 2.00 ? For

(b) 20 mL 0.22 M CH3COOH + 30 mL 0.15 M NaOH

H3PO4, K a1 = 6.67 × 10–3

51. (S)

Which of the following mixtures is (are) buffer ?

(c) A 0.10 M NaHCO3 solution

Given : H3PO4 + OH– (from NaOH) o H2PO4– + H2O

(d) 15 mL 0.12 M CH3NH2 + 10 mL 0.07 M HCl 46. (S)

47. (S)

48. (S)

Calculate the [NH 3]/[NH 4+] ratio in an ammoniaammonium chloride buffer with a pH of 9.00. (Kb = 1.8 × 10–5 for ammonia) (a) 0.56 : 1.00

(b) 0.74 : 1.00

(c) 0.86 : 1.00

(d) 1.12 : 1.00

How many milliliters of 0.250 M NH4Cl would have to be added to 450 mL of 0.350 M KOH in order to produce a buffer solution with a pH of 9.10 ? (K b (NH 3 ) = 1.8 × 10–5) (a) 450 mL

(b) 600 mL

(c) 750 mL

(d) 350 mL

A buffer solution of pH = 9.00 is made by dissolving ammonium chloride and ammonia in water. How many moles of ammonium chloride must be added to 1.0 L of 0.25 M ammonia to prepare this buffer ?

(a) 200

(b) 400

(c) 600

(d) 800

52. (I)

A buffer solution is formed by mixing 100 mL 0.01 M CH3COOH with 200 mL 0.02 M CH3COONa. If this buffer solution is made to 1.0 L by adding 700 mL of water, pH will change by a factor of

53. (S)

What volume of 0.40 M NH3 solution must be added to 1.0 L of 0.10 M NH4Cl solution to give a buffer having pH of 10.00 ? For NH3, Kb = 1.8 × 10–5

54. (S)

NH3 (aq)  H 2 O U NH 4 (aq)  OH  (aq); K b =1.8 × 10–5

(a) 1.4 L

(b) 1.1 L

(c) 0.97 L

(d) 0.61 L

For the overall reaction

H 2S(aq) U 2H  (aq)  S2 (aq) , the value of K a is 1×10–22. The Ksp for ZnS is 1.2 × 10–23. What would be the maximum concentration of Zn+2(aq) ion in a 0.01 M solution of H2S that has its pH adjusted to 5 by mixing a strong acid with the H2S ?

(a) 0.25 mol

(b) 0.45 mol

(a) 1.2 × 10–9 M

(b) 2.4 × 10–6 M

(c) 0.65 mol

(d) 2.2 mol

(c) 3.8 × 10–4 M

(d) 1.8 × 10–1 M

C-18

IONIC EQUILIBRIUM 55. (S)

10–3 M H3PO4 (pH = 7) is used in fertilisers as an aqueous soil digesting. Plants can absorb zinc in water soluble form only. Zinc phosphate is the source of zinc

For the overall reaction H2S(aq) U 2H+(aq) + S2– (aq), the value of KA is 1 × 10–22. The Ksp for FeS is 4 × 10–19. What would be the maximum concentration of Fe+2 (aq) ion in a 0.1 M solution of H2S that has has its pH adjusted to 2 by mixing a strong acid with the H2S ? (a) 4 × 10–38 M

(b) 6 × 10–10 M

(c) 6 × 10–6 M

(d) 4.0 M

and PO34 ions in the soil. Ksp of zinc phosphate = 9.1 × 10–33. 59. (C)

Polyprotic Acids and Bases 56. (S)

H3 PO 4  H 2 O U H3O   H2 PO 4 ; pK1 = 2.15

60. (C)

H 2 PO4  H 2O U H3O   HPO 24 pK2 = 7.20

57. (S)

(a) 10–3 M

(b) 1.2 × 10–4 M

(c) 2.2 × 10–4 M

(d) 1.1 × 10–10 M

[Zn2+] ion in the soil is (a) 2.9 × 10–11 M

(b) 4.0 × 10–10 M

(c) 3.0 × 10–6 M

(d) 9.1 × 10–5 M

Hence pH of 0.01 M NaH2PO4 is :

Salt Hydrolysis

(a) 9.35

(b) 4.675

61. (S)

(c) 2.675

(d) 7.350

Which of the following statements is correct ? (a) There will be no change in pH of solution (i) and (ii)

very large ; K a 2 = 1.20 × 10–2. [HSO4–], M

[SO 24 ], M

(a) 0.250

0.250

0.000

(b) 0.250

0.250

0.0120

(c) 0.500

0.000

0.500

(d) 0.261

0.239

0.0110

[H3O+], M

(b) The pH of solution (i) will remain the same but pH of solution (ii) will increase (c) The pH of solution (i) will remain same but of solution (ii) will decrease (d) The pH of solution (ii) will remain same but of solution (i) will increase 62. (S)

H 2 CO3  H 2 O U H3 O  HCO3 (Ka = 4.30 ×10–7) (b) 2.46 × 10–4 M

(c) 4.06 × 10–11 M

(d) 6.84 × 10–6 M

The correct order of increasing [H3O+] in the following aqueous solution is : (a) 0.01 M H2S < 0.01 M H2SO4 < 0.01 M NaCl < 0.01 M NaNO3

What is the equilibrium [OH–] in 0.1413 M H2CO3 ?

(a) 7.07 × 10–14 M

A 100 mL portion of water is added to each of the following two solutions. (i) 100 mL of 0.02 M KCl (ii) 100 mL of 0.02 M HCl

Calculate the molar concentrations of H3O+, HSO4–, and SO42– in a 0.250 M solution of H2SO4. Assume K a1 is

58. (S)

[PO34 ] ion in the soil with pH = 7, is

(b) 0.01 M NaCl < 0.01 M NaNO3 < 0.01 M H2S < 0.01 M H2SO4 (c) 0.01 M H2S < 0.01 M NaNO3 = 0.01 M NaCl < 0.01 M H2SO4

Comprehension

(d) 0.01 M H2S < 0.01 M NaNO3 < 0.01 M NaCl < 0.01 M H2SO4

H3PO4 is a tribasic acid with pK a1 , pK a 2 and pK a 3 2.12, 7.21, and 12.32, respectively. It is used in fertiliser productions and its various salts are used in food, dtergent, toothpaste, and in metal treatment.

63. (S)

pH of water is 7. When a substance Y is dissolved in water, the pH becomes 11. The substances Y is a salt of : (a) weak acid and weak base

Small quantities of H3PO4 are used in imparting the sour or tart taste to soft drinkes, such as Coca Cola, and beers, in which H3PO4 in present 0.05% by weight (density = 1.0 g mL–1).

C-19

(b) strong acid and strong base (c) strong acid and weak base (d) weak acid and strong base

IONIC EQUILIBRIUM Degree of hydrolysis for a salt of strong acid and weak base is :

64. (M)

69. (X)

Column-I

(a) independent of dilution

(Salt)

(b) increase with dilution (c) increase with decrease in Kb of the bases (d) decreases with decrease in temperature A 0.1 M sodium acetate solution was prepared. The –10 Kh = 5.6 × 10

65. (M)

–

–3

–

–6

(p) Only cation hydrolysis

(B) CH3COONa

(q) Only anion hydrolysis

(C) NaClO4

(r) Both cation and anion

70. (X)

(s) No hydrolysis

Match the Column-I with Column-II :

Column-I

(c) The [OH ] concentration is 7.48 × 10 M (d) The pH is approximately 8.88

Column-II

(A)NH4Cl

(p) No hydrolysis

(B) NaCl

(q) h

Kh / C

(C) CH3COONa

(r) h

Kw CK b

(D) CH3COONH4

(s) h

Kh

Match the following : Column I

(A) Degree of hydrolysis for salts of strong

Column II (P) the hydrolysis of anion which is irreversible.

acid and weak base (B) Hydrolysis constant

(C) Na2O in water is basic due to

+

NaOH + H

(Q) is not possible because

where, C = Concentration of Salt ; Kh = hydrolysis constant

of the reaction of strong base with strong acid.

Kw = Ionic product of water ; Kb = Dissociation constant of weak base

(R) is affected when

71. (A)

temperature is changed and concentration is changed.

(D) Na + H2O +

(S) is independent of volume of solution taken but depends upon temperature.

Assertion : Aqueous solution of ammonium carbonate is basic.

(a) A

(b) B

(c) C

(d) D

72. (M)

73. (S)

Assertion : H3BO3 is aprotic acid. Reason : Borax is salt of H3BO3 and NaOH ; its aqueous solution is alkaline in nature. (a) A

(b) B

(c) C

(d) D

Assertion : When aqueous, solution of CH3COONH4 is diluted, then its degree of hydrolysis increases. Reason : Ammonium acetate is the salt of weak acid and weak base, its degree of hydrolysis does not depend on the concentration.

Reason : Acidic or Basic nature of a salt solution of a salt of weak acid and weak base depends on Ka and Kb of the acid and base forming it.

68. (A)

(Nature of hydrolysis)

(A)NH4CN

(D) Fe(NO3)2

(b) The [OH ] concentration is 7.48 × 10 M

67. (A)

Column-II

hydrolysis

–5

(a) The degree of hydrolysis is 7.48 × 10

66. (X)

Match the Column-I with Column-II :

C-20

(a) A

(b) B

(c) C

(d) D

Which among the following salts will give basic solution on hydrolysis ? (a) NaH2PO4

(b) NH4Cl

(c) NaCl

(d) K2CO3

Calcium lactate is a salt of weak acid i.e., lactic acid having general formula Ca(Lac)2. Aqueous solution of salt has 0.3 M concentration. pOH of solution is 5.60. If 90% of the salt is dissociated then what will be the value of pKa ? (a) 2.8 – log (0.54)

(b) 2.8 + log (0.54)

(c) 2.8 + log (0.27)

(d) None of these

IONIC EQUILIBRIUM 74. (S)

75. (S)

76. (M)

If 0.1 mol of salt is added to 1L water, which of these salts is expected to produce the most acidic solution ? (a) NaC2H3O2

(b) NH4NO3

(c) CuSO4

(d) AlCl3

80. (S)

–11

(a) 6.3 × 10 (c) 1.6 × 10 81. (S)

Hydrolysis constants of two salts KA and KB of weak –8 –6 acids HA and HB are 10 and 10 . If the dissociation –2 constant of third acid HC is 10 . The order of acidic strengths of three acids will be :

–5

–5 –4

(c) 6.75 × 10

(a) HA > HB > HC

(b) HB > HA > HC

Solubility Equilibria

(c) HC > HA > HB

(d) HA = HB = HC

82. (S)

83. (S)

(c) pH of a 0.10 M aqueous solution of NaA at 25ºC will be 9. –4

77. (I)

If the equilibrium constant of the reaction of a weak 9 acid HA with a strong base is 10 , then pH of a 0.10 M NaA solution is

78. (A)

Assertion : If HA and HB are two weak non-basic acids with Ka(HA) < Kb(HB), then, the aqueous solution of NaA will have higher pH than pH of aqueous solution of NaB of same concentration.

84. (S)

85. (M)

(b) B

(c) C

(d) D

–2

(d) 5.38 × 10

The solubility of silver benzoate (C6H5COOAg) in H2O

(a) S1 > S2 > S3 > S4

(b) S4 > S3 > S2 > S1

(c) S2 > S3 > S4 > S1

(d) S3 > S2 > S4 > S1

Ksp of Mg (OH)2 is 1 × 10–12. 0.01 M MgCl2 will be precipitating at the limiting pH : (a) 8 (b) 9 (c) 10 (d) 12 The solubility products of MA, MB, MC and MD are 1.8 × 10–10, 4 × 10–3, 4 × 10–8 and 6 × 10–5 respectively. If a 0.01 M solution of MX is added dropwise to a mixture containing A–, B–, C– and D– ions then the one to be precipitated first will be : (a) MA (b) MB (c) MC (d) MD A solution containing a mixture of 0.05 M NaCl and 0.05 M Nal is taken. (K sp of AgCl = 10 –10 and Ksp of AgI = 4 × 10–16). When AgNO3 is added to such a solution:

Reason : Conjugate base of a weaker acid is stronger than the same of stronger acid. (a) A

–3

(b) 3.52 × 10

and S4 respectively. The decreasing order of solubility is

–5

(d) If Kb of a weak base BOH is 10 at 25ºC, equilibrium constant for neutralization of HA with BOH at 25ºC –5 will be 10

–6

(d) 1.6 × 10

and in a buffer solution of pH = 2, 3, and 4 are S1, S2, S3

The equilibrium constant (Kc) for the reaction of a weak 9 acid HA with strong base NaOH is 10 at 25ºC. Which of the following are correct deduction ?

(b) pH of a 0.01 M aqueous solution of HA at 25ºC will be 3.5

–10

(b) 6.3 × 10

If pKb for fluoride ion at 25ºC is 10.83, the ionisation constant of hydrofluoric acid in water at this temperature is (a) 1.74 × 10

(a) The ionization constant (Ka) at 25ºC is 10 for HA.

79. (A)

A solution of 0.1 M NaZ has pH = 8.90. The Ka of HZ is

(a) the concentration of Ag+ required to precipitate Cl– is 2 × 10–9 mol/L (b) the concentration of Ag+ required to precipitate I– is 8 × 10–15 mol/L

Assertion : Knowing K a 2 of CO2 can determine pH of an aqueous solution of Na 2 CO 3 of known concentration.

86. (M)

Reason : Only first hydrolysis of CO32 ion is significant because for most of the dibasic acids,

K a1 !! K a2 .

(c) AgCl and AgI will be precipitate together (d) first AgI will be precipitated Which of the following is (are) correct when 0.1 L of 0.0015 M MgCl2 and 0.1L of 0.025 M NaF are mixed together ? (Ksp of MgF2 = 3.7 × 10–8) (a) MgF2 remains in solution

(a) A

(b) B

(c) C

(d) D

(b) MgF2 precipitates out (c) MgCl2 precipitates out (d) Cl– ions remain in solution

C-21

IONIC EQUILIBRIUM

Comprehension

91. (I)

Assume that Ag ( NH 3 ) 2 is the only complex formed

Solubility of a salt is defined as the maximum amount of a salt which can be dissolved in a given amount of solvent at a particular temperature. Unit of solubility is mole per litre of solution. The solubility of a salt of weak acid with strong base (CH3COOAg) in acidic buffer can be calculated as –

Given : The dissociation constant for

Ag( NH 3 ) 2

+

CH3COO (aq) (S – X)

+

H

+

+

92. (I)

HN3

X

–

(S  X) [H  ] X

The solubility of LiOH in a solution with pH = 8 is –12 (Ksp = 1.8 × 10 ) –18

(a) 1.8 × 10 –6

88. (C)

89. (C)

Comprehension The product of the concentrations of the ions of an electrolyte raised to power of their coefficients in the balanced chemical equation in the solution at any concentration. Its value is not constant and varies with change in concentration. Ionic product of the saturated solution is called solubility product Ksp.

–6

(b) 1.8 × 10

–4

(c) 10

(d) 1.8 × 10

The solubility of AgCl is maximum in (a) 0.1 M NaCl

(b) 0.1 M HCl

(c) 55.5 M H2O

(d) None of these

(i) When Kip = Ksp, the solution is just saturated and no precipitation takes place.

The solubility of CaCO3 is 7 mg/L. Calculate the solubility product of BaCO3 from this information and from the fact that when Na2CO3 is added slowly to a solution containing equimolar concentration of Ca 2+

(ii) When Kip < Ksp, the solution is unsaturated and precipitation will not take place.

2+ 2+

and Ba , no precipitate is formed until 90% of Ba

(iii) When Kip > Ksp, the solution is supersaturated and precipitation takes place.

has been precipitated as BaCO3. –10

(a) 4.9 × 10

–10

(c) 6.9 × 10 90. (X)

+

H + N 3 .

Find the concentration of Ag ions, if excess of solid AgN3 is added to a solution maintained at ph = 4. The –5 ionisation constant Ka of HN3 is 2.0 × 10 . The solubility of AgN 3 in pure water is found to be –3 5.4 × 10 M at 25ºC.

Dissociation constant of weak acid

87. (C)

.

+

Ksp = [Ag ] [CH3COO ] = S (S – X)

(K a )

–13

HN3 (hydroazoic acid) is a weak acid dissociating as :

ZZZX YZZ Z CH3COOH

(from acidic buffer)

–8

(As in 10–3 M)

S S = Solubility of salt

–

+

Ag + 2NH3; Kd = 6.0 × 10 and

Ksp (AgBr) = 5.0 × 10

CH3COOAg YZZ ZZZX Z CH3COO (aq) + Ag (aq) S–X

How much AgBr could dissolve in 1.0 L of 0.4 M NH3 ?

–5

(b) 5.9 × 10

93. (C)

Which of the following is most soluble ?

–5

(d) 8.3 × 10

(a) Bi2S3 (Ksp = 1× 10–70) (b) MnS (Ksp = 7 × 10–16)

Match the following :

Column I +Y X

–X Y

+Y X

–X Y

+Y X

–X Y

(A) [A ] [B ] > Ksp (B) [A ] [B ] < Ksp

(c) CuS (Ksp = 8 × 10–37) (d) Ag2S (Ksp = 6 × 10–51)

Column II (P) Precipitation just starts

94. (C)

(Q) Buffer capacity

The concentration of Ag+ ions in a given saturated solution of AgCl at 25ºC is 1.06 × 10–5 g ion per litre. The solubility product of AgCl is :

(C) [A ] [B ] = Ksp

(R) The solid AXB4 will precipitate out

(a) 0.353 × 10–10

(b) 0.530 × 10–10

(D) d(n)/dpH

(S) No precipitation

(c) 1.12 × 10–10

(d) 2.12 × 10–10

C-22

IONIC EQUILIBRIUM 95. (C)

96. (I)

97. (A)

If the solubility of Li3Na3 (AlF6)2 is x mol L–1, then its solubility product is equal to : (a) 12x3

(b) 18x3

(c) x8

(d) 2916x8

101. (C) The factor by which the solubility of AgCN is increased in the above solution as compared to its solubility in pure water is

–3

–9

M(OH)x has Ksp of 4 × 10 and its solubility of 10 M. The value of x is ______ . –6

2+

Assertion : On miximg 500 mL of 10 M Ca ion and –6 – 500 mL of 10 M F ion, the precipitate of CaF2 will be –18 obtained Ksp (CaF2) = 10 .

98. (S)

(b) B

(c) C

(d) D

9§W· (a) 10 ¨ ¸ ©M¹

5§W· (c) 10 ¨ ¸ ©M¹

5

7§W· (b) 10 ¨ ¸ ©M¹

3§W· (d) 10 ¨ ¸ ©M¹

5

6

(b) 1.6 × 10

(d) 6.2 × 10

–10

–1

100. (C) The solubility of AgCN (mol-L ) in the above solution is –8

(a) 1.58 × 10 (c) 2 × 10

–5

(d) 12 g

(b) 6.2 × 10 (d) 4 × 10

–15

(a) 2.9

(b) 3.9

(c) 4.4

(d) 4.9 –12

–

9

(c) 1.3 g

–8 –15 For H2S : K a1 = 9.2 × 10 , and K a 2 = 1.2 × 10

5

The concentration ratio of HCN to CN in the solution is

(c) 1.6 × 10

(b) 43 g

MnS(s) U Mn 2  S2 , Ksp = 3.6 × 10

104. (S) For Ag 2 CO 3 , K sp = 6.2 × 10 , For AgCl, K sp –10 = 2.8 ×10 . Solid Ag2CO3 and solid AgCl are added to a beaker containing 1.00 M Na2CO3(aq). Under these 2– – conditions the [CO3 ] = 1.00 M. Calculate the [Cl ] in solution when equilibrium is established. –4

Answer the following questions :

–7

(a) 0.0013 g

2+

The solubility product constant of AgCN is 2.5×10– 16 . The acid dissociation constant of HCN is 6.2 × 10 –10 . Now solid AgCN is dissolved in a buffer solution of pH 3.

(a) 6.25 × 10

(d) 5000

103. (S) What is the highest pH at which 0.050 M Mn remains entirely in a solution that is saturated with H2S, at a concentration of, [H2S] = 0.10 M. Given for the reaction,

Comprehension

99. (C)

(c) 1250

–10

Solubility of calcium phosphate (molecular mass, M) in water is W g per 100 mL at 25ºC. Its solubility product at 25ºC will be approximately : 5

(b) 100

102. (S) How many grams of AgCl (Ksp = 1.8 × 10 ) will dissolve –5 in 1.0 L of 6.0 M NH3 (Kb = 1.8 × 10 ). The Kformation + 7 for Ag(NH3)2 is 1.7 × 10 .

Reason : If Ksp is less than ionic product, precipitate will be obtained. (a) A

(a) 10

–6

(a) 1.1 × 10

(b) 1.26 × 10

(c) 0.15

(d) 2.8 × 10

–6

105. (S) Given the following K sp values : for M(OH) 4 , –19 –14 Ksp(M) = 4.0 × 10 ; for Z(OH)2, Ksp(Z) = 1.0 × 10 . One mole of each of the above solids is placed in a beaker containing 1 (one) litre of pure water. These solids go into equilibrium with their ions. Calculate the pH required for the metal ion concentrations to be equal, 4+ 2+ [M ] = [Z ]. (a) 12.6

(b) 9.6

(c) 6.7

(d) 11.8

–10

–7

C-23

IONIC EQUILIBRIUM 106. (I)

107. (I)

Solubility product constant of a springly soluble salt –12 MCl2 is 4 × 10 at 25ºC. Also, at 25ºC, solubility of 8 MCl2 is an aqueous solution of CaCl2 is 4 × 10 times less compared to its solubility in pure water. Hence, concentration (molarity) of CaCl2 solution is

Which of the following is/are correct for the given curve ? (a) The solution being titrated is a base and the titrant is an acid (b) The pH at the equivalence point must be less than 7

The solubility product constant of a metal carbonate –12 MCO3 is 2 × 10 at 25ºC. A solution is 0.1 M in M(NO3)2 and it is saturated with 0.01 M CO2. Also the –7

ionization constant of CO2 are K a1 = 4 × 10 and K a 2 –11

= 5 × 10 at 25ºC. The minimum pH that must be maintained to start any precipitation is

(c) The titrated solution is strong acid (d) The titrated solution is strong base 110. (A) Assertion : The pH of the solution at the mid point of the weak acid strong base titration becomes equal to the pKa of the acid. Reason : The molar concentrations of proton acceptor and proton donor become equal at mid point of a weak acid.

Indicators and Titrations 108. (M) Titration curve for a weak acid with a strong base is

Which of the following s/are correct for the above curve ?

(a) A

(b) B

(c) C

(d) D

111. (I)

The equivalence point in a titration of 40.0 mL of a solution of a weak monoprotic acid occurs when 35.0 mL of a 0.10 M NaOH solution has been added. The pH of the solution is 5.5 after the addition of 20.0 mL of NaOH solution. What is the dissociation constant of the –6 acid ? (in 10 )

112. (I)

When a 40 mL of a 0.1 M weak base, BOH is titrated with 0.10 M HCl, the pH of the solution at the end point is 5.5. What will be the pH if 10 mL of 0.10 M NaOH is added to the resulting solution ?

113. (X) Match the Column-I with Column-II :

(a) At the half-neutralization point pH = pKa

Column-I

(b) The pH is greater than 7 at the equivalence point

(A) has highest pH at end

(c) The colour change in basic medium (d) The solution at the half neutralization is buffer solution 109. (M) The titration curve given below involves 1.0 M solutions of an acid and a base.

point when titrated against

Column-II (p) CH3COOH Ka = 1.8 × 10

–5

standard NaOH (B) has lowest pH at the end point when titrated against

(q) HCN Ka = 5 × 10

–10

standard NaOH (C) release maximum heat when neutralized with

(r) HF Ka = 5 × 10

–4

NaOH (D) release equal amount of heat when neutralized with strong acid or base

C-24

(s) NH4OH Kb = 1.8 × 10

–5

IONIC EQUILIBRIUM 114. (X) Match the Column-I with Column-II :

118. (C) Determine the pH after 20.0 mL of 0.100 M NaOH has been added.

Column-I

Column-II

(Titration)

(Indicator used)

(A) Strong acid versus

(p) Methyl orange (3 – 4.4)

strong base (B) Weak acid versus strong base (r) Phenolphthalein (8 – 10)

weak base (D) Weak acid versus weak base 115. (X) Match the Column-I with Column-II : Column-II (pH range)

(A) Phenolphthalein

(p) 4.2 – 6.3

(B) Litmus

(q) 3.1 – 4.4

(C) Methyl red

(r) 8.3 – 10.0

(D) Methyl orange

(s) 5.0 – 8.0

(c) 5.50

(d) 6.18

(a) 5.50

(b) 7.00

(c) 8.41

(d) 9.24

120. (C) Determine the pH after a total of 75.0 mL of 0.100 M NaOH has been added.

(s) No suitable indicator

Column-I (Indicator)

(b) 5.33

119. (C) Determine the pH after a total of 50.0 mL of 0.100 M NaOH has been added.

(q) Methyl red (4.3 – 6.3)

(C) Strong acid versus

(a) 3.25

(a) 8.41

(b) 9.24

(c) 11.24

(d) 11.32

Comprehension 100 mL of a solution of X is titrated with a 0.1 M solution of Y giving the following titration curve :

116. (S) During the titration of weak diprotic acid H2A against strong base NaOH, the pH of the solution half-way to the first equivalence point and that at first equivalence point are given respectively by : (a) pK a1 and pK a 2

(b) pK a1 and

pK a1  pK a 2 2

(c) pKa and (pK a1  pK a 2 ) (d)

CK a1 and

pK a1  pK a 2 2

117. (S) 0.2 g sample of benzoic acid, C6H5COOH is titrated with 0.12 M Ba (OH)2 solution. What volume of Ba(OH)2 solution is required to reach the equivalence point ? Answer the following four questions based on the information provided.

–1

Molar mass of benzoic acid is 122 g mol (a) 6.82 mL

(b) 13.6 mL

(c) 17.6 mL

(d) 35.2 mL

121. (C) Y is a

Comprehension Consider the titration of a diprotic acid (H 2 A ;

K a1 = 3.1 × 10 –6, K a 2 = 4.8 × 10–12) with a solution

(a) strong acid

(b) strong base

(c) weak base

(d) weak acid

122. (C) X is a

of NaOH. Specifically, you titrate 50.0 mL of 0.100 M H2A with 0.100 M NaOH. Answer the following question based on this titration experiment.

C-25

(a) strong acid

(b) strong base

(c) weak base

(d) weak acid

IONIC EQUILIBRIUM 123. (C) The initial concentration of X is (a) 0.10 M

(b) 0.02 M

(c) 0.05 M

(d) 0.01 M

124. (C) The approximate value of ionization constant (Ka/Kb) of X is (a) 5 × 10 (c) 2 × 10

–9 –4

(b) 5 × 10 (d) 3 × 10

–6 –5

125. (S) Following figure represents simulated titration curves for solutions of four acids titrated with the same standard base solution. Based on these titration curves, we can accurately predict that

(a) Curve 4 represents the smallest concentration and the weakest acid. (b) Curve 1 represents the largest concentration and the strongest acid. (c) The dissociation constant for the acid represented –4 –4 by curve 3 is about 1 × 10 (i.e., Ka = 1 × 10 ). (d) All the responses above are correct.

C-26

IONIC EQUILIBRIUM

Answer Key Advanced Objective Questions 1. (a)

2. (a,c)

3. (d)

4. (d)

5. (a)

6. (a)

11. (c,d)

12. (a)

13. (d)

14. (a)

15. (0006)

21. (c)

22. (a,b)

23. (a,b,c) 24. (d)

31. (a)

32. (d)

33. (c)

41. (a)

42. (d)

51. (d)

8. (a)

9. (a)

10. (d)

16. (a,b,c,d) 17. (a)

18. (a)

19. (c)

20. (b)

25. (c)

26. (d)

27. (a)

28. (c)

29. (d)

30. (a)

34. (d)

35. (d)

36. (c)

37. (b,c)

38. (a,b,d) 39. (d)

40. (d)

43. (b)

44. (b,d)

45. (a,c,d) 46. (a)

47. (b)

48. (b)

49. (a)

50. (d)

52. (0000)

53. (a)

54. (a)

55. (d)

57. (d)

58. (c)

59. (c)

60. (a)

61. (b)

62. (b)

63. (d)

64. (b,c,d) 65. (a,c,d)

67. (a)

68. (b)

69. (A-r), (B-r), (C-s), (D-p) 75. (c)

56. (b)

66. A o R, B o S, C o P, D o Q

70. (A-q,r), (B-p), (C-q), (D-s)

76. (a,b,c,d) 77. (0009) 78. (a)

7. (a)

79. (a)

80. (c)

71. (d)

72. (a,d)

73. (a)

74. (b)

81. (c)

82. (c)

83. (b)

84. (a)

85. (a,b,d) 86. (b,d)

87. (b)

88. (c)

89. (a)

90. A o R, B o S, C o P, D o Q

91. (0001)

92. (0000)

93. (b)

94. (c)

95. (d)

96. (0002)

97. (d)

98. (b)

101. (c)

102. (b)

103. (d)

104. (a)

105. (d)

106. (0002) 107. (0004) 108. (a,b,c,d)

99. (b)

100. (c)

109. (a,b,c) 110. (a)

111. (0004)

112. (0009) 113. (A–q), (B–r), (C–r), (D–p,s)

114. (A–p,q,r), (B–r), (C–p,q), (D–s) 115. (A–r), (B–s), (C–p), (D–q)

116. (b)

120. (d)

117.(a)

118. (b)

119. (d)

121. (a)

Dream on !!

[\]^[\]^

C-27

122. (c)

123. (b)

124. (c)

125. (d)

SOME BASIC CONCEPTS OF CHEMISTRY

SOME BASIC CONCEPTS OF CHEMISTRY EXERCISE - 3: ADVANCED OBJECTIVE QUESTIONS 1. 2. 3. 4.

All questions marked “S” are single choice questions All questions marked “M” are multiple choice questions All questions marked “C” are comprehension based questions All questions marked “A” are assertion–reason type questions (A) If both assertion and reason are correct and reason is the correct explanation of assertion. (B) If both assertion and reason are true but reason is not the correct explanation of assertion. (C) If assertion is true but reason is false. (D) If reason is true but assertion is false.

5.

All questions marked “X” are matrix–match type questions

6.

All questions marked “I” are integer type questions

Atoms

Molecules

1. (S) If we consider that 1/6 in place of 1/12, mass of carbon atom is taken to be the relative atomic mass unit, the mass of one mole of a substance will

5. (S)

the number of moles of CO2 left are

(a) decrease twice (b) increase two fold

6. (S)

(c) remain unchanged (d) be a function of the molecular mass of the substance 2. (A) Assertion : Both 138 g of K2CO3 and 12 g of carbon have same number of carbon atoms. Reason : Both contains 1 g atom of carbon which contains 6.022 × 1023 carbon atoms. (a) A

(b) B

(c) C

(d) D

7. (S)

8. (M)

3. (A) Assertion : 1 Avogram is equal to 1 amu. (b) B

(c) C

(d) D

4. (X) Column I

9. (M)

(P) 0.5 NA electrons

(B) 1.2 g of Mg2+

(Q) 15.9994 amu

(c) 0.288 × 10–3

(d) 1.68 × 10–2

A gaseous mixture contains oxygen and nitrogen in the ratio of 1 : 4 by weight. Therefore, the ratio of their number of molecules is (a) 1 : 4

(b) 1 : 8

(c) 7 : 32

(d) 3 : 16

A compound possesses 8% sulphur by mass. The least molecular mass is (a) 200

(b) 400

(c) 155

(d) 355

8 g O2 has same number of molecules as that in : (b) 7 g CO (d) 22 g CO2

Which of the following have same number of atoms ? (b) 0.1 mol of NH (d) 22.4 L of Cl2 at STP

10. (A) Assertion : Number of molecules present in SO2 is twice the number of molecules present in O 2. Reason : Molecular mass of SO2 is double to that of

(C) Exact atomic weight of (R) 0.2 mole atoms mixture

O2.

of oxygen isotopes (D) 0.9 ml of H2O

(b) 28.8 × 10–3

(a) 6.4 g of O2 (c) 4.0 g of He

Column II

(A) 5.4 g of Al

(a) 2.85 × 10–3

(a) 14 g CO (c) 11 g CO2

Reason : Avogram is reciprocal of Avogadro’s number. (a) A

If 1021 molecules are removed from 200mg of CO2, then

(S) 0.05 moles

C-28

(a) A

(b) B

(c) C

(d) D

SOME BASIC CONCEPTS OF CHEMISTRY

Stoichiometric Calculations

Or KClO {2Al.

11. (S) P and Q are two elements which forms P 2Q3 and PQ2. If 0.15 mole of P 2Q3 weights 15.9g and 0.15 mole of PQ 2 weights 9.3g, the atomic weight of P and Q is (respectively) : (a) 18, 26

(b) 26,18

(c) 13, 9

(d) None of these

I : CaO + 3C o CaC2 + CO II : CaC2 + 2H2O o Ca(OH)2 + C2H2 CaC2 (calcium carbide) is prepared in step I. It is used to prepare acetylene (C2H2) in step II. Suppose we want to

12. (S) 1 mole of oxalic acid is treated with conc. H2SO4. The resultant gaseous mixture is passed through a solution of KOH. The mass of KOH consumed will be (where KOH absorbs CO2.) (COOH)2

CO + CO2 + H2O

2 KOH + CO2

K2CO3 + H2O

(a) 28 g

(b) 56 g

(c) 84 g

(d) 112 g

determine amount of CaO that can give enough CaC2 to converted required amount of C2H2. Amount of CaO is determined in step I and then amount of C2H2 in step II. We can relate CaO and C2H2 stoichiometrically by writing net reaction which is CaO + 3C + 2H2 O o Ca(OH)2 + C2H2 + CO

13. (M) 0.2 mole of K3PO4 and 0.3 mole of BaCl2 are mixed in 1 L of solution. Which of these is/are correct ?

Thus, CaO { C2H2 14. (C) NX is produced by the following step of reactions

(a) 0.2 mole of Ba3(PO4)2 will be formed

M + X2 o M X2

(b) 0.1 mole of Ba3(PO4)2 will be formed

MX2 + X2 o M3X8

(c) 0.6 mole of KCl will be formed

M3X8 + N2CO3 o N X + CO2 + M3O4

(d) 0.3 mole of KCl will be formed How much M (metal) is consumed to produce 206 gm of

Comprehension Often more than one reaction is required to change starting materials into the desired product. This is true for many reaction that we carry out in the laboratory and for many industrial process. These are called sequential reactions. The amount of desired product from each reaction is taken as the starting material for the next reaction.

NX. (Take At. wt of M = 56, N=23, X = 80) (a) 42 gm (c)

14

gm

(b) 56 gm (d)

7 gm 4

15. (C) The following process has been used to obtain iodine from oil-field brines in California.

I : 2KClO3 o 2KCl + 3O2

NaI + AgNO o AgI + NaNO

II : 4Al + 3O2 o 2Al2 O KClO3 decomposes in step I to give O2, which in turn, is used by Al to form Al2O3 in step II. First we determine

AgI + Fe o FeI2 + Ag FeI2 + Cl2 o FeCl3 + I2

O2 formed in step I and then Al used by this O2 in step II. Both reactions can be added to determine amount of KClO3 that can give required amount of O2 needed for Al.

AgNO required per hour will be

Net : 2KClO3 + 4Al o 2KCl + 2Al2O

(a) 170 kg

(b) 340 kg

Thus, 2KClO {4Al

(c) 255 kg

(d) 510 kg

C-29

If 381 kg of iodine is produced per hour then mass of

[atomic mass Ag– 108, I– 127, Fe–56, N–14, Cl–35.5]

SOME BASIC CONCEPTS OF CHEMISTRY 16. (C) 120 gm Mg was burnt in air to give a mixture of MgO and Mg3N2. The mixture is now dissolved in HCl to form

18. (S) If 7.0 moles of Y is placed in a container and allowed to react with X until equilibrium is reched according to the reaction :

MgCl2 and NH4Cl, if 107 grams NH4Cl is produced. The

2X + Y o 2Z

reaction are follows I.

It is found that the equilibrium mixture contains 8.0 moles of X and 5.0 moles of Y. How many moles of X were present in the original container ?

1 Mg + O2 o MgO , 2

II. 3Mg  N2 o Mg 3 N2

(a) 10

(b) 12

III. MgO + 2HCl o MgCl2 + H2O,

(c) 14

(d) 16

19. (S) Consider the given reversible reaction at equilibrium

IV. Mg3 N2 + 8HCl o 2NH4Cl + 3MgCl2 Then the moles of MgCl2 formed is :

2NO + Cl2 o 2ClNO(g)

(At. wt.

Suppose that 0.30 mol NO, 0.20 mole of Cl2 and 0.50 mole of ClNO were placed in a 25.00L vessel and allowed to reach the equilibrium. At equilibrium, the concentration of ClNO was found to be 0.024 molar. Molar concentration of NO present at equilibrium is

Mg = 24, N = 14, Cl = 35.5) (a) 3 moles

(b) 6 moles

(c) 5 moles

(d) 10 moles

17. (X) On the left column, some reactions are indicated and on the right column, properties of reactions are described. Match them appropriately, and select the correct code. Column I (A)

(B)

(C)

Igniting MnO2 in air converts it quantitatively to Mn3O4. A sample of pyrolusite is of the following composition : MnO2 = 80%, SiO2 and other inert constituents = 15%

(q) Mass of reactant = Mass

S (1.0g) + O2 (1.0 g)

and rest bearing H2O. The sample is ignited to constant weight. What is the % of Mn in the ignited sample ?

of product

22. (I)

A mixture contains equi-molar quantities of carbonates of two bivalent metals. One metal is present to the extent of 13.5% by weight in the mixture and 2.50 gm of the mixture on heating leaves a residue of 1.18 gm. Calculate the % age by weight of the other metal.

23. (I)

A 0.01 moles of sample of KClO was heated under such conditions that a part of it decomposed according to the equation :

(r) Stoichiometric amounts of

 o SO 2

(d) 0.01 M

21. (I) reagent.

 o N 2 F4

(c) 0.008 M

A mixture of FeO and Fe3O4 when heated in air to a constant weight, gains 5% of its weight. Find the percentage of Fe3O4.

N 2 (5.00g)  H 2 (1.00g) (p) First reactant is the limiting

N 2 (3g)  F2 (10g)

(b) 0.006 M

20. (I)

Column II

 o NH

(a) 0.004 M

reactants.

(s) Second reactant is the limiting reactant. p

q

r

s

(a) 2KClO3 o 2KCl + 3O2 and the remaining undergoes change according to the equation :

(a) B

A

A

C

(b) 4KClO3 o 3KClO4 + KCl

(b) B

C

B

A

If the amount of O2 evolved was 134.4 mL at S.T.P.,

(c) B

C

C

A

calculate the % age by weight of KClO4 in the residue.

(d) C

A

B

C

C-30

SOME BASIC CONCEPTS OF CHEMISTRY 24. (M) Three metals of alkaline earth metal group (A, B, and C) when reacted with a fixed volume of liquid Br2 separately gave a product (metal bromides) whose mass is plotted against the mass of metals taken as shown in the figure.

27. (M) Which of the following statements is/are correct ? The following reaction occurs : ' o CCl4  S2Cl 2 2Al  3MnO 

108.0g of Al and 213.0g of MnO was heated to initiate the reaction . (Mw of MnO = 71, atomic weight of Al = 13) (a) Al is present in excess. (b) MnO is present is excess. (c) 54.0g of Al is required. (d) 159.0g of MnO is in excess. From the plot, predict what relation can be concluded between the atomic weights of A, B, and C ? (a) C > B

Percent Purity 28. (S) To 1 L of 1.0 M impure H2SO4 sample, 1.0 M NaOH solution was added and a plot was obtained as follows :

(b) B > A (c) C < A < B (d) Data is insufficient to predict 25. (I)

One commercial system removes SO2 emission from smoke at 95ºC by the following set of reaction : SO 2 (g)  Cl2 (g)  o SO 2 Cl 2 (g) SO 2Cl2 (g)  H 2 O(l )  o H 2SO 4  HCl

o C SO 4  H 2O H 2SO 4  C (OH) 2  How many grams of CaSO4 may be produced from 3.78g of SO2 ?

The % purity of H 2 SO 4 and the slope of curve, respectively, are :

26. (M) Which of the following statements is/are correct ? 1.0g mixture of CaCO3(s) and glass beads liberate 0.22g

(a) 75%, –1/2

(b) 75%, –1

of CO2 upon treatment with excess of HCl. Glass does not react with HCl.

(c) 50%, –1/3

(d) 50%, –1/2

Percent Yield

CaCO3  2HCl  o CO 2  H 2O  CaCl2

[Mw CaCO3 = 100. Mw of CO2 = 44, [Atomic weight of Ca = 40]

29. (S) In the preparation of iron from haematite (Fe2O3) by the reaction with carbon

(a) The weight of CaCO3 in the original mixture is 0.5g. (b) The weight of calcium in the original mixture is 0.2g. (c) The weight percent of calcium in the original mixture is 40% Ca. (d) The weight percent of Ca in the original mixture is 20% Ca.

C-31

Fe 2O3  C  o Fe  CO 2 How much 80% pure iron could be produced from 120 kg of 90% pure Fe2O3 ? (a) 94.5 kg

(b) 60.48 kg

(c) 116.66 kg

(d) 120 kg

SOME BASIC CONCEPTS OF CHEMISTRY 36. (C) Volume of acid that contains 63g pure acid is.

30. (S) NH3 is formed in the following steps : I. Ca + 2C o CaC2

50% yield

(a) 100 mL

(b) 40.24 mL

II. CaC2 + N2 o CaCN2 + C

100% yield

(c) 63.38 mL

(d) 70.68 mL

III. CaCN2 + 3H2O o 2NH3 + CaCO3 50% yield

37. (C) Volume of water required to make 1N solution from 2 mL conc. HNO3.

To obtain 2 mol NH3, calcium required is : (a) 1 mol

(b) 2 mol

(a) 29.56 mL

(b) 30.56 mL

(c) 3 mol

(d) 4mol

(c) 28.56 mL

(d) 31.56 mL

Strength : Mass Percent 31. (S) If 100 ml of H2SO4 (A) and 100 ml of H2O (B) are mixed. Then the mass per cent of H2SO4 would be (Given density

38. (S) An aqueous solution of glucose (C6H12O6) is 0.01 M. To 200 mL of this solution, which of the following should be carried out to make it 0.02 M ?

of H2SO4 = 0.9 g/ml; density of H2O = 1.0 g/ml)

I. Evaporate 50 ml of solution

(a) 60

(b) 50

II. Add 0.180 gm of glucose

(c) 47.36

(d) 90

III. Add 50 mL of water

32. (S) If 100 mL of H2SO4 and 100 mL of H2O are mixed, the

The correct option is :

mass percent of H2SO4 in the resulting solution is (d H2SO4

0.09g mL1 , d H2O

1.0g mL1 )

(a) 90

(b) 47.36

(c) 50

(d) 60

(a) I

(b) II

(c) I, II

(d) I, II, III

39. (S) Equal volumes of 0.200 M HCl and 0.400M KOH are mixed. The concentrations of the ions in the resulting solution are :

Strength : Molality

(a) [K+] = 0.40M, [Cl–] = 0.20 M, [H+] = 0.20 M

33. (A) Assertion : Molality and mole fraction units of concentration do not change with temperature.

(b) [K+] = 0.20 M, [Cl–] = 0.10 M, [OH–] = 0.10 M

Reason : These concentration units are defined in terms of mass rather in terms of volume and mass is independent of temperature. (a) A

(b) B

(c) C

(d) D

(c) [K+] = 0.10 M, [Cl–] = 0.10 M, [OH–] = 0.10 M (d) [K+] = 0.20 M, [Cl–] = 0.10 M, [OH–] = 0.20 M 40. (M) You are provided with 1 M solution of NaNO3 whose density = 1.25 g/ml

34. (M) Select dimensionless quantity(ies) : (a) vapour density

(b) molality

(a) The percentage by mass of NaNO3 = 6.8

(c) specific gravity

(d) mass fraction

(b) The percentage by mass of H2O = 93.2

Comprehension

(c) The molality of the solution is 10.72 –1

HNO3 used as a reagent has specific gravity of 1.42g mL and contains 70% by strength HNO3. 35. (C) Normality of acid is. (a) 16.78

(b) 15.78

(c) 14.78

(d) 17.78

(d) The solution has 0.2 moles of NaNO3. 41. (A) Assertion : In laboratory, reagents are made to a specific molarity rather molality. Reason : The volume of sulution is easier to measure than its mass.

C-32

(a) A

(b) B

(c) C

(d) D

SOME BASIC CONCEPTS OF CHEMISTRY COMPREHENSION

Strength : Mole Fraction

The analytical molarity of a solution gives the total number of moles of a solute in one litre of the solution. The equilibrium molarity represents the molar concentration of particular species in a solution at equilibrium. In order to specify the equilibrium molarity of a particular species it is necessary to know how the solute behaves when it is dissolved in a solvent. e.g., if analytical molarity of HCl is 0.1 M then equilibrium molarity of NaOH equal to zero because HCl is completely dissociated. 42. (C) Calculate the analytical molarity of Cl– ion in solution which is prepared by mixing 100 ml of 0.1 M NaCl and 400 ml of 0.01 M BaCl2. (a) 0.018 M

(b) 0.036 M

(c) 0.084 M

(d) 0.046 M

47. (M) The mole fraction of NaCl in aqueous solution is 0.2. The solution is (a) 13.9 m (b) Mole fraction of H2O is 0.8 (c) acidic in nature (d) neutral

Strength : Variation 48. (M) When 100 ml of 0.1 M KNO3, 400 ml of 0.2 M HCl and 500 ml of 0.3 M H2SO4 are mixed. Then in the resulting solution (a) The molarity of K+ = 0.01 M (b) The molarity of SO42– = 0.15 M (c) The molarity of H+ = 0.38 M

43. (C) The molarity of 68 % of H2SO4 whose density is 1.84 g/ cc is (a) 12.76 M

(b) 6.84 M

(c) 18.4 M

(d) 6.8 M

44. (C) HCl is 80% ionised in 0.01 M aqueous solution. The equilibrium molarity of HCl in the solution is

(d) The molarity of NO3– = 0.01 M and Cl– = 0.08 M 49. (A) Assertion : Molality of solution is independent of temperature while mole fraction depends on temperature. Reason : Normality is the ratio of moles of solute and volume of solution while mole fraction is the ratio of moles of solute and weight of solvent present in solution.

(a) 0.002

(b) 0.06

(a) A

(b) B

(c) 0.02

(d) 0.008

(c) C

(d) D

45. (M) Which of the following statements is/are correct ? 20.0 mL of 6.0 M HCl is mixed with 50.0 mL of 2.0 M Ba(OH)2, and 30 mL of water is added.

to V2 by adding solvents, its molarity before dilution M1 and after dilution M2 are related as :

remaining in solution is 0.8

(a) The concentration M. (b) The concentration of M.

remaining in solution is 1.2

(c) The concentration of Ba 1.0M

2+

(d) 80 mmoles of

50. (A) Assertion : When a solution is diluted from volume V1

M1V1 = M2V2 Reason : During dilution, moles of the solute remains conserved.

remaining in solution is

(a) A

(b) B

(c) C

(d) D

51. (A) Assertion : For a binary solutiokn of two liquids, A and B, with the knowledge of density of solution, molarity can be converted into molality.

is in excess.

46. (M) The density of a solution of H2SO4 is 1.84 gm/ml and it contain 93% H2SO4 by volume. Then (a) Molarity of H2SO4 is 10.42 (b) Mass of H2O = 91 gm (c) Mass of 100 gm solution = 184 gm (d) None of the above

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Reason : Molarity is defined in terms of volume and molality in terms of mass, and mass and volume are related by density. (a) A

(b) B

(c) C

(d) D

SOME BASIC CONCEPTS OF CHEMISTRY 52. (I)

53. (I)

54. (I)

50 mL of 1 M HCl, 100 mL of 0.5 M HNO3, and x mL of 5 M H2SO4 are mixed together and the total volume is made upto 1.0 L with water. 100 mL of this solution exactly neutralises 10 mL of M/3 Al2(CO3)3. Calculate the value of x. How many mL of a solution of concentration 100 mg Co2+ per mL of a solution of concentration 20 mg Co2+ per mL.

(d) 100 mL of 2.0 M HCl +200 mL of 1.0 M NaOH

[Na † ] = 0.4 M

+150 mL of 4.0 M CaCl2

[Cl 4 ] = 2.8 M

+50 mL of H2O 58. (S) 100 mL of mixture of NaOH and Na2SO4 is neutralised

A solution contains 75 mg NaCl per mL. To what extent must it be diluted to give a solution of concentration 15 mg NaCl per mL of solution.

by 10 mL of 0.5 M H2SO4. Hence, NaOH in 100 mL solution is

Strength : Stoichiometric Calculations 55. (S) How much NaNO3 must be weighed out to make 50 ml of an aqueous solution containing 70 mg of Na+ per ml? (a) 12.394 g

(b) 1.29 g

(c) 10.934 g

(d) 12.934 g

59. (S)

(b) 49.13%

(c) 50%

(d) None of these

Column II

(a) 11.1 g CaCl2 and 29.25g

(p) [Ca2+] = 0.8 M

of NaCl are diluted

[Na † ] = 1.2 M

with water to 100 mL

[Cl 4 ] = 2.8 M

(b) 3.0 L of 4.0 M NaCl and

(q) [Ca2+] = 0.001 M

4.0 L of 2.0 M CaCl2 are

[Na † ] = 0.005 M

combined and diluted

[Cl4 ] = 0.007 M

[Na † ] = 1.8 M

4.0 M CaCl2

[Cl4 ] = 5.0 M

(d) None

BrO34  5Br 4 o Br2  3H2 O

(a) 6.0 × 10–4

(b) 1.2 × 10–4

(c) 9.0 × 10–3

(d) 1.8 × 10–3

60. (S) 1 g alloy of Cu and Zn reacted with excess of dil. H2SO4 to give H 2 gas which occupies 60 ml at STP. The percentage of Zn in the alloy (Given only Zn reacts with H2SO4) (a) 17%

(b) 34%

(c) 83%

(d) 40%

61. (S) A solution of NaOH is prepared by dissolving 4.0 g of NaOH in 1 L of water. Calculate the volume of the HCl gas at STP that will neutralize 50 mL of this solution. (a) 224 mL

(b) 56 mL

(c) 112 mL

(d) 448 mL

62. (M) 11.2 L of a gas at STP weighs 14 g. The gas could be :

(r) [Ca2+] = 1.6 M

is added to 200 mL of

(c) 0.6 g

solution that contains excess of H† ions, the moles of Br2 formed are

to 10.0 L (c) 3.0 L of 3.0 M NaCl

(b) 0.4 g

Molar Volume of Gas based Calculations

57. (X) Match the solution mixtures given in column I with the concentrations given in column II. Column I

(a) 0.2 g

1f 50 mL 0.1 M BrO34 is mixed with 30 mL of 0.5 M Br 4

56. (S) 11.4 gm of a mixture of butene, C4H8 and butane C4H10, was burned in excess oxygen. 35.2 gm of CO2 and 16.2 gm of H2O were obtained. Calculate the percentage by mass of butane in original mixture. (a) 50.87%

(s) [Ca2+] = 1.2 M

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(a) N2

(b) CO

(c) NO2

(d) N2O

SOME BASIC CONCEPTS OF CHEMISTRY

Empirical Formula

Laws of Chemical Combination

63. (M) An oxide of nitrogen has 30.43% nitrogen (At. wt. of N=14) and its one molecule weight 1.527 × 10–22 g. Which of the following statement regarding the oxide is (are) true ?

69. (S) Two elements X and Y have atomic weights of 14 and 16. They form a series of compounds A, B, C D and E in which for the same amount of element X, Y is present in the ratio 1 : 2 : 3 : 4 : 5. If the compound A has 28 parts by weight of X and 16 parts by weight of Y, then the compound C will have 28 parts by weight of X and

(a) Its empirical formula is N2O (b) Its empirical formula is NO2.

(a) 32 parts by weight of Y

(c) Its molecular formula is N2O4.

(b) 48 parts by weight of Y

(d) Its molecular formula is N4O2.

(c) 64 parts by weight of Y

Comprehension

(d) 80 parts by weight of Y

A crystalline hydrated salt on being rendered anhydrous loses 45.6% of its weight. The percentage composition of anhydrous salt is : Al = 10.5%, K = 15.1 %, S = 24.8% and oxygen = 49.6% Answer the following four questions based on these information. [Molar masses are : Al = 27, K = 39, S = 32]

70. (S) One part of an element A combines with two parts of B (another element). Six parts of element C combine with four parts of element B. If A and C combine together, the ratio of their masses will be governed by : (a) law of definite proportions (b) law of multiple proportions

64. (C) What is the empirical formula of the salt ? (a) K 2 AlS2O7

(b) K 2 Al2S2O7

(c) KAlS2O8

(d) K 3AlS2O12

(c) law of reciprocal proportions (d) law of conservation of mass 71. (S) Zinc sulphate contains 22.65% Zn and 43.9% H2O. If the law of constant proportions is true, then the mass of zinc required to give 40g crystals will be :

65. (C) What is the empirical formula of the hydrated salt ? (a) K 2 AlS2O7 .10H 2 O

(b) K 2 Al2S2O7 .16H 2 O

(c) K 3AlS2O12 .8H 2O

(d) KAlS2O8 .12H 2O

66. (C) If 50 g of the above hydrated salt is dissolved in 150 gram of water, molality of the resulting solution will be (a) 0.7

(b) 0.6

(c) 0.5

(d) 0.4

(c) 16.98%

(d) 18.49 %

(b) Na2CO3.7H2O

(c) Na2CO3.5H2O

(d) Na2CO3.3H2O

(d) 906 g

(c) constant proportions (d) reciporcal proportions 73. (S) Potassium combines with two isotopes of chlorine (35Cl and 37Cl) respectively to form two samples of KCl. Their formation follows the law of :

68. (S) When a hydrate of Na2CO3 is heated until all the water is removed, it loses 543 per cent of its mass. The formula of the hydrate is (a) Na2CO3.10H2O

(c) 0.906 g

(a) conservation of mass(b) multiple proportions

67. (S) The percentage of Fe in Fe in Fe0.93O1.00 is (b) 84.2%

(b) 9.06 g

72. (S) 3 g of a hydrocarbon on combustion in excess of oxygen produces 8.8g of CO2 and 5.4 g of H 2O. The data illustrates the law of :

3+

(a) 15.0%

(a) 90.6 g

(a) constant proportions (b) multiple proportions (c) reciprocal proportions (d) none of these.

C-35

SOME BASIC CONCEPTS OF CHEMISTRY

Principle of Atom Conservation

80. (A) Assertion : 1mole of H2SO4 is neutralised by 2 moles of NaOH but 1 equivalent of H2SO4 is neutralised by 1

74. (S) 2.76 g of silver carbonate on being strongly heated yields a residue weighing

75. (I)

equivalent of NaOH.

(a) 2.16 g

(b) 2.48 g

Reason : Equivalent weight of H2SO4 is half of its

(c) 2.32 g

(d) 2.64 g

molecular weight while equivalent weight of NaOH is 40.

Igniting MnO2 in air converts it quantitatively to Mn3O4. A sample of pyrolusite is of the following composition : MnO2 = 80%, SiO2 and other inert constituents = 15%, and rest bearing H2O. The sample is ignited to constant weight. What is the percent of Mn in the ignited sample ?

(NH 4 )2 SO 4 Fe 2 (SO 4 )3 . 24H 2O can be made from the

77. (I)

(c) 0.33 × 10–4 mol

(d) 2 × 10–4 mol

(c) C

(d) D

Reason : 1/2 mole H2 has produced when 1 mole of H+(aq) accepted 1 mole of e–.

sample of Fe containing 0.0056 g of it ? (b) 0.5 × 10–4 mol

(b) B

81. (A) Assertion : Equivalent volume of H2 is 11.2 L at 1 atm and 273 K.

76. (S) How many moles of ferric alum

(a) 10–4 mol

(a) A

(a) A

(b) B

(c) C

(d) D

82. (A) Assertion (A) : The equivalent mass of an element is variable.

A sample of a mixture of CaCl2 and NaCl weighing 4.22g

Reason (R) : It depends on the valency of the element.

was treated to precipitate all the Ca as CaCO3, which was then heated and quantitatively converted to 0.959g of CaO. Calculate the percentage of CaCl2 in the mixture.

(a) A

(b) B

(c) C

(d) D

83. (S) N2 + 3H2 o 2NH3

(Ca = 40, O = 16, C = 12 and Cl = 35.5)

Molecular weight of NH 3 and N 2 are x 1 and x 2 , respectively. Their equivalent weights are y1 and y2, respectively. Then (y1 – y2) is

Equivalent Concept 78. (S) A metal oxide has the formula Z2O3. It can be reduced by hydrogen to give free metal and water. 0.16 gm of the metal oxide requires 6 mg of hydrogen for complete reduction. The atomic weight of the metal is : (a) 27.9

(b) 159.6

(c) 79.8

(d) 55.8

§ 2x1  x 2 · (a) ¨ ¸ 6 © ¹

(b) (x1 – x2)

(c) (3x1 – x2)

(d) (x1 – 3x2)

84. (S) The vapour density of a chloride of an element is 39.5. The Ew of the elements is 3.82. The atomic weight of the element is

79. (M) For the reaction H3PO4 + Ca(OH)2 o CaHPO4 + 2 H2O

(a) 15.28

(b) 7.64

1 mol

(c) 3.82

(d) 11.46

1 mol

85. (M) Which of the following statements regarding the compound AxBy is/are correct ?

Which are true statements? (a) Equivalent weight of H3PO4 is 49

(a) 1 mole of AxBy contains 1 mole of A and 1 mole B

(b) Resulting mixture is neutralised by 1 mol of KOH (c) CaHPO4 is an acidic salt

(b) 1 equivalent of AxBy contains 1 equivalent of A and 1 equivalent of B

(d) 1 mol of H3PO4 is completely neutralized by 1.5 mol

(c) 1 mole of AxBy contains x moles of A and y moles of B

of Ca(OH)2

(d) equivalent weight of AxBy = equivalent weight of B

C-36

SOME BASIC CONCEPTS OF CHEMISTRY 92. (X) Match the items given in column I with those in column II.

86. (M) Which of the statements are true ? (a) The equivalent weight of Ca3(PO4)2 is Mw/6.

Column I

(b) The equivalent weight of Na3PO4. 12H2O is Mw/3.

(a) 9.8% H2SO4 by weight (density = 1.8g mL )

(d) The equivalent weight of potash alum K2SO4Al2(SO4)3. 24H2O is Mw/8.

Normality

(b) 9.8 % H3PO4 by weight

(q) 1.2 M

(density = 1.2g mL–1) (c) 1.8 NA molecules of

HNO 3 are mixed together and solution made to one

(d) 250 mL of 4N NaOH

litre. The normality of the resulting solution is (a) 0.20 N

(b) 0.10 N

(c) 0.50 N

(d) 0.025 N

(b) 0.0115 N

(c) 0.0125 N

(d) 0.046 N

93. (S) 10 mL of 0.2 N HCl and 30 mL of 0.1 N HCl together exactly neutralises 40 mL of solution of NaOH, which is also exactly neutralised by a solution in water of 0.61 g of an organic acid.What is the equivalent weight of the organic acid ?

(d) 183

(c) % of Mg present in the sample is 12%

normality of the resulting solution will be

(d) 4 N

(c) 122

(b) Mass of Mg sample unreacted is 0.88 gm

mixture was made 1000 ml by adding water. The

(c) 3 N

(b) 91.5

(a) Mass of Mg present in the sample is 0.12 gm

HNO 3 were mixed together and the volume of the

(b) 2 N

(a) 61

94. (M) 1 gm Mg sample is treated with 125 ml 0.1 N HCl and the excess of HCl is neutralised by 50 ml 0.5 N NaOH completely. The correct statement is/are :

89. (S) 50 ml of 10 N H2SO4, 25 ml of 12 N HCl and 40 ml of 5N

(a) 1 N

(d) Mass of impurities present in the sample is 0.88 gm.

–1

90. (S) Which of the following 1 g L solution has the highest

95. (X) Match the Column Column (a) 20 ml (N) HCl reacts

normality ? (b) H2SO4

(c) HCl

(d) HNO3

(b) 10 ml

91. (A) Assertion :- 0.1 M H3PO3 (aq) solution has normality

N HCl reacts 2

with 50 ml

equal to 0.3N when completely reacted with NaOH. Reason : H3PO3 is dibasic acid.

(c) 50 ml

(b) B with 100 ml (d) 100 ml

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N NaOH. 50

N HCl reacts 2

with 50 ml

left = 0 (q) No. of molecules of HCl

N NaOH. 10

N HCl reacts 10

(d) D

Column (p) No. of molecules of HCl

N NaOH. 5

with 50 mL

(a) NaOH

(s) 1.10 m

+ 250 mL of 1.6 M Ca(OH)2

88. (S) 0.115 g of pure sodium metal was dissolved in 500 ml distilled water. The normality of the above solution, whose resulting volume is 400 mL, would be (a) 0.010 N

(r) 1.8 Equivalents

HCl is 500 mL

87. (S) 10 mL of N/2 HCl, 20 mL of N/2 H2SO4 and 30 mL N/3

(c) C

(p) 3.6 N

–1

(c) The equivalent weight of K2SO4 is Mw/2.

(a) A

Column II

N NaOH. 10

left = 6.02 × 1021 (r)

No. of molecules of HCl left = 2.71 × 1022

(s) No. of molecules of HCl left = 1.8 × 1021

SOME BASIC CONCEPTS OF CHEMISTRY

96. (M) An aqueous solution of phosphoric acid (H3PO4) being titrated has molarity equal to 0.25 M. Which of the following could be normality of this solution ? (a) 0.25 N

(b) 0.50 N

(c) 0.75 N

(d) 1.00 N

97. (M) An aqueous solution of 6.3g of a hydrated oxalic acid (H2C2O4.xH2O) is made up to 250 mL. The 40 mL of 0.10 N NaOH was required to completely neutralize 10mL of the above prepared stock solution. Which of the following statements(s) about is (are) correct ? (a) The acid is dehydrate. (b) Equivalent weight of the hydrated acid is 45. (c) Equivalent weight of the anhydrous acid is 45. (d) 20 mL of the same stock would require 40 mL of 0.10 M Ca(OH)2 solution for complete neutralization.

C-38

SOME BASIC CONCEPTS OF CHEMISTRY

ANSWER KEY Advanced Objective Questions 1. (c)

2. (a)

3. (c)

4. (A) o (R) (B) o (P, S), (C) o (Q), (D) o (P),(S)

5. (a)

6. (c)

7. (b)

8. (bc)

9. (ab)

10. (d)

11. (b)

12. (d)

13. (bc)

14. (a)

15. (d)

16. (c)

17. (c)

18. (b)

19. (c)

20. (0080)

21. (0059)

22. (0014)

23. (0060)

24. (a,b)

25. (0008)

26. (abd)

27. (ac)

28. (b)

29. (a)

30. (d)

31. (c)

32. (b)

33. (a)

34. (a,c,d)

35. (b)

36. (c)

37. (a)

38. (c)

39. (d)

40. (a,b)

41. (a)

42. (b)

43. (a)

44. (a)

45. (abcd)

46. (abc)

47. (abd)

48. (a,b,c,d)

49. (d)

50. (a)

51. (d)

52. (0010)

53. (0002)

54. (0005)

55. (d)

56. (a)

57. (a o q), (b o p), (c o r), (d o s)

58. (b)

59. (c)

60. (a)

61. (c)

62. (ab)

63. (bc)

64. (c)

65. (d)

66. (b)

67. (a)

68. (a)

69. (b)

70. (c)

71. (b)

72. (a)

73. (d)

74. (a)

75. 59.36%

76. (b)

77. 45.04%

78. (d)

79. (abcd)

80. (b)

81. (a)

82. (a)

83. (a)

84. (b)

85. (b,c,d)

86. (abcd)

87. (d)

88. (c)

89. (a)

90. (c)

91. (d)

93. (c)

94. (abcd )

92. (a – p, s; b – p, q, s; c – p, r; d – r) 96. (abc)

97. (acd)

Dream on !!

[\]^[\]^

C-39

95. (a – q; b – q; c – s; d – r)

CHEMICAL EQUILIBRIUM

EXERCISE - 4: PREVIOUS YEAR IIT QUESTIONS 1.

6.

The oxidation of SO2 by O2 to SO3 is an exothermic reaction. The yield of SO3 will be maximum if (1981)

For the reversible reaction, N2(g) + 3H2(g)

(a) temperature is increased and pressure is kept constant (b) temperature is reduced and pressure is increased (c) both temperature and pressure are increased

1.44 u 10 5 (a)

(d) both temperature and pressure are reduced 2.

(a) total pressure

(c) (1981)

(b) catalyst

7.

(c) the amounts of H2 and I2 present (d) temperature

(b)

(8.314 u 773)  2

NH3

(0.082 u 773)

1.44 u10 5 2

(d)

(0.082 u 773)  2

When two reactants, A and B are mixed to give products C and D, the reaction quotient Q, at the initial stages of the reaction (2000) (a) is zero

Pure ammonia is placed in a vessel at a temperature where its dissociation constant (D) is appreciable. At equilibrium N2 + 3H2

(b) decreases with time

(c) is independent of time (d) increases with time 8.

(1984)

(a) Kp does not change significantly with pressure

At constant temperature, the equilibrium constant (Kp) for 2NO2 is expressed the decomposition reaction N2O4

(c) concentration of NH3 does not change with pressure

by K = (4x2P) (1 – x2), where P = presssure, x = extent of decomposition. Which one of the following statements is true ? (2001)

(d) concentration of hydrogen is less than that of nitrogen

(a) Kp increases with increase of P

For the reaction ;

(b) Kp increases with increase of x

(b) D does not change with pressure

CO (g) + H2O(g)

CO2 (g) + H2 (g),

(c) Kp increases with decrease of x

at a given temperature, the equilibrium amount of CO2(g) can be increased by (1998)

(d) Kp remains constant with change in P and x 9.

(a) adding a suitable catalyst

(c) decreasing the volume of the container

For the chemical reaction X3Y (g)

the amount of X3Y at equilibrium is affected by

2NO2(g)

At a fixed temperature, the volume of the reaction container is halved. For this change, which of the following statements hold true regarding the equilibrium constant (Kp) and degree of dissociation (D) ?

(d) increasing the amount of CO(g)

3X (g) + Y (g)

Consider the following equilibrium in a closed container N2O4 (g) 2

(b) adding an inert gas

5.

(0.082 u 500)  2 1.44 u 10 5

2HI (g)

The equilibrium constant Kp changes with

4.

1.44 u10 5

For the reaction : H2(g) + I2(g)

3.

2NH3(g)

at 500ºC, the value of Kp is 1.44 × 10–5 when partial pressure is measured in atmospheres. The corresponding value of Kc, with concentration in mole litre–1 is (2000)

(a) Neither Kp nor D changes

(1999)

(b) Both Kp and D changes

(a) temperature and pressure (b) temperature only

(c) Kp changes but D does not change

(c) pressure only

(d) Kp does not change but D changes

(d) temperature, pressure and catalyst

D-1

CHEMICAL EQUILIBRIUM 10.

N2 + 3H3

14.

2NH3

The forward reaction at constant temperature is favoured by (1991)

(a) The equilibrium will shift to forward direction because according to IInd law of thermodynamics the entropy must increase in the direction of spontaneous reaction

(a) introducing chlorine gas at constant volume (b) introducing an inert gas at constant pressure (c) increasing the volume of the container

2G NH3

(d) introducing PCl5 at constant volume

where G is Gibbs free energy per mole of the gaseous species measured at that partial pressure. The condition of equilibrium is unaffected by the use of catalyst, which increases the rate of both the forward and backward directions to the same extent

Fill in the Blanks 15.

16.

A ten-fold increase in pressure on the reaction, N2(g) + 2NH3 (g) at equilibrium, results in ................ 3H2 (g) in Kp. (1996)

17.

For a gaseous reaction 2B  o A, the equilibrium constant Kp is ........... to/than Kc.

(d) Catalyst will not alter the rate of either of the reaction.

Objective Questions (One or more than one correct option) For the gas phase reaction C2H4 + H2

True/False

C2H6 ('H = – 32.7 kcal)

carried out in a vessel, the equilibrium concentration of C2H4 can be increased by (1984) (a) Increasing the temperature

18.

When a liquid and its vapour are at equilibrium and the pressure is suddenly decreased, cooling occurs. (1984)

19.

If equilibrium constant for the reaction, A2 + B2

(b) decreasing the pressure

2AB, is K, then for the backward reaction AB

(c) removing some H2

When NaNO3 is heated in a closed vessel, oxygen is liberated and NaNO2 is left behind. At equilibrium (1986) (a) addition of NaNO2 favours reverse reaction

(1984)

20.

Catalyst makes a reaction more exothermic.

(1987)

21.

The rate of an exothermic reaction increases with increasing temperature. (1993)

(b) addition of NaNO3 favours forward reaction

Subjective Questions

(c) increasing temperature favours forward reaction

22.

(d) increasing pressure favours reverse reaction SO2 (g) + Cl2(g) is The equilibrium SO2Cl2(g) attained at 25ºC in a closed container and an inert gas, helium is introduced. Which of the following statements are correct ? (1989)

1 2

1 1 A 2  B2 , the equilibrium constant is . 2 K

(d) adding some C2H6

13.

For a given reversible reaction at a fixed temperature, equilibrium constant Kp and Kc are are related by .............. (1994)

(c) The catalyst will increase the rate of forward reaction by E

12.

PCl3(g) + Cl2(g)

Which is correct statement if N2 is added at equilibrium condition ? (2006)

(b) The condition for equilibrium is G N 2  3G H 2

11.

For the reaction, PCl5 (g)

One mole of nitrogen is mixed with three moles of hydrogen in a four litre container. If 0.25 per cent of nitrogen is converted to ammonia by the following reaction N2(g) + 3H2 (g)

2NH3 (g), then

calculate the equilibrium constant, Kc in concentration units. What will be the value of Kc for the following equilibrium ?

(a) Concentration of SO2, Cl2 and SO2Cl2 change (b) More chlorine is formed

1 3 N2 (g) + H 2 (g) 2 2

(c) Concentration of SO2 is reduced (d) All the above are incorrect

D-2

NH3 (g)

(1981)

CHEMICAL EQUILIBRIUM 23.

PCl5(g) 24.

29.

One mole of Cl2 and 3 moles of PCl5 are placed in a 100 litre vessel heated at 227ºC. The equilibrium pressure is 2.05 atmosphere. Assuming ideal behaviour, calculate the degree of dissociation for PCl5 and Kp for the reaction PCl3 (g) + Cl2 (g)

Hydrogen is introduced until the total pressure of the system is 8.5 atm at equilibrium and 0.08 mole of methanol

The equilibrium constant of the reaction

is formed. Calculate (i) Kp and Kc and (ii) the final pressure if the same amount of CO and H2 as before are used, but with no catalyst so that the reaction does not take place. (1993)

At a certain temperature, equilibrium constant (Kc) is 16 for the reaction ;

30.

The progress of reaction, A

SO 2 (g) + NO 2 (g)

CH3OH(g)

CO (g) + 2H2(g)

(1984)

2AB (g) at 100ºC is 50. If a one litre A2(g) + B2(g) flask containing one mole of A2 is connected to a two litre flask containing two moles of B2, how many mole of AB will be formed at 373ºC ? (1985) 25.

0.15 mole of CO taken in a 2.5 L flask is maintained at 750 K along with a catalyst so that the following reaction can take place :

SO3(g) + NO(g)

nB

with time, is represented in fig. given below. Determine

If we take one mole of each of all the four gases in a one litre container, what would be the equilibrium concentration of NO and NO2 ? (1987) 26.

N2O4 is 25% dissociated at 37ºC and one atmosphere pressure. Calculate (i) K P and (ii) the percentage dissociation at 0.1 atmosphere and 37ºC. (1988)

27.

The equilibrium constant Kp of the reaction, 2SO2(g) + O2 (g)

2SO3 (g) (i) the value of n

is 900 atm at 800 K. A mixture containing SO3 and O2 having initial pressure of 1 and 2 atm respectively is heated at constant volume to equilibrate. Calculate the partial pressure of each gas at 800 K. (1989) 28.

(ii) the equilibrium constant, K and (iii) the initial rate of conversion of A.

For the reaction CO(g) + 2H2 (g)

31.

The degree of dissociation is 0.4 at 400 K and 1.0 atm for PCl3 + Cl2. Assuming the gaseous reaction PCl5 ideal behaviour of all the gases, calculate the density of equilibrium mixture at 400 K and 1.0 atm (relative atomic mass of P = 31.0 and Cl = 35.5) (1998)

32.

When 3.06 g of solid NH4SH is introduced into a two litre evacuated flask at 27ºC, 30% of the solid decomposes into gaseous ammonia and hydrogen sulphide.

CH3OH(g)

hydrogen gas is introduced into a five litre flask at 327ºC, containing 0.2 mole of CO(g) and a catalyst, until the pressure is 4.92 atm. At this point 0.1 mole of CH3OH(g) is formed. Calculate the equilibrium constant, Kp and Kc. (1990)

(1994)

(i) Calculate Kc and Kp for the reaction at 27ºC. (ii) What would happen to the equilibrium when more solid NH4SH is introduced into the flask ? (1999)

D-3

CHEMICAL EQUILIBRIUM

ANSWER KEY Previous Year IIT Questions 1. (b)

3. (a)

4. (d)

5. (a)

11. (a,b,c,d) 12. (c,d)

13. (d)

14. (bcd)

15. Kp = Kc (RT)'n

16. no change

17. smaller

18. (T)

20. (F)

21. (F)

22. 0.468 L2 mol–2, 0.68

23. 0.33

26. 0.266 atm; 63%

27.

2. (d)

19. (F)

85 175 atm , atm 87 87 –5

28. Kp = 0.11 atm–2 , Kc = 277.77 M–2

6. (d)

29. 220

–2

32. 8.1 × 10 , 4.84 × 10 , (ii) No effect

Dream on !!

[\]^[\]^

D-4

7. (d)

8. (d)

24. 1.76

30. (i) n = 2, (ii) 1.2

9. (d)

31. 4.54g/L

10. (b)

IONIC EQUILIBRIUM

EXERCISE - 4: PREVIOUS YEAR IIT QUESTIONS 1.

2.

The pH of 10–8 M solution of HCl in water is (b) –8

(c) between 7 and 8

(d) between 6 and 7 +

(d) KNO3 (aq) + NaCl (aq) = KCl (aq) + NaNO3 (aq) 9.

(d) 10

Of the given anions, the strongest base is

(c) ClO

(c) 2Na (s) + 2H2O(l) = 2 NaOH (aq) + H2(g)

–8

 3

(d) ClO

(d) phenophthalein (8 to 9.6)

 4

10.

An acidic buffer solution can be prepared by mixing the solution of (1981) 11.

(b) ammonium chloride and ammonium hydroxide

(a) BF3

(b) AlCl3

(c) BeCl2

(d) SnCl4

The conjugate acid of NH 2 is (a) NH3

(b) NH2OH

(d) sodium chloride and sodium hydroxide

(c) NH 4

(d) N2H4

(1985)

(1985)

–10

The precipitate of CaF2 (Ksp = 1.7 × 10 ) is obtained, when equal volumes of the following are mixed (1982) (a) 10–4 M Ca2+ + 10–4 M F–

The pKa of acetyl salicylic acid (aspirin) is 3.5. The pH of gastric juice in human stomach is about 2–3 and the pH in the small intestine is about 8. Aspirin will be (1988)

(b) 10–2 M Ca2+ + 10–3 M F–

(a) unionised in the small intestine and in the stomach

(c) 10–5 M Ca2+ + 10–3 M F–

(b) completely ionised in the small intestine in the small intestine

2+

–5

12.

–

(d) 10 M Ca + 10 M F

(c) ionised in the stomach and almost unionised in the small intestine

A certain buffer solution contains equal concentration of X– and HX. The Kb for X– is 10–10. The pH of the buffer is

(d) ionised in the small intestine and almost unionised in the stomach

(1984)

7.

The compound that is not a Lewis acid is

(c) sulphuric acid and sodium sulphate

–3

6.

(b) methyl red (5 to 6)

(c) bromothymol blue (6 to 7.5)

(a) solution of acetate and acetic acid

5.

The best indicator for detection of end point in titration of a weak acid and a strong base is (1985) (a) methyl orange (3 to 4)

(1981)

(b) ClO 2

(a) ClO–

(1985)

(b) AgNO3 (aq) + HCl (aq) + AgCl (s) + HNO3 (aq)

–1

(b) 10–12

(c) 10

4.

–6

At 90ºC, pure water has [H3O ] as 10 mol L . What is the value of Kw at 90ºC ? (1981) –14

An example of a reversible reaction is

(a) Pb(NO3)2 (aq) + 2NaI (aq) = PbI2 (s) + 2NaNO3 (aq)

(a) 8

(a) 10–6

3.

8.

(1981)

(a) 4

(b) 7

(c) 10

(d) 14

13.

precipitation of AgCl (Ksp = 1.8 × 10–10) will occur only with (1988)

A certain weak acid has a dissociation constant of 1.0 × 10–4. The equilibrium constant for its reaction with a strong base is (1984) (a) 1.0 × 10–4

(b) 1.0 × 10–10

(c) 1.0 × 1010

(d) 1.0 × 10–14

When equal volumes of the following solutions are mixed,

(a) 10–4 M (Ag+) and 10–4 M (Cl–) (b) 10–5 M (Ag+) and 10–5 M (Cl–) (c) 10–6 M (Ag+) and 10–6 M (Cl–) (d) 10–10 M (Ag+) and 10–10 M (Cl–)

D-5

IONIC EQUILIBRIUM 14.

15.

16.

Which of the following is the strongest acid ? (a) ClO3 (OH)

(b) ClO2 (OH)

(c) SO(OH)2

(d) SO2 (OH)2

(1989)

21.

Hg2+ is treated with 10–16 M sulphide ion. If Ksp of MnS, FeS, ZnS and HgS are 10 –15, 10 –23 , 10 –20 and 10 –54 respectively, which one will precipitate first ? (2003)

Amongst the following hydroxides, the one which has the lowest value of Ksp at ordinary temperature (about 25ºC) is (1990) (a) Mg(OH)2

(b) Ca(OH)2

(c) Ba(OH)2

(d) Be(OH)2

22.

Which of the following solutions will have pH close to 1.0 ? (1992) (a) 100 Ml of (M/10) HCl + 100 mL of (M/10) NaOH

17.

A solution which is 10–3 M each in Mn2+, Fe2+, Zn2+ and

(a) FeS

(b) MgS

(c) HgS

(d) ZnS

HX is a weak acid (Ka = 10–5). It forms a salt NaX (0.1 M) on reacting with caustic soda. The degree of hydrolysis of NaX is (2004) (a) 0.01%

(b) 0.0001%

(c) 0.1%

(d) 0.5%

(b) 55 mL of (M/10) HCl + 45 mL of (M/10) NaOH

The pH of 0.1 M solution of the following salts increases in the order (2004)

(c) 10 mL of (M/10) HCl + 90 mL of (M/10) NaOH

(a) NaCl < NH4Cl < NaCN < HCl

(d) 75 mL of (M/5) HCl + 25 mL of (M/5) NaOH

(b) HCl < NH4Cl < NaCl < NaCN

The solubility of A2X3 is y mol dm–3. Its solubility product is (1997)

(c) NaCN < NH4Cl < NaCl < HCl

23.

(d) HCl < NaCl < NaCN < NH4Cl 4

4

(a) 6y

(b) 64y

24. 5

5

(c) 36y 18.

(d) 108y

The pH of 0.1 M solution of the following salts increase in the order (1999)

(a) 0.0001%

(b) 0.01%

(a) NaCl < NH4Cl < NaCN < HCl

(c) 0.1%

(d) 0.15%

(b) HCl < NH4Cl < NaCl < NaCN

25.

(c) NaCN < NH4Cl < NaCl < HCl

19.

A 0.004 M solution of Na2SO4 is isotonic with 0.010 M solution of glucose at same temperature. The percentage dissociation of Na2SO4 is (2004)

(d) HCl < NaCl < NaCN < NH4Cl

(a) 25%

(b) 50%

For a sparingly soluble salt ApBq, the relationship of its solubility product (Ls) with its solubility (S) is (2001)

(c) 75%

(d) 85%

p+q

(a) Ls = S

p

. p .q

q

(c) Ls = Spq . pp.qq 20.

A weak acid HX has the dissociation constant 1 × 10–5M. It forms a salt NaX on reaction with alkali. The precentage hydrolysis of 0.1 M solution of NaX ? (2004)

p+q

(b) Ls = S

q

. p .q

26.

p

(d) Ls = Spq. (p.q)(p+q)

Identify the correct order of solubility of Na2S, CuS and ZnS in aqueous medium (2002) (a) CuS > ZnS > Na2S

(b) ZnS > Na2S > CuS

(c) Na2S > CuS > ZnS

(d) Na2S > ZnS > CuS

D-6

CH3NH2 (0.1 mole, Kb = 5 × 10–4) is added to 0.08 moles of HCl and the solution is diluted to one litre, resulting hydrogen ion concentration is (2005) (a) 1.6 × 10–11

(b) 8 × 10–11

(c) 5 × 10–5

(d) 8 × 10–2

IONIC EQUILIBRIUM 27.

Ag+ + NH3

[Ag(NH3)]+ ; k1 = 3.5 × 10–3

31.

[Ag (NH3)2]+ ;

[Ag(NH3)]+ + NH3

2 M weak monoacidic base (Kb = 1 × 10–12 at 5

25ºC) is titrated with

k2 = 1.7 × 10–3 then the formation constant of [Ag(NH3)2]+ is

28.

2.5 mL of

2 M HCl in water at 25ºC. The 15

concentration of H+ at equivalence point is (Kw = 1 × 10–14 at 25ºC) (2008)

(2006)

(a) 6.08 × 10–6

(b) 6.08 × 106

(a) 3.7 × 10–13 M

(b) 3.2 × 10–7 M

(c) 6.08 × 10–9

(d) None of these

(c) 3.2 × 10–2 M

(d) 2.7 × 10–2 M

Objective Questions (One or more than one correct option)

The species present in solution when CO2 is dissolved in water are (2006)

32.  3

(a) CO2, H2CO3, HCO , CO

Which of the following statement (s) is (are) correct ?

2 3

(1998) –8

(a) The pH of 1.0 × 10 M solution of HCl is 8

(b) H 2CO3 , CO32 (c) HCO3 , CO32 29.

N2 + 3H3

(b) The conjugate base of H 2 PO 4 is HPO 24 (d) CO 2 , H 2CO3

(c) Autoprotolysis constant of water increases with temperature

2NH3

(d) When a solution of a weak monoprotic acid is titrated against a strong base, at half-neutralisation point pH =

Which is correct statement if N2 is added at equilibrium condition ? (2006)

§1· ¨ ¸ pK a ©2¹

(a) The equilibrium will shift to forward direction because according to IInd law of thermodynamics the entropy must increase in the direction of spontaneous reaction (b) The condition for equilibrium is G N 2  3G H 2

33.

(1999) (a) sodium acetate and acetic acid in water

2G NH3

(b) sodium acetate and HCl in water

where G is Gibbs free energy per mole of the gaseous species measured at that partial pressure. The condition of equilibrium is unaffected by the use of catalyst, which increases the rate of both the forward and backward directions to the same extent

30.

A buffer solution can be prepared from a mixture of

(c) ammonia and ammonium chloride in water (d) ammonia and sodium hydroxide in water

Fill in the Blanks The conjugate base of HSO4 in aqueous solution is

(c) The catalyst will increase the rate of forward reaction by E

34.

(d) Catalyst will not alter the rate of either of the reaction.

35.

Solubility product constant (Ksp) of salts of types MX, MX2 and M3X at temperature ‘T’ are 4.0 × 10–8, 3.2 × 10–14

An element which can exist as a positive ion in acidic solution and also as a negative ion in basic solution is said to be ............... (1984)

36.

Silver chloride is sparingly soluble in water because its lattice energy is greater than ................ energy. (1987)

37.

o I3 , the Lewis acid is ............ In the reaction I– + I2 

.................

and 2.7 × 10–15, respectively. Solubilities (mol, dm–3) of the salts at temperature ‘T’ are in the order (2008) (a) MX > MX2 > M3X

(b) M3X > MX2 > MX

(c) MX2 > M3X > MX

(d) MX > M3X > MX2

(1982)

(1997) 38.

D-7

(CH3OH2)+ is ............... acidic than ( CH3 NH3 ).

(1997)

IONIC EQUILIBRIUM

True/False

is solution ? The dissociation constant of formic acid is

39.

Aluminium chloride (AlCl3) is a Lewis acid because it can donate electrons. (1982)

40.

Solubility of sodium hydroxide increases with increase in temperature. (1985)

41.

2.4 × 10–4 and the degree of dissociation of sodium formate is 0.75. (1985) 48.

Calculate its solubility (in g/litre) in 0.02 M Mg(NO3)2 solution. (1986)

The following species are in increasing order of their acidic property : ZnO, Na2O2, P2O5, MgO (1985)

49.

Subjective Questions 42.

How many moles of sodium propionate should be added to 1 L of an aqueous solution containing 0.020 moles of propionic acid to obtain a buffer solution of pH 4.75 ? What will be pH if 0.010 moles of HCl are dissolved in the above buffer solution ? Compare the last pH value with the pH of 0.010 M HCl solution. Dissociation constant of propionic acid, Ka at 25ºC is 1.34 × 10–5.

43.

44.

45.

(ii) 0.1 M each of acetic acid and acetate ion ? Assume the total volume is one litre. Ka for acetic acid = 1.8 × 10–5. 50.

(1981)

20 mL of 0.2 M sodium hydroxide is added to 50 mL of 0.2 M acetic acid solution to give 70 mL of the solution. What is the pH of this solution ? Calculate the additional volume of 0.2 M NaOH required to make the pH of the solution 4.74. (Ionisation constant of CH3COOH = 1.8 × 10–5). (1982)

(1987)

Freshly precipitated aluminium and magnesium hydroxides are stirred vigorously in a buffer solution containing 0.25 mol/L of NH4Cl and 0.05 M of ammonium hydroxide calculate the concentration of alminimum and magnesium ions in solution Kb[NH4OH] = 1.8 × 10

–5

–12

Ksp[Mg(OH)2] = 8.9 × 10

Ksp[Al(OH)3] = 6 × 10

Give reason for the statement that “the pH of an aqueous solution of sodium acetate is more than seven”. (1982)

–32

(1989)

After making the necessary approximations, calculate :

What is the pH of 1.0 M solution of acetic acid ? To what volume must one litre of this solution be diluted so that the pH of the resulting solution will be twice the original value ?

(i) pH

Given : Ka = 1.8 × 10–5.

51.

The dissociation constant of a weak acid HA is 4.9 × 10–8.

52.

(1990)

The solubility product of Ag2C2O4 at 25ºC is 1.29 × 10

–11

3 –3

mol L . A solution of K2C2O4 containing 0.1520 mole in 500 mL water is shaken at 25ºC with excess of Ag2CO3 till the following equilibrium is reached

A solution contains a mixture of Ag+ (0.10 M) and Hg2+ (0.10 M) which are to be separated by selective precipitation. Calculate the maximum concentration of iodide ion an which one of them gets precipitated almost completely. What percentage of that metal ion is precipitated ? (1984)

Ag2CO3 + K2C2O4

Ag2C2O4 + K2CO3

At equilibrium, the solution contains 0.0358 mole of K2CO3. Assuming the degree of dissociation of K2C2O4 and K2CO3 to be equal, calculate the solubility product of Ag2CO3. (1991)

Ksp : AgI = 8.5 × 10–17, HgI2 = 2.5 × 10–26 47.

What is the pH of the solution when 0.2 mole of hydrochloric acid is added to one litre of a solution containing (i) 1 M each of acetic acid and acetate ion ?

(ii) OH– concentration in a decimolar solution of the acid. Water has a pH of 7. (1983) 46.

The solubility of Mg(OH)2 in pure water is 9.57 × 10–3 g/litre.

53.

The concentration of hydrogen ions in a 0.20 M solution of formic acid is 6.4 × 10–3 mol/L. To this solution, sodium formate is added so as to adjust the concentration of sodium formate to one mole per litre. What will be the pH of this solution ? The dissociation constant of formic acid

D-8

A 40 mL solution of a weak base, BOH is titrated with 0.1N HCl solution. The pH of the solution is found to be 10.04 and 9.14 after the addition of 5.0 mL and 20.0 mL of the acid respectively. Find out the dissociation constant of the base. (1991)

IONIC EQUILIBRIUM 54.

60.

The solubility product (Ksp) of Ca(OH) 2 at 25ºC is

What is the pH of a 0.50 M aqueous NaCN solution ?

–5

4.42 × 10 . A 500 mL of saturated solution of Ca(OH)2 is mixed with equal volume of 0.4 M NaOH. How much Ca(OH)2 in milligrams is precipitated ? (1992) 55.

61.

5 M NaHCO3 solution should be mixed with a 10 mL sample

to form NH3 and H2O at 25ºC is 3.4 × 1010 L/mol/s. Calculate

of blood which in 2 M in H2CO3, in order to maintain a pH

the rate constant per proton transfer from wate to NH3.

(1993)

(1996)

An aqueous solution of a metal bromide MBr2 (0.05 M) is saturated with H2S. What is the minimum pH at which MS will precipitate ?

and K1 = 10–7 and K2 = 1.3 × 10–13, for H2S.

62.

A sample of AgCl was treated with 5.00 mL of 1.5 M Na2CO3 solution to give Ag2CO3. The remaining solution contained 0.0026 g of Cl– ions per litre. Calculate the solubility product of AgCl. (1997)

(1993)

[Ksp (Ag2CO3) = 8.2 × 10–12]

For the reaction,

63.

[Ag(CN)2]–

Ag+ + 2CN–

An acid type indicator, HIn differs in colour from its conjugate base (In–). The human eye is sensitive to colour differences only when the ratio [In–]/[HIn] is greater than 10 or smaller than 0.1. What should be the minimum change in the pH of the solution to observe a complete colour

The equilibrium constant, at 25ºC, is 4.0 × 10–19. Calculate the silver ion concentration in a solution which was originally 0.10 M in KCN and 0.03 M in AgNO3. (1994) 58.

The ionisation constant of NH 4 in water is 5.6 × 10–10 at 25ºC. The rate constant for the reaction of NH 4 and OH–

Ksp for MS = 6.0 × 10–21, conc. of saturated H2S = 0.1 M

57.

change ? (Ka = 1.0 × 10–5)

(1997)

The progress of reaction, A

64.

nB

The solubility of Pb(OH)2 in water is 6.7 × 10–6M. Calculate the solubility of Pb(OH)2 in a buffer solution of pH = 8.

with time, is represented in fig. given below. Determine

(1999)

(i) the value of n

(iii) the initial rate of conversion of A.

65.

The average concentration of SO2 in the atmosphere over a city on a certain day is 10 ppm, when the average temperature is 298 K. Given that the solubility of SO2 in water at 298 K is 1.36653 mol/L and pKa of H2SO3 is 1.92, estimate the pH of rain on that day. (2000)

66.

500 mL of 0.2 M aqueous solution of acetic acid is mixed with 500 mL of 0.2 M HCl at 25ºC. (i)

(ii) the equilibrium constant, K and

59.

(1996)

The pH of blood stream is maintained by a proper balance of H2CO3 and NaHCO3 concentrations. What volume of

of 7.4 ? (Ka for H2CO3 in blood is 7.8 × 10–7) 56.

(pKb of CN– = 4.70)

Calculate the degree of dissociation of acetic acid in the resulting solution and pH of the solution.

(ii) If 6 g of NaOH is added to the above solution, determine the final pH. [Assume there is no change in

(1994)

Calculate the pH of an aqueous solution of 1.0 M ammonium formate assuming complete dissciation

volume on mixing : Ka of acetic acid is 1.75 × 10–5 mol L–1.] (2002)

(pKa of formic acid = 3.8 and pKb of ammonia = 4.8) (1995)

D-9

IONIC EQUILIBRIUM 67.

(a) In the following equilibrium N2O4 (g)

2NO2 (g)

when 5 moles of each are taken, the temperature is kept at 298 K the total pressure was found to be 20 bar. Given that

'G of (N 2 O4 ) 100kJ 'G of (NO2 ) 50kJ (i) Fina 'G of the reaction (ii) The direction of the reaction in which the equilibrium shifts (b) A graph is plotted for a real gas which follows van der Waals’ equation with pVm taken on Y-axis and p on X-axis. Find the intercept of the line where Vm is molar volume. (2004) 68.

0.1 M of HA is titrated with 0.1 M NaOH, calculate the pH –6

at end point, Given, Ka (HA) = 5 × 10 and D < < 1. (2004) 69.

The dissociation constant of a substituted benzoic acid at –4

25ºC is 1.0 × 10 . The pH of 0.01 M solution of its sodium salt is (2009)

D-10

IONIC EQUILIBRIUM

Answer Key Previous Year IIT Questions 1. (d)

2. (b)

3. (a)

4. (a)

5. (b)

6. (a)

7. (c)

8. (d)

9. (d)

10. (c)

11. (a)

12. (d)

13. (a)

14. (a)

15. (d)

16. (d)

17. (d)

18. (b)

19. (a)

20. (d)

21. (c)

22. (a)

23. (b)

24. (b)

25. (c)

26. (b)

27. (a)

28. (a)

29. (b)

30. (d)

31. (d)

32. (bc)

33. (abc)

34. SO2

35. amphoteric

36. hydration

37. I2

38. more

39. (F)

40. (F)

41. (F)

45. (i) 4.15, (ii) 1.43 × 10–10m

46. 5 × 10–13M, 99.83

47. 4.20

–4

48. 8.7 × 10 g/lit –11

49. 4.5686 ; 1

52. 9.675 × 10

53. Kb = 1.8 × 10

58. (i) n = 2, (ii) 1.2

59. 6.50

65. 4.86

–15

50. 1.28 × 10 –5

m, 0.68 M

54. 747.4 mg 60. 11.5 –4

66. (i) 1.75 × 10

(ii) 4.75

55. 80mL 5

61. 6.12 × 10

4

51. 2.3724 ; 2.78 × 10 lit. 56. 1 62. 2 × 10

57. 7.50 × 10 –8

67. (a) (i) O, (ii) Forward direction

69. (8)

Dream on !!

[\]^[\]^ D-11

–18

+

M Ag

64. 1.2 × 10–3 M 68. (9)

SOME BASIC CONCEPTS OF CHEMISTRY

EXERCISE - 4: PREVIOUS YEAR IIT QUESTIONS 1.

2.

Dissolving 120g of urea (mol. wt. 60) in 1000g of water gave a solution of density 1.15 g/mL. The molarity of the solution is (2011) (a) 1.78 M

(b) 2.00 M

(c) 2.05 M

(d) 2.22 M

Given that the abundances of isotopes

54

Fe,

56

Fe and

6.

7.

6.3g of oxalic acid dihydrate have been dissolved in water to obtain a 250 mL solution. How much volume of 0.1 N NaOH would be required to neutralise 10 mL of this solutions ? (2001) (a) 40 mL

(b) 20 mL

(c) 10 mL

(d) 4 mL

The normality of 0.3 M phosphorous acid (H3PO3) is

Fe are 5%, 90% and 5%, respectively, the atomic mass 57 of Fe is (2009) (a) 55.85

(b) 55.95

(c) 55.75

(d) 56.05

(1999)

8. 3.

Mixture X = 0.02 mole of [Co(NH3)5SO4] Br and 0.02

(a) 0.1

(b) 0.9

(c) 0.3

(d) 0.6

The weight of 1 × 1022 molecules of CuSO4. 5H2O is (1991)

mole of [Co(NH3)5Br] SO4 was prepared in 2 L solution. 1 L of mixture X + excess of AgNO 3 solution o Y 1 L of mixture X + excess of BaCl2 solution o Z 9. Number of moles of Y and Z are

4.

(a) 0.01, 0.01

(b) 0.02, 0.01

(c) 0.01, 0.02

(d) 0.02, 0.02

Which has maximum number of atoms ? (a) 24g of C (12) (c) 27g of Al (27)

5.

(2003)

10.

(d) 108g of Ag (108)

(a) 6.023 × 10

(c)

6.023 u 1054 9.108

1 u 1031 (b) 9.108

(d)

(c) 4.159 g

(d) none of the three

The sulphate of a metal M contains 9.87% of M. This

(a) 40.3

(b) 36.3

(c) 24.3

(d) 11.3

If 0.5 mol of BaCl2 is mixed with 0.2 mol of Na3PO4, the maximum number of moles of Ba3 (PO4)2 that can be formed is (1981)

How many moles of electron weighs one kilogram ?

23

(b) 415.9g

sulphate is isomorphous with ZnSO4.7H2O. The atomic weight of M is (1991)

(2003)

(b) 56g of Fe (56)

(a) 41.59 g

11.

1 u 108 9.108 u 6.023

D-12

(a) 0.7

(b) 0.5

(c) 0.30

(d) 0.10

The total number of electrons present in 18 ml of water (density of water is 1 g ml–1) is (1980) (a) 6.02 × 1023

(b) 6.02 × 1023

(c) 6.02 × 1024

(d) 6.02 × 1025

SOME BASIC CONCEPTS OF CHEMISTRY 12.

29.2% (w/W) HCl stock solution has density of 1.25 g mL–1. The molecular weight of HCl is 36.5g mol–1. The volume (mL) of stock solution required to prepare a 200 mL solution 0.4 M HCl is (2013)

13.

20% surface sites have adsorbed N2. On heating N 2 gas evolved from sites and were collected at 0.001 atm and 298 K in a container of volume is 2.46 cm3. Density of surface sites is 6.023 × 1014/cm2 and surface area is 1000 cm2, find out the number of surface sites occupied per molecule of N2.

14.

(2005)

In a solution of 100 mL 0.5 M acetic acid, one gram of active charcoal is added, which adsorbs acetic acid. It is found that the concentration of acetic acid becomes 0.49 M. If surface area of charcoal is 3.01 × 10 2m2, calculate the area occupied by single acetic acid molecule on surface of charcoal. (2003)

15.

Calculate the molarity of water if its density is 1000 kg/m3.

(2003)

16.

3 g of a salt of molecular weight 30 is dissolved in 250 g of water. The molality of the solution is : .......... (1983)

17.

The weight of 1×1022 molecules of CuSO4 ˜ 5H 2O is : .......

18.

(1991)

A sugar syrup of weight 214.2 g contains 34.2 g of sugar (C 12H22O11). Calculate : (i) molal concentration and (ii) mole fraction of sugar in the syrup. (1988)

D-13

SOME BASIC CONCEPTS OF CHEMISTRY

ANSWER KEY Previous Year IIT Questions 1. (c)

2. (b)

3. (a)

4. (a)

5. (d)

6. (a)

9. (c)

10. (d)

11. (c)

12. (8)

13. (2)

14. 5 × 10–19 m2

16. 0.4m

17. 4.14g

18. (i) 0.56, (ii) 0.0099

Dream on !!

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D-14

7. (d)

8. (c) 15. 55.55M