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CONTENTS S.NO.

TOPIC

PAGE NO.

1.

SALT

2

2.

IONIC DISSOCIATION

2

3.

OSTWALD’S DILUTION LAW

2

4.

STRENGTHS OF ACIDS AND BASES

4

5.

SELF IONIZATION OF WATER

6

6.

HYDROGEN ION / HYDROXYL ION CONC. OF ACID / BASE

6

7.

pH SCALE

7

8.

COMMON ION EFFECT

8

9.

BUFFER SOLUTIONS

8

10.

BUFFER ACTION

9

11.

SOLUBILITY

12

12.

HYDROLYSIS OF SALT

14

13.

DEGREE OF HYDROLYSIS

14

14.

THEORY OF INDICATOR

18

1

Ionic Equilibrium

IONIC EQUILIBRIUM 1. SALT A substance which ionizes in water to produce ions other than H+ and OH– is called a salt. Types of Salts: Neutral Salts : Those salts whose aqueous solutions neither turn blue litmus red nor red litmus blue are called neutral salts. These are prepared by the neutralization of strong acid and strong base. e.g. NaCl, K2SO4, KNO3 etc. Basic Salts : Those salts whose aqueous solution turn red litmus blue are called basic salts. These are formed by the neutralization of strong bases with weak acids. e.g. Na2CO3, CH3COONa Mixed salts : Salts formed by the neutralization of one acid by two bases or one base by two acids are called mixed salts. e.g. CaOCl2 Double Salts : Acompound of two salts whose aqueous solution shows the tests for all constituent ions is called double salt e.g. Mohr Salt FeSO4. (NH4)2 SO4. 6H2O PotashAlum K2SO4.Al2 (SO4)3. 24H2O. Complex Salts : Acompound whose solutions does not give test for the constituent ions is called a complex salt. e.g. K4 [Fe(CN)6] Li (AlH4)

2. IONIC DISSOCIATION The process in which molecules (acids, bases, and salts) when dissolved in water or when melted break into ions is called ionic dissociation. Electrolytes. Hence substances which dissociate into ions in aqueous solutions are called electrolytes. e.g. NaCl, NaNO3, HCl, K2SO4 etc. Strong Electrolyte And Weak Electrolyte and Non Electrolytes: Those electrolytes which dissociate almost completely into ions are known as strong electrolytes. e.g. HCl, HBr, HI, HClO4, NaCl, Na2SO4, KNO3 etc. Those electrolytes which dissociate partially are called weak electrolytes. e.g. H3PO4, HF, H2CO3,HCN, CH3COOH, NH4OH, etc. .Those electrolytes which do not dissociate into ions in aqueous solutions are called non electrolytes. Molecules of the substances which do not dissociate into ions in aqueous solutions are called non- electrolytes. e.g. sugar, urea, etc.

3. OSTWALD’S DILUTION LAW Ostwald’s pointed out that like chemicalequilibriuminionic equilibriumwe can applylawof mass action. An equilibriumbetweenionized and unionized molecules. Considera binary electrolyte having conc. C and degree of dissociation is. A   B 0 0 C C(1 – ) C  C AB At time = 0 At time = t 

K aq





[A ][B ]

1   1

[AB]



C  C

C(1 )



C

2

1 

, for a weak electrolyte

2

Ionic Equilibrium

K eq  C 2 ,  

K eq

C If 1 mole ofAB is present in ‘V’litre of solution. C

1 V

  K eq  V  C  C

Conc. of A+ = Conc. B–

(i) (ii)

K eq C

 K eq C

Limitation : This is only for weak electrolytes not for strong electrolytes. This law is not applicable for strong electrolyte because strong electrolytes are almost completelyionC ized at all dilution and hence  does not give accurate results.  Dissociation of Weak Acids and Bases : Dissociation of Weak acids : Consider the dissociation of a weak acid HAin water, represented by the equation – H3O + A HA + H2O C

Initial conc. moles/litre

HA

Final concentration

(C – C)

0

0

in moles/lit.

H+ + A–

C C in moles/lit.

C α2 Ka = 1 α



This equation is referred to as Ostwald dilution law. In case of weak acids the degree of dissociation  is very small, therefore (1 – ) may be taken to be equal to unity Hence

Since Also as  =

Ka  C

VK a

This shows that decree of dissociation is inversely proportional to square root of concentration and directly proportional to square root of dilution of the solution. Dissociation of Weak Base in Water : The dissociation of a weak base can be represented in the same manner as a weak acid. e.g. (initial)

C BOH

Final concentration in moles/lit. C – C

O

O

B + OH–

C

C.

3

Ionic Equilibrium

Where C = Initial conc. in moles/litre  = Degree of dissociation 

Kb =

As the base is weak   1 Kb = C 2    = Kb / c Hence,

CB  COH  CBOH

C 2α2 Cα2  C(1α) 1α

1 –  = 1.



Kb C

[OH–] = C = C × [OH–] =



Kb C

4. STRENGTHS OF ACIDS AND BASES (i) (ii)

(iii) (iv)

The strength of an acid/base depend upon the number H+/OH– present in solution. Since  = V  K a or  = V  Kb   increases when V increase Ionization increases with dilution, hence number of H+ /OH– increases.  At infinite dilution the ionization of all acids and bases tends to become almost equal and all acids and bases behave equally strong at infinite dilution. All mineral acids or bases which ionize fully at all dilutions are considered as strong acids. While acids/bases like CH3COOH/NH4OH respectively which ionize to a less extent are called weak acids/bases. The relative strenths is generally compared in terms of their dissociation constants. We known Ka1 = C1 1 2 1  1 Ka2 = C2 2 2 2  1 K a1 C1 K a2 C2

=

α1  α2

(v) Ex.1 Sol.

K a1 C 2 α  1 α2 K a2 C1

K a1 Ka 2

α1  α2

K b1 C 2  K b2 C1

α1  α2

K b1 K b2

when C1  C2

when C1  C2 when C1  C2 when C1  C2

Strength of all strong acids/bases in water is same. This is called levelling effect

Calculate the degree of ionization of 0.01 M solution of HCN, Ka of HCN is 4.8 × 10–10. Also calculate hydronium ion concentration. The ionization of HCN may be represented as, HCN(aq)  H 2O()



CN– - (aq)  H 3O (aq)

If degree of ionization of HCN is  then equilibrium concentration of given species are

4

Ionic Equilibrium

[HCN]  C(1 – ) [CN – ]  C [H3O  ]  C where C = concentration of HCN. Applying law of chemical equilibrium C2 [CN – ][H 3O ] (C)(C)   C(1– ) (1 – ) [HCN] Since  is very small as compared with unity therefore, 1 –  in the denominator may be taken as 1. Ka 

K a  C 2 

Ka 4.8 10–10  2.2 10–4 .  C 0.01

[H3O ]  C  0.01 2.2 10 –4  2.2 10 –6 mol L–1. Ex.2 Sol.

The Ka for formic acid and acetic acid are 2 × 10–4 and 2 × 10–5 respectively. Calculate the relative strength of acids with same molar concentration– Relative strength of weak acid 𝑐 2 𝑘𝑎1 𝑎1 ∗ = 1 𝑐 𝑘𝑎2 𝑎2 Ka

1

 Relative strength =

K a2

( C1 = C2)

2 𝑥 10−4 2 𝑥10−5

Relative strength for HCOOH to CH3COOH = Ex.3

10 : 1

Calculate the degree of ionization of 0.02 M acetic acid if its K a  1.8  10 –5 . What would be the degree of ionization if the solution also contains 0.01 M sodium acetate ?

Sol.

CH 3COOH(aq)

CH COO3– (aq)  H  (aq)

The degree of ionization of this weak acid can be calculated by the approximate relation : Ka 1.810 –5  0.03  310 –2 .  C 0.02 Now let us calculate the degree of ionization when the solution also contains 0.01 M sodium acetate. Sodium acetate being a strong electrolyte would be completely ionized in solution. Let x mol L–1 of acetic acid be ionized. 

CH 3COOH(aq) (0.02– x )M

CH3 COO – (aq)  H  (aq) xM

xM

CH 3COONa(aq)   CH3 COO– (aq)



Na  (aq) 0.01 M

5

Ionic Equilibrium

[H ]  x mol L–1 [CH3COO –]  (x  0.01) mol L–1  0.01 mol L–1 [ x is very small as compared to 0.01] [CH3COOH]  (0.02 – x) mol L–1  0.02 mol L–1 Ka 

[H  ][CH3COO – ] [CH3COOH]

1.810–5 

(x)(0.01) (0.02) –5

1.8 10  0.02  3.6 10–5 M 0.01 –5 x 3.6 10  1.810 –3  Degree of ionization,   0.02 0.02 Thus, it may noted that the degree of ionization of acetic acid has decreased from 310–2 to 1.8 10–3 due to the presence of sodium acetate. x

5. SELF IONIZATION OF WATER Water can behave both as an acid as well as a base. This behaviour of water is due to ionization ofwater to form protons and hydroxyl ions. H2O Hence K = 

H+ + OH– C H  COH – CH 2 O

K CH2O = CH+ COH– CH2O = Kw = CH+ COH– K Kw is the Ionic Product of water and may be defined as the product of concentration of ions [(H+) and (OH–)] ions. Its value depends only on temperature and is found to be 1 × 10–14 at 25ºC At Kw = 0.11 × 10–14 mol 2/lit 2 OºC Kw = 0.30 × 10–14 mol2/lit2 10ºC 25ºC Kw = 1 × 10–14 mol2 /lit2 –14 2 2 100ºC Kw = 7.5 × 10 mol /lit For pure water CH+ = COH– = 10 –7 mol/lit at 25ºC Degree of dissociation of pure water at 25ºC For pure water CH+ = COH–. Also at room temperature CH+ COH– = 10–14 mol2 /lit2  CH+ = 10–7 mol/lit  C = 10–7 mol/lit 107 107  = 1.8 × 10–9 C 55.6 Hence degree of dissociation = 1.8 × 10–9  % degree of dissociation = 1.8 × 10–7 =

6. HYDROGEN ION / HYDROXYL ION CONC. OF ACID/ BASE (i)

In case of strong acids (or bases) concentration in water solution is taken as equal to normality of the acid/base since they ionize completely. [H+] = Normality of acid = Molarity × Basicity [OH–] = Normality of base = Molarity × Acidity

6

Ionic Equilibrium

(ii)

In case of weak acids/Bases, the H+/OH– concentration is less than normal and may be calculated by using Oswald's dilution law. [H+] =

K aC

[OH–] =

= C = N ×  = C = N × 

K bC

7. pH SCALE : It may be defined in number of ways. (i) The pH value of a solution is equal to the negative power to which 10 must be raised in order to express [H+] concentration [H+] = 10–pH. (ii) It can also be defined as the negative logarithmof its [H+] ion concentrations pH = –log[H+] (iii) pH values do not give instantaneous idea above the relative strengths of the solution pH calculation of solution of a mixture of two weak Acids in water : Let two weak acids be HA and HB and their conc. are C1 and C2, 1 is the degree of dissociation of HA in presence of HB (due to common ion effect) and 2 be degree of dissociation of HB in presence of HA. In aqueous solution of HAand HB following equilibrium exists. conc. at eq.

+

A– C1  1

+ (C1 1 + C2 2)

B– C2  2

HA+ H 2O(l)

H3O+

C1(1 – 1)

C11 + C22

HB + H2O(l)

H3O+

C2(1 – 2) 







K a[HA] 

[H3O ][A ] [C11  C12 ][C12 ]  C1 (1  1 ) [HA]

K a[HB] 

[H3O ][B ] [C11  C22 ][C22 ]  [C2 (1  2 )] [HB]

pH  – log[H  ]  – log[C11  C2 2 ] pH of a DibasicAcid and PolyproticAcid : Let’s take the eg. of a dibasic acid H2A.Assuming both dissociation is weak. Let the initial conc. of H2A is C and 1 and 2 be degree of dissociation for first and second dissocation. HA– C1 (1 – 2)

H2A

C(1 –1) HA– C1 (1 – 2) 

+

H+ + C1 + C1 2

H+ C1 + C1 2 A– – C1 2.



Ka1 

[HA ][H ] [H 2A]

Ka1 

[C1 (1 2 )][C1 C1 2 ] C(1  1 )

7

Ionic Equilibrium

Ka 2 

Ex.4 Sol.

Ex.5 Sol.





[H ][A ] [C1  C12 ][C12 ]  [C1 (1  2 )] [HA  ]

After solving for 1 and 2. We can calculate the H+ conc. [H+] = C1 + C1 2 pH = – log [C1 + C1 2] pOH SCALE : It is defined as the negative logarithms of hydroxyl ions concentration. pOH = –log (OH–). Also it is known that [H+] [OH–] = 10–14 = Kw –log[H+] (+) –log[OH–] = 14 = pK w pH + pOH = pKw = 14.  The pH of a 0.05 M solution of H2SO4 in water is nearly? pH= – log10 H + The concentration of H+ ions is expressed in gm equivalent Molarity of H2SO4 = 0.05 Normality = 0.05 × 2 = 0.1  pH = –log 0.1  pH = 1 or Calculate the pH of solution having H+ ion concentration of 5 × 10–4 mole/litre [H+] ion concentration = 5 × 10–4 mole/litre pH = – log [5 × 10–4] = – (log 5 + log 10–4) = – 0.7 + 4 = 3.3

8. COMMON ION EFFECT When a solution of weak electrolyte is mixed with another electrolyte which provides one or more ion common with the weak electrolyte the dissociation of weak electrolyte is suppressed. This suppression of dissociation of weak electrolyte on addition of a common ion is called common ion effect. For example if we consider dissociation of CH3COOH in the presence of CH3COONa we get following situation : – + CH3COOH ⇌ CH3COO + H CH3COONa CH3COO– + Na + In this case the CH3COO– ion contributed by CH3COONa suppresses the dissociation of CH3COOH. This suppression of dissociation of CH3COOH is called common ion effect.

9. BUFFER SOLUTIONS A buffer is a solution which resists any change in its pH value on either (a) dilution or (b) addition of acid/base. The process by which the added H+/OH– are removed to maintain the pH of solution, is known as buffer action. Types of Buffer : 1. Simple buffers : Asalt of weak acid and weak base in water e.g. (a) NH4CN or CH3COONH4. (b) Proteins and amino acids. (c) Amixture of an acid salt and normal salt of a poly basic acid e.g. NaHCO3 + Na2CO3 Na2 HPO4 + Na3PO4.

8

Ionic Equilibrium

2.



Mixed Buffer : These are of two types (a) Acid Buffer Solution : An acidic buffer solution of a weak acid and its salt with strong base. e.g. CH 3COOH + CH3COONa – + CH3COOH CH3COO + H (Weakly ionized) – CH3COO + Na+ (Highly ionized) CH3COONa The sodium acetate, being a salt ionizes completely to form CH3COO– and Na+ ion. But acetic acid being a weak acid is ionized to a less extent.Also its ionizations is further suppressed by the acetate ion from sodium acetate. Let C moles/litre be the concentration of CH3COOH taken and C1 be the concentration of CH3COONa. The degree of dissociation of CH3COOH is 𝛼 in the presence of sodium acetate Initial conc. (mol/lit.) C 0 0 – CH3COOH H+ CH3COO + ⇌ C Final conc. (mol/lit.) (C – C) C Initial conc.(mol/lit.)

Final conc.(mol/lit.) Hence in the solution

C1 CH3COONa 0

0 0 – + CH3COO + Na C1 C1

CH3COOH concentration = (C – C) CH3COO– concentration = C + C1 H+ concentration = CH+ = C. But since  is very small [CH3COOH] = (C – C)  C. [CH3COO-] = C + C1  C1 Ka =

CCH 3COO– CH  C CH3COOH

=> CH+ = Ka ×



pH = – log CH+ = PKa + log



pH = pKa + log [Salt]

C C1

C1 C

[Acid]

This equation is called Henderson’s Equation.

10. BUFFER ACTION : When a few drops of acid is added then the H+ ions from the acid combines with excess of CH3COO– ions to form CH3COOH. Hence there is no rise in [H+ ] ion concentrations but due to the consumptions – – of CH3COO the concentration of CH3COO decreases and the concentration of CH3COOH increases hence the ratio of

[Salt] [Acid]

decreases slightly. Thus the pH change is minimal, meaning the solution has

resisted the change in pH. In other wards one can saythat the pH change which is very minimal is not due to change in the concentration of [H+] but due to change in buffer capacity of solution. On the other hand when NaOH is added, the [OH–] ions of the base reacts with the unionized CH3COOH to form acetate ion and water. OH– + CH3COOH  CH3COO– + H 2O Ex.6 Sol.

Calculate the ratio of pH of a solution containing 1 mole of CH COONa + 1 mole of HCl per litre and 3 of other solution containing 1 mole CH3COONa + 1 mole of acetic acid per litre. Case I. pH when 1 mole CH3COONa and 1 mole HCl are present. 0 0 1 1 CH COOH + NaCl CH3COONa + HCl   3 After reaction 1 1 0 0

Before reaction

9

Ionic Equilibrium



[CH3COOH] = 1 M



[H+] = C. = C



pH1 = –



pH2 = – log Ka + log

 K a   (K .C)  (K ) a a  C   

 C=1

1

log Ka 2 Case II : pH when 1 mole CH3COONa and 1 mole of CH3COOH; a acidic buffer solution forms [Salt] = 1 M, [Acid] = 1 M  [Salt] [Acid]

pH2 = –log Ka 

pH1 1  pH 2 2

Basic Buffers : Abasic buffer solution consists of a mixture of a weak base and its salt with a strong acid. e.g. mixture of NH4OH and NH4Cl.

⇌

NH4OH

NH4+ + OH –

Weakly ionized Highlyionized NH4Cl NH4+ + Cl– + The NH4 ions of NH4Cl suppress the ionization of NH4OH due to common ion effect. Let the concentrations of NH4OH taken be C moles/lit. and -be the degree of ionization after adding NH4Cl. Let C1 be the concentration of NH4Cl taken. Initial conc. (mol/lit.) C 0 0 + NH4OH NH4 OH – Final conc. (mol/lit.) (C – C) C C C1 Initial conc.(mol/lit.) 0 0 + NH4 + Cl– NH4Cl 0 Final conc.(mol/lit.) C1 C1 The concentration of [NH4+] = C1 + C  C1 [  1] [NH4OH] = C – C  C [  1] [OH–] = C. Kb =

C NH  C OH – 4

C NH 4 OH

 COH– = Kb C NH



C NH 4 OH C NH 4  C



pOH = pKb + log C = pKb + log C1 NH 4 OH



pOH = pKb + log [Base]

4

[Salt]

BufferAction : When few drops of base say NaOH is added then the OH– ions added react with NH 4+ to form

10

Ionic Equilibrium

[NH 4 ]

–

NH4OH and thus the concentration of [OH ] remains unchanged. But the ratio of [NH OH] changes. 4 Thus the change in pH is very small and that too due to change in buffer capacity. On the other hand when a few drops of acid (Say HCl) is added, then the [H+] ions of acid combine with excess of NH4OH to form H2O and NH4+ ions. i.e. NH4OH + H+ NH4+ + H2O. + Thus the addition of acid does not increase the H ions but since the concentration of NH4 OH decreases [NH  ]

4 and [NH4+] ion conc. increases, the ratio [NH OH] increases and thus pH changes infinitively. 4

Ex.7

Abuffer of pH 9.26 is made by dissolving x moles of ammonium sulphate and 0.1 mole of ammonia into 100 mL solution. If pKb of ammonia is 4.74, calculate value of x.

Sol.

(NH4)2SO4

2 NH4 + SO42 

Thus, every one mole of (NH4)2SO4 gives two moles of NH4 . 

 

millimoles of NH3, (NH4OH) = 100  0.1 = 10 millimol millimoles of (NH4)2SO4 = 100  x = 100 x 1000 millimol millimoles of NH4 = 200  x = 200 x millimol pH = 9.26 pOH = 14 – 9.26 = 4.74 

pOH  pK b  log 4.74 = 4.74 + log

[NH 4 ] [ NH4OH] 200x 10

log 20x = 0 20 x = 1 x=

1 = 0.05. 20

Buffer Capacity : It is defined as the number of moles of the acid/base added to the buffer solution to produce a change in pH by one unit. Buffer Capacity =

number of moles of acid/base added change in pH

In general Buffer Capacity would be maximum when both components are present in equimolar proportions. Agiven acid must have a pH ranging between pKa + 1 to pKa – 1 if it is to be used in buffer solution. The best buffer will have the acid with pH = pKa. Ex.8 Sol.

What amount of sodium propanoate should be added to one litre of an aqueous solution containing 0.02 mole of propanoic acid (Ka = 1.0 × 10–5 at 25ºC) to obtain a buffer solution of pH 6. Using the expression pH = pKa + log

[Salt] [Acid]

11

Ionic Equilibrium [Salt]

we get, 6 = –log(1.0 ×10–5)+ log [0.02 M] [Salt]

Which gives 6 = 5 + log [0.02 M] [Salt] [0.02 M]

= 10 or [Salt] = 0.2 M

Ex.9

What will be the pH of the solution, if 0.01 mole of HCl is dissolved in a buffer solution containing 0.03 mole of propanoic acid (Ka = 1.0 × 10–5) and 0.02 moles of salt, at 25ºC.

Sol.

pH = pKa + log

[Salt] [Acid] (0.02  0.01)

= –log(1.0 × 10–5) + log (0.03  0.01) 1

= 5 + log  4  = 5 – 0.6 = 4.4   Ex.10 What amount of HCl will be required to prepare one litre of a buffer solution (containing NaCN and HCN) of pH 10.4 using 0.01 mole of NaCN. Given Kion (HCN) = 4.1 × 10–10 Sol. The addition of HCl converts NaCN into HCN. Let x be the amount of HCl added. We will have. [NaCN] = (0.01 – x) [HCN] = x Substituting these values along with pH and Ka in the expression. pH = – log Ka + log

[Salt] [Acid]

We get 10.4 = –log[4 × 10–10] + log 0.01 x x

or

10.4 = 9.4 + log

or

log

or

0.01 x x

or

x = 9.9 × 10–4 M

0.01 x x

0.01 x x

=1

= 10  11x = 10–2

11. SOLUBILITY It is the amount of the solute in gram that can be dissolved in 100 gm of a solvent to obtain a saturated solutions at a particular temperature. However solubility can also be expressed in moles/litre. Solubility in gm/lit.

Solubility of solution in moles/litre = Molecular weight of solute The factors affecting solubility are. (a) Nature of solvent (b) Nature of solute (c) Temperature (d) Pressure

12

Ionic Equilibrium

Solubility Product : If a sparingly soluble salt is added in water, very little amount of it dissolves in water, and thus solution becomes saturated, but a highly soluble salt saturation is reached on dissolving more of salt. For all salts at saturation an equilibrium is achieved between the undissolved salt and the ions insolutions. B+(aq) + + A– (aq)

BA(s) Applying law of mass action we get K=

[A – ] [B  ] [BA]

K[BA] = [B+] [A–]



Since only little of salt dissolves so the concentration of salt remains constant K[AB] = K × Constant = Ksp = [B+] [A–]  For any salt BxAy xBy+ (aq) + yAx–(aq) Ksp = [By+ ]x [Ax–] y Ksp = Kc This expression shows that in a saturated solution of a sparingly soluble salt the ionic product is equal to the solubility product. Ksp = Solubility Product It is defined as the product of the molar concentrations of its ion raised to the power equal to its number of ions present at equilibrium representing the ionization of one molecule of salt at a given temperature. Points to Remember: When Ksp = Ki solution saturated (1) When Ki < Ksp unsaturated solution (2) (3) When Ki > Ksp super saturated solution. Hence precipitation occurs to keep Ki = Ksp Relationships Between Solubility and Solubility Product: (1) Let the solubility of a salt Bx Ay in water be s moles/lit. Thus at equilibrium BxAy xBy+ (aq) + yAx–(aq) At equilibrium xs mol/lit ys moles/lit. Ksp = [By+ ]x [Ax–] y  (xs)x (ys)y = Ksp = xxyy sx + y Hence for salts of type MA (AgCl, BaSO4, etc.) 

s=

Ksp = 4 s 3



s=

Ksp = 27 s 4



s = 4 K sp /27

Ksp = s2 For M2Atype of salts

K sp

3

K sp /4

For MA3 Ex.11 A salt M2 X3 dissolves in water such that its solubility is x g. mole/litre. What is KSP of salt ? Sol. Solubility of M2X3 = x gm mole/litre 2M+3 + 3X –2 M2 X3 +3 [M ] = 2x  [X–2] = 3x Solubility product KSP = (2x) 2.(3x) 3 =108 x 5 Ex.12 The solubility of AgCl in water, in 0.02 M CaCl2, in 0.01M NaCl and in 0.05 M AgNO3 are S0, S1,S2,S3 respectively. What is the relationships between these quantities ?

13

Ionic Equilibrium

Sol.

(B) Solubility = 

S1 =

Solubility Product Concentration of Common ion

K SP 0.02 K SP

= 50 KSP

S2 = 0.01 = 100 KSP K sp = 20 KSP S = 3

So,

0.05

S2 > S1 > S3 Again solubility will be greatest in water. So, S0 > S2 > S1 > S3

12. HYDROLYSIS OF SALT : It is the process involving action of water on a salt to form a mixture of acid and alkali. The hydrolysis of a salt is reverse of neutrilization. Let there be a salt BA. The hydrolysis of such a salt can be represented as H2O + AB water salt

HA+ BOH acid base

[HA] [BOH]



K = [AB][H O] 2 

Kh =

[HA] [BOH] [AB]

K[H2O] =

[HA] [BOH] [AB]

where Kh = hydrolysis constant.

The hydrolysis constant is dependent on nature of acid or base which is formed as a result of hydrolysis. Degree of Hydrolysis : It is defined as the fraction of the total salt, which is hydrolysed at equilibrium. Anionic Hydrolysis : It is the Hydrolysis of salts of Weak acids and Strong Base e.g. CH3COONa, Na2CO3, K2CO3, KCN, Na2S etc. Let the salt hydrolysis be represented as BA + H2O (1)

BOH + HA B +A– BA SinceA– is a strong base is under hydrolysis according to the equation +

HA + OH–

A– + H 2O

This is called anionic hydrolysis Kh =

[HA] [OH – ] [A – ]

Multiplying and dividing RHS by H+ we get Kh =

     K w  [H  ][OH ], K a  [H ][A ]   [HA]   

[HA] [OH – ][H  ] K w  Ka [A – ][H  ]

Kw

Hence hydrolysis constant = Kh = K 

a

1

Kh  K i.e. hydrolysis constant varies inversely with dissociation constant of acid. a

13. DEGREE OF HYDROLYSIS : Let C moles/Lof salt be taken then C moles/Lof cation and anion will be formed respectively. BA C

B+ C mole/L

+

A– C mole/L.

14

Ionic Equilibrium

Let ‘h’be the degree of hydrolysis ofA–  Ch moles/lit. of A–will take part in hydrolysis and Ch moles/lit. of HAand Ch moles/L of [OH–] will be formed. C

O

O

A– + H2O

(C – Ch)

At equilibrium conc. in moles/L. 

Ch

[HA][OH ]

Kh =

=

–

[A ]

When h