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• Aniruddha Kawade

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[Type text] Electrolytes: Whose aqueous solution conduct electricity. Electrolytes are classified into two groups. 1.

Strong Electrolytes: Are those whose ionization is almost 100%, in aqueous solution

generally all ionic compounds are strong electrolytes e.g. NaCl, NaNO3, HCl, KCl. 2.

Weak Electrolytes: Whose degree of dissociation is < 10-15%. Generally covalent

compound are weak electrolytes. e.g. NH 4 OH, CH 3COOH . Strong and Weak electrolytes is a value term, it only depends upon degree of ionization. Some times covalent compounds acts as a strong electrolytes in highly diluted solution. Ka = C ∝2 Ka ∝= = Ka V C ionized molecules ∝= Total molecules “The fraction of total no. of molecules of electrolytes dissolved, that ionizes at equilibrium is called degree of ionization or degree of dissociation.” For: unionized molecules

ionized molecules

∝ Kb, pH < 7, acidic solution iii) if Ka < Kb then pH > 7, alkaline solution In the hydrolysis of salt of weak acid and a weak base such as NH4CN, CH3COONH4. Both the ions are hydrolysied, if we assume K a ≅ K b , then the hydrolysis of the cation and anion of the salt occur approximately to equal extent for a salt which has K a < K b , it would be expected at the first glance that CN − ions hydrolysed to a much greater extent than NH +4 ions. However, the hydrolysis of CN– ions produced OH– ions according to the equation. CN − + H 2 O

HCN + OH −

which can react with NH +4 ions as OH − + NH +4

NH 4 OH

This latter reaction causes equilibrium in the form reaction to be displaced to the right. Because OH– ions are removed from the solution. Also the production of OH– by the former reaction displaces the latter reaction to the right. Therefore the hydrolysis of one ion drags the hydrolysis of the other ion along so that both the hydrolysis are fairly extensive not too far in extant from each other so it is fairly safe to assume that [HCN] = [NH4OH], even in the case of the salt where K a ≠ K b .

Case IV : Salts of strong Acid + Strong Base e.g., NaCl, KNO3 , … etc.

[Type text]

[Type text] This category of salts does not undergo salt hydrolysis

Solubility and solubility Product A solution which remain in contact with excess of the solute is said to be saturated. The amount of a solute, dissolved in a given volume of a solvent (in 1 litre) to form a saturated solution at a given temperature, it termed as the solubility of the solute in the solvent at that temperature.

Molar Solubility: No. of moles of solute dissolved in per litre of solution Solubility Product: In a saturated solution of a salt, there exists a dynamic equilibrium b/w the excess of the solute and ions furnished by that parts of the solute which has gone in solution. The solubility product of a sparingly soluble salt is given as product of the conc. of the ions raised to the power equal to the no. of times the ion occur in the equation after the dissociation of the electrolyte. AxBy xAy+ + yBx–

K sp = [A y + ]x [Bx − ]y Let the solubility of AxBy is S then

K sp = [xS]x [yS]y K sp = x x .y y [Sx + y ] The principle of solubility product is applicable for sparingly soluble salt.

Common-ion Effect on Solubility The common ion presence in the solution decrease the solubility of a given compound e.g. The solubility of BaSO4 in Na2SO4 solution is smaller than that in an aqueous solution. Consider saturated solution of AgCl. If a salt having either of the ion common to AgCl say KCl is added to it, then AgCl(g) + aq.

Ag + + Cl –

KCl + aq.  → K + + Cl – For AgCl

K SP = [Ag + ][Cl – ]

[Cl – ] Increases in solution due to presence of KCl and thus to have K SP constant, [Ag + ] will decrease or AgCl will precipitate out from solution, i.e., solubility of AgCl will decrease with increasing concentration of KCl in solution. Let 0.1 M KCl(aq.) solution with AgCl(aq.) . If solubility of AgCl is s mol litre –1 , then, For AgCl

K SP = [Ag + ][Cl – ] K SP = s(s + 0.1)

[Type text]

[Type text] s being small in comparison to 0.1 and thus may be neglected therefore, K K SP = s × 0.1 or s AgCl = SP 0.1 where s is solubility of AgCl in presence of 0.1 M KClaq.

Ionic Product For a solution of a salt at a specified concentration, the product of the concentration of the ions, each raised to the proper power, is called as the ionic product for a saturated solution in equilibrium with excess of solid, the ionic product is equal to solubility product. At equilibrium, ionic product = solubility product If ionic product is less than solubility product it means solution is unsaturated means more salt can be dissolve in it. If ionic product greater than solubility it means solution is holding more salt than can dissolve in it, therefore ppt started till, until or unless ionic product becomes equal to Ksp.

[Type text]

[Type text]

Preferential Precipitation of Salts Frequently, a solution contains more than one ion capable of forming a ppt. with another ion which is added to the solution. e.g., in a solution containing Cl–, Br– and I–, if Ag+ ions are added then out of the three, the least soluble silver salt is ppt first. If the addition of Ag+ ions is continued, eventually a stage is reached when the next lesser soluble salt starts ppt along with the least soluble salt and so on if the stocihiometry of the ppted salts is the same, then the salt with the minimum Ksp or minimum solubility will ppted first followed by higher Ksp. If the stoichiometry of the ppted salts is not the same, then with Ksp alone, we can’t predict which ion will ppted first. e.g. a solution containing Cl– and CrO 4–2 both of these ions form ppt with Ag+ though the Ksp (AgCl) > Ksp (Ag2CrO4). Yet it is AgCl (less soluble) which ppted first when Ag+ ions added to the solution. In order to predict which ion (Cl– or CrO −42 ) ppt first. We have to calculate the conc. of Ag+ ion needed to start ppt through the Ksp and given conc. of Cl– and CrO −42 , if the conc. of Ag+ ions needed to start the ppt of CrO −42 is larger than that of Cl–. Hence as AgNO3 is added to the solution, the minimum of the two conc. of Ag+ to start the ppt will be reached first and thus the corresponding ion (Cl– in this case) will be ppted in preference to the other. During the course of ppt conc. of Cl– decreases and conc. of Ag+ increases when its’s conc. become equals to the conc. required (of Ag+) for CrO −42 . At this stages the whole of Cl– ions have been ppted the addition of more of AgNO3 causes the ppt of both the ions together. i)

Solubility of a salt of weak acid and strong base in Basic Buffer suppresses than pure water due to common ion effect. But in acidic buffer solution soubility increase than pure water.

ii)

Solubility of salt of weak acid and weak base in pure water: Let the solubility of salt be S, and y mol/litre is the amount of salt getting hydrolysed. CH3COONH4 CH3COO– + NH +4 … (1) –

CH3COO

+

S–y NH +4 + H2O

S–y S–y K sp = (S − y)(S − y) = (S − y) 2 Due to hydrolysis of salt from equation (2) [CH3COOH][NH 4 OH] y.y Kh = = − + [CH3COO ][NH 4 ] (S − y)(S − y) 2

 y  Kh =   S− y  and we also know that Kw Kh = Ka Kb

[Type text]

S–y CH3COOH + NH4OH y

y

… (2)

[Type text]

Solubility of a salt of weak acid and weak base in acidic buffer Let the solubility of salt be S and y be the amount of weak acid being formed. CH3COO– + NH +4

CH3COONH4

S–y –

y

H  → CH3COOH +

CH3COO

+

S–y

(from Acidic Buffer) y

… (2)

K sp = [CH3COO− ][NH 4+ ] = [S – y] [y] = y [S – y] for equation (2) 1

= K′a =

K a (CH3COOH)

[CH 3COOH] y = − + [CH3COO ][H ] (S − y)(H + )

Solubility of CH3COONH4 in acidic buffer would be higher than in pure water

Solubility of a salt of weak acid and weak base in basic buffer CH3COO– + NH +4

Similarly CH3COONH4

y + 4

NH + OH

−

S–y

NH4OH

S – y (from buffer) y K sp (CH3COONH 4 ) = [CH3COO− ][NH 4+ ] K sp = y(S – y) 1 [NH 4 OH] y = K′b = = + − Kb [NH 4 ][OH ] (S − y)(OH − ) The solubility of CH3COONH +4 in basic buffer would be higher than pure water. Illustration 12:

A 100.0 mL sample is removed from a water solution saturated with CaSO4 at 25°C. The water is completely evaporated from the sample and a deposit of 0.24 g CaSO4 is obtained. What is Ksp for CaSO4 at 25°C ?

Solution:

CaSO4(s)

Ca2+ (aq) + SO24 − (aq), Ksp = ?

Data shows that the solubility of CaSO4 is 0.24 g per 100 mL.

[Type text]

0.24 1000 × mol L−1 = 0.01765 M 100 136

∴

[CaSO4] =

∴

[Ca2+] = [ SO24 − ] = 0.01765 M .

∴

Ksp = [Ca2+] [SO 24− ] = (0.01765)2 = 3.115 × 10–4.

[Type text]

Solubility of AgCl in an aqueous solution containing NH3 Let the amount of NH3 initially be `a’ M. if the solubility of salt be `b’ mole/ litre. Ag+

AgCl (s) At time = 0 at time = t

b Ag+

+

2NH3

+

Cl–

0 0 b–y y + Ag (NH3 ) 2 (aq.)

b–y a –2y y where y is the amount of Ag+ which reacted with NH3. K sp = [Ag + ][Cl− ] = (b − y)y Kf =

[Ag(NH3 ]+2 ] y = + 2 [Ag ][NH 3 ] (b − y)(a − 2y)

After knowing the value of Ksp and Kf the value of solubility can be calculated.

Acid-base Indicators An acid & base indicator are substance which changes it’s colour within limits with variation in pH of the solution to which it is added. Indicators, in general are either organic weak acid or weak bases with a characteristics of having different colours in the ionized and unionized form e.g. phenolphthalein is a weak acid (ionized form is pink and unionized form is colourless).

Acidic Indicator Action (e.g. HPh) HPh (Phenolpthalein) is a colourless weak acid HPh H + + Ph − Colourless

K IN =

(Pink )

+

−

[H ][Ph ] [HPh]

If the solution is acidic, the H+ by the acid increases and since Kin is constant and it does not depend upon the concentration so Hph also increases means equilibrium will shift towards left means solution remain colourless. By addition of alkali, OH– will be furnished and that OH– will combines with H+ of HPh to form water and equilibrium will moves towards right and therefore solution becomes pink. Thus HPh appears colourless in acidic and pink in alkaline solution pH range of HPh is (8.3 –10). pH = pK IN + log

[I−n ] [HI n ]

The colour of the indicator changes from colour A to colour B at a particular point known as end point of indicator. At this point [HI n ] = [I n– ] means pH = pK IN (at this point half of indicator is in the acid form and half in the form of its conjugate form.

[Type text]

[Type text]

Indication (Basic) action of MeOH (Methyl Orange) When MeOH is dissolved in water and undergoes dissociation to a small extent. The undissociated molecules are yellow while dissociated Me+ are red in colour. MeOH yellow

Me + + OH − red

colourless

If the solution is acidic, the H+ furnished by the acid combines with OH– ions furnished by the indicators to form undissociated water. This shifts the equilibrium towards right giving red coloured solution. Therefore in acid solution, this indicator gives red colour. In the presence of alkali, OH– increases and due to common ion effect the dissociation of MeOH surpress means equilibrium will shifts towards left. Hence the solution in alkaline medium remains yellow in colour. Colour of solution depends upon relative amount of ionized form to unionized form (ratio of Me+/MeOH). In general pH range of indicator lies B/w pK in − 1 to pK in + 1 Case 1 :

pH = pK in − 1

I−n = 0.1 = 10% [HIn] Percentage ionization of indicator would be I −n 0.10 HI n 1 × 100% = × 100% = × 100 = 9.1% − [I n ] + [HI n ] 0.10 HI n + (HI n ) 11 Means

In fact, pH = pKin – 1 is the maximum pH upto which the solution has a distinct colour characteristic of HIn. At pH smaller than this value, more of the indicator is present in the unionized form. Thus at pH ≤ pKin –1, the solution has a colour characteristic of HIn. ii)

at pH = pK in + 1

[I −n ] = 10 [HIn] Percentage of ionization of indicator is

Mean

[I −n ] 10 [HIn] 1000 × 100% = × 100% = = 91% − [I n ] + [HIn] 10 [HIn] + [HIn] 11 Thus most of the indicator is present in the ionized form I −n and solution gets the colour characteristic. In fact pH = pKin + 1 is the minimum pH upto which the solution has a distinct [Type text]

[Type text] characteristic of I −n . At pH greater than this value, still more of the indicator is present in the ionized form. Thus at pH ≥ pK in + 1 , the solution has a colour characteristics of I −n . Illustration 13: An indicator is a weak acid and pH range of its colour is 3.1 to 4.5. If the neutral point of the indicator lies in the centre of the hydrogen ion concentrations corresponding to the given pH range, calculate the ionization constant of the indicator. Solution:

pH = – log [H3O+], or log [H3O+] = – pH ∴ [H3O+] = antilog of (–pH) for pH = 3.1 [H3O+]1 = antilog of (–3.1) = antilog of (4.9) = 7.94 × 10−4 for pH = 4.5 [H3O+]2 = antilog of (–4.5) = antilog of (5.5) = 3.16 ×10 −5 Since neutral point lies at the centre of the hydrogen ion concentration in the given pH range, hence [H3O+] at the neutral point [H3O+] = =

[H 3O + ]1 + [H3O + ]2 2

7.94 × 10−4 + 3.16 × 10−5 = 4.13 × 10−4 M 2

Let indicator be HIn behaving as weak acid, then HIn + H2O H3O+ + In– [H O + ][In − ] K In = 3 [ionization constant of indicator is KIn] [HIn] = [H3O+] {since at neutral point [In–] = [HIn]} = 4.13 × 10–4.

Ostwald’s Theory According to this theory: a) The colour change is due to ionization of the acid-base indicator. The unionized form has different colour than the ionized form. b) The ionization of the indicator is largely affected in acids and bases as it is either a weak acid or a weak base. In case, the indicator is a weak acid, its ionization is very much low in acids due to common H+ ions while it is fairly ionized in alkalies. Similarly if the indicator is a weak base, its ionization is large in acids and low in alkalies due to common OH– ions. Considering two important indicators phenolphthalein (a weak acid) and methyl orange (a weak base), Ostwald theory can be illustrated as follows:

Phenolphthalein: It can be represented as HPh. It ionizes in solution to a small extent as: HPh

Colourless

[Type text]

H + + Ph − Pink

[Type text]

[H + ][Ph − ] Applying law of mass action, K = [HPh] The undissociated molecules of phenolphthalein are colourless while Ph– ions are pink in colour. In presence of an acid, the ionization of HPh is practically negligible as the equilibrium shifts to left hand side due to high concentration of H+ ions. Thus, the solution would remain colourless. On addition of alkali, hydrogen ions are removed by OH– ions in the form of water molecules and the equilibrium shifts to right hand side. Thus, the concentration of Ph– ions increases in solution and they impart pink colour to the solution. H 3+ O + In −

HIn + H 2 O

`Acid form '

`Base form '

Conjuage acid-base pair

[In − ][H 3+ O] [HIn] ; K In = Ionization constant of indicator, [H 3+ O] = K In × [HIn] [In − ] [HIn] pH = – log10 [H 3+ O] = – log10 [Kin] – log10 [In − ] K In =

pH = pKIn + log10

[In − ] (Handerson equation for indicator) [HIn]

At equivalence point; [In − ] = [HIn] and pH = pKIn

Methyl orange: It is a weak base and can be represented as MeOH. It is ionized in solution to give Me+ and OH– ions. MeOH Yellow

Me+ + OH − Re d

Applying law of mass action

[Me+ ][OH − ] K= [MeOH] In presence of an acid, OH– ions are removed in the form of water molecules and the above equilibrium shifts to right hand side. Thus, sufficient Me+ ions are produced which impart red colour to the solution. On addition of alkali, the concentration of OH– ions increases in the solution and the equilibrium shifts to left hand side (due to common ion effect), i.e., the ionization of MeOH is practically negligible. Thus, the solution acquires the colour of unionized methyl orange molecules, i.e., yellow. This theory also explains the reason why phenolphthalein is not a suitable indicator for titrating a weak base against strong acid. The OH– ions furnished by a weak base are not sufficient to shift the equilibrium towards right hand side considerably, i.e., pH is not reached to 8.3. Thus, the solution does not attain pink colour. Similarly, it can be explained why methyl orange is not a suitable indicator for the titration of weak acid with strong base.

[Type text]

[Type text]

Quinonoid Theory: According to this theory: a) The acid-base indicators exist in two tautomeric forms having different structures. Two forms are in equilibrium. One form is termed benzenoid form and the other quinonoid form.

b) The two forms have different colours. The colour change is due to the interconversion of one tautomeric form into other. c) One form mainly exists in acidic medium and the other in alkaline medium. Thus, during titration the medium changes from acidic to alkaline or vice-versa. The change in pH converts one tautomeric form into other and thus, the colour change occurs. Phenolphthalein has benzenoid form in acidic medium and thus, it is colourless while it has quinonoid form in alkaline medium which has pink colour. OH

O OH

C

−

C

H+

O

OH

C

OH

COO−

O

Methyl orange has quinonoid form in acidic solution and benzenoid form in alkaline solution. The colour of benzenoid form is yellow while that of quinonoid form is red. CH 3 −

O3S

NH

N

N

CH 3

Quinonoid form — Acidic solution (red) OH H+ −

O3S

N

N

CH3 N

CH3

[Type text]

[Type text]

Solved Problems Objective Problem 1: For preparing a buffer solution of pH 6 by mixing sodium acetate and acetic

acid, the ratio of concentration of salt and acid (Ka = 10–5) should be (a) 1:10 (b) 10:1 (c) 100:1 (d) 1:100. Solution:

pH = pKa + log = 5 + log ∴ (b)

[salt] [acid]

10 if [salt]/[acid] = 10:1, then pH = 6 1

Problem 2: The concentration of hydroxyl ion in solution left after mixing 100 mL of 0.1 M

MgCl2 and 100 mL of 0.2 M NaOH (Ksp of Mg(OH)2 = 1.2 × 10–11] is (a) 2.8 × 10–3 (b) 2.8 × 10–2 (c) 2.8 × 10–4 (d) 2.8 × 10–5. Solution:

MgCl2 + 2NaOH  → Mg(OH)2 + 2NaCl mM before 10 20 0 0 mM after 0 0 10 20 reaction thus, 10 m mole of Mg(OH)2 are formed. The product of [Mg2+] [OH–]2 is therefore 2

 10   20  −4  200  ×  200  = 5 × 10 Which is more than    

Ksp of Mg(OH)2. Now solubility (S) of Mg(OH)2 can be derived by Ksp = 4S3 ∴ S=

3

K sp 4

= 1.4 ×10 –4 ∴ [OH–] = 2S = 2.8 × 10–4

∴ (c) Problem 3: 1 ml of 0.1M HCl is added into 99 ml of water. Assume volumes are additive,

what is pH of resulting solution. (a) 3 (c) 2 Solution:

[Type text]

nHCl = MV = .1M × 10–3 L = 10–4 mol Vfinal = (1+99) ml = 100 ml = .1L [HCl] = n/v = 10–3 M = [H3O+] pH = –log [H3O+] = – log 10–3 = 3

(b) 1 (d) 4

[Type text] ∴ (a) Problem 4: What is pH of 0.02 M solution of ammonium chloride at 25°C? Kb(NH3)

= 1.8 × 10–5. (a) 5.477 (c) 7

Solution:

(b) 8.523 (d) 4.8732

For a salt of weak base and strong acid, at 25°C 1 1 pKb – log C 2 2 1 1 = 7 – (– log 1.8 × 10–5) – log 0.02 = 5.477 2 2

pH = 7 –

∴(a)

Problem 5: The pH of 0.1M CH3COOH is 2.873. What is pH of 0.1M NH4OH.

Ka(CH3COOH) = 1.8 × 10–5 and Kb(NH4OH) = 1.8 × 10–5 (a) 11.127 (b) 2.873 (c) 7 (d) 9.53

Solution:

Since Ka(CH3COOH) = Kb(NH3 and concentration are equal and so pH (CH3COOH) = pOH (NH4OH) pH = 2.873 ∴ pH = 14 – pOH = 14 – 2.873 = 11.127 ∴ (a)

Problem 6: A solution of HCl is diluted so that its pH changes by 0.3. How does

concentration of H+ ion change? (a) 0.5 times of initial value (b) 10–3 times increases Solution:

(b) 0.3 times of initial value (d) None

Let H+ ion concentration changes by x factor. Q pH = – log [H3O+] ∴ pH + ∆pH = – log {x(H3O+]} = –log x – log [H3O+] or ∆pH = – log x = 0.3 ∴ x = 0.5 ∴ (a)

0 Problem 7: Liquid NH3 ionises to a slight extent. At –60 C its ionic product

K NH3 =  NH 4+   NH 2−  = 10 −30 The number of NH 2− ions present per ml. of pure liquid NH3 are (a) 300 ions (b) 400 ions (c) 600 ions (d) 500 ions. Solution:

[Type text]

(c) K = [NH4+][NH2-] = x2 x = 10-15 = [NH2-]

[Type text] [NH2-] = 10-15 × ≈ 600 ions /mol

1 × 6.023 × 10 23 106

Problem 8: To a 50 ml of 0.1 M HCl solution, 10 ml of 0.1 M NaOH is added and the

resulting solution is diluted to 100 ml. What is change in pH of the HCl solution? (a) 4.398 (b) .398 (b) 0.1M (d) None. Solution:

Before adding HCl solution pH = 1 [Q [HCl] = [H3O]+ = 10–1 M] nHCl (initially) = MV = 0.1 M × 0.05 L = 5 × 10–3 mol nNaOH added = MV = 0.1 M × 0.01 L = 1 × 10–3 mol HCl + NaOH → NaCl + H2O 1 × 10–3 mol 0 t=0 5 × 10–3 mol 4 × 10–3 mol 0 Vfinal = 100 ml = 0.1 L n 4 × 10 –3 mol [HCl] = = = 4 × 10–2 M V 0.1L + pH = – log [H3O ] = 2 – log4 = 2 – 2 log2 = 2 – 2 × 0.301= 2 – 0.602 = 1.398 Increase in pH = (1.398 – 1) = 0.398 ∴(b)

Problem 9: What amount of solid sodium acetate be added into 1 litre of the 0.1 M

CH3COOH solution so that the resulting solution has pH almost equal to pKa (CH3COOH) = 4.74 (a) 12gm (b) 5 gm (b) 10 gm (d) 14.924 gm. Solution:

[Type text]

Since the resulting solution be acidic buffer, one may use Henderson equation. [CH 3COO – ] pH = pK a + log [CH 3COOH] Let n mol of CH3COONa be added to do so n mol or, pH = 4.74 + log vL 0.1 mol vL n or, 5 = 4.74 + log 0.1 n or, log = 0.26 0.1 n = antilog 0.26 = 1.8197 or, 0.1 ∴ n = 0.18197 mol ≅ 0.182 mol

[Type text] Amount of sodium acetate = 0.182 × 82 gm = 14.924 gm ∴(d) Problem 10: To a 100 ml solution of 0.1 M CH3COONa and 0.1 M CH3COOH, 0.4 gm of

solid NaOH was added. Assuming volume remains constant, calculate the change in pH value? Given that pKa (CH3COOH) = 4.74. (a) 0.125 (b) 0.225 (b) 0.01 (d) 0.872. Solution:

Before NaOH addition, pH = pKa = 4.74 [Since [CH3COO–] = [CH3COOH]] The following reaction occurs due to NaOH addition. H3CCOOH + NaOH → H3CCOONa + H2O t = 0 0.01 mol 0.001 mol 0 – 0.001 mol – 0.001 mol 0.001 mol –––––––––––––––––––––––––––––––––––– (0.01 – 0.001) mol 0.001 mol nH3 CCOONa = 0.01 mole After reaction, nCH3 COOH = 0.009 mol nH

3 CCOO

−

= (0.01 + 0.001) mol = 0.011 mol

[CH 3COO − ] [CH 3COOH] 0.011/ V 11 = 4.74 + log = 4.74 + log = 4.74 + 0.0872 0.009 / V 9

pH = pKa + log

change in pH = log

11 = 0.872 9

∴(d) Problem 11. The expression for the solubility product of Al2(SO4)3 is

(a) Ksp = [Al3+] [SO42-] (b) Ksp = [Al3+]3 [SO42-]2

(b) Ksp = [Al3+]2 [SO42-]3 (b) Ksp = [Al3+]2 [SO42-]2

[MP PET 1999]

Solution:

(b) Solubility of Al2(SO4)3 Al2(SO4)3 2Al+++ + 3SO4− − Ksp = [Al3+]2 [SO42-]3

Problem 12. ON addition of amount chloride to a solution of ammonium hydroxide

(a) Dissociation of NH4OH increases (b) Concentration of OH− decrease (c) concentration of OH− decreases (d) Concentration of NH4+ and OH− increases [CPMT 1999, MP PMT 1989]

Solution: [Type text]

(c) Due to common ion effect.

[Type text]

Problem 13. The solubility product of a salt having general formula MX2, in water is :

Solution:

4 × 10-12. The concentration of M2+ ions in the aqueous solution of the salt is (b) 1.0 × 10-4 M (a) 2.0 × 10-6 M (c) 1.6 × 10-4 M (d) 4.0 × 10-10 M [AIEEE 2005] 2+ − (b) MX2 M + 2X s 2s Ksp = (2S)2 (S) = 4S3 K sp 4 × 10−12 ⇒ S = 23 =3 = 1.0 ×10 −4 M 4 4

Problem 14. If the solubility product Ksp of a sparingly soluble salt MX2 at 250C is 1.0 × 10-11,

the solubility of the salt in mole litre-1 at this temperature will be (a) 2.46 × 1014 (b) 1.36 × 10-4 (c) 2.60 × 10−7 (d) 1.20 × 10-10

[RPMT 2000]

Solution:

M+ + 2X−; Ksp = 4S3 (s) (2s)2

(b) MX2 S= 2

3

1× 10−11 = = 1.35 × 10−4 4 4

K sp

3

Problem 15. Which of the following will occur if a 0.1 M solution of a weak acid is diluted to

0.01 M at constant temperature (a) [H+] will decrease to 0.01 M (b) pH will decrease (c) Percentage ionization will increase (d) Ka will increase [UPSEAT 2001, 02]

Solution :

(b)

Problem 16. 0.1 mole of CH3NH2 (Kb = 5 × 10 ) is mixed with 0.08 mole of HCl and diluted -4

to one litre. What will be the H+ concentration in the solution ? (a) 8 × 10-2 M (b) 8 × 10-11M (c) 1.6 × 10-11 M (d) 8 × 10-5 M [IIT 2005]

Solution :

[Type text]

(b) CH3NH2 + HCl  → CH3NH3+Cl− 0.1 0.08 0 0.02 0 0.08 (Basic buffer solution) 0.08 pOH = pKb + log = pKb + 0.602 0.02

[Type text] = 3.30 + 0.602 = 3.902 ∴ pH = 10.09 [H+] = 7.99 × 10-11 ≈ 8 × 10-11 M Problem 17. When solid potassium cyanide is added in water then

(a) pH will increase (c) pH will remain the same

Solution :

(a) KCN + H2O weak acid.

(b) pH will decrease (d) Electrical conductivity will not change [CPMT 2002; BHU 2002]

KOH + HCN .

KOH is a strong base and HCN is

Problem 18. At 25 C, the dissociation constant of a base BOH is 1.0 × 10 . The 0

-12

concentration of Hydroxyl ions in 0.01 M aqueous solution of the base would be (b) 1.0 × 10-5 mol L-1 (a) 2.0 × 10-6 mol L-1 (c) 1.0 × 10−6 mol L-1 (d) 1.0 × 10-7 mol L-1 [CBSE PMT 2005]

Solution :

(d) initial At. eq.

B+ 0 Cα

BOH C C - Cα

+

OH− 0 Cα

C 2α 2 = Cα 2 assuming α meq of strong acid.

2.

(d)

3.

(c) 5.74 = 4.74 + log

x , calculating x = 0.02, hence (NH4)2SO4 required = 0.01 5 × 10 −3

Subjective Problem 1: What amount of sodium propanoate should be added to one litre of an aqueous

solution containing 0.02 mol of propanoic acid to obtain a buffer solution of pH 4.75 ? What will be the pH if 0.01 mol of hydrogen chloride is dissolved in the above buffer solution ? Compare the last pH value with the pH of 0.01 molar HCl solution. Dissociation constant of propanoic acid at 250C is 1.34 × 10-5.

Solution :

Using the expression [ salt ] pH = pKa + log [ acid ] We get

4.75 = -log(1.34 × 10-5) + log

which gives

4.75 = 4.87 + log

or

[ Salt ]

[ salt ]

[ salt ] [ acid ]

0.02 M

= 0.76 or [salt] = 1.52 × 10-2 M

0.02 M Hence, Amount of sodium propanoate to be added = 1.52 × 10-2 mol The addition of 0.01 mol of hydrogen chloride convert the equivalent amount of sodium propanoate into propanoic acid .

[Type text]

[Type text] Hence, we will have (0.0152 - 0.01) mol L-1 pH = 4.87 + log = 4.87 + log(0.173) (0.02+0.01)mol L-1 = 4.87 – 0.76 = 4.11 The pH of 0.01 molar HCl solution would be pH = -log (0.01) = 2 Problem 2: The solubility of Mg(OH)2 in pure water is 9.57 × 10

-3

g L-1. Calculate its

solubility (g L-1) in 0.02 M Mg(NO3)2

Solution :

We have Molar mass of Mg(OH)2 = 58 g mol-1 Concentration of Mg(OH)2 in pure water =

9.57 × 10−3 gL−1 = 1.65 × 10−4 mol L−1 −1 58 g mol

If follows that Mg(OH)2 (s)

Mg2+ (aq) + 2OH− (aq) 1.65 × 10-4 mol L-1 2 × 1.65 × 10-4 mol L-1 Hence, Ksp (Mg(OH)2) = [Mg2+] [OH−]2 = (1.65 × 10-4) (2 × 1.65 × 10-4)2 (mol L-1)3 = 1.80 × 10-11 (mol L-1)3 Now in 0.02 M Mg(NO3)2 solution. Concentration of Mg2+ ions = 0.02 M The concentration of OH− that can exist in 0.02 M Mg(NO3)2 solution is 1/ 2

 K sp   [OH ] =    Mg 2+     −

1/ 2

 1.80 × 10 −11 (mol L−1 )3  =  0.02 mol L−1  

= 3 × 10 −5 mol L−1

Concentration of Mg(OH)2 in 0.02 M Mg(NO3)2 solution =

1 × 3 × 10−5 mol L−1 2

Solubility of Mg(OH)2 in 0.02 M Mg(NO3)2 solution = 1 −5 −1  −1  × 3 × 10 mol L  (58 g mol ) 2  -4 -1 = 8.7 × 10 g L . Problem 3: 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 1.3653 mol litre–1 and the pKa of H2SO3 is 1.92, estimate the pH of rain on that day.

Solution :

[Type text]

10 = 10 −5 10 6 molar conc. of SO2 in presence of water = (Amount of SO2 × solubility of SO2 in water) [SO2] = 10-5 × 1.3653 mol/L = 1.3653 × 10-5 M Amount of SO2 in atmosphere =

[Type text] Now H2SO3 H+ + HSO3− [1.3653 × 10-5] – x x x as eq. 2 x ka = = 10 −1.92 (1.3653 × 10 −5 − x solving x = 1.364 × 10-5 ∴ pH = -log(1.364 × 10-5) = 4.865 Problem 4: A solution contains 0.1 M H2S and 0.3 M HCl. Calculate the conc. of S –

HS ions in solution. Given K a1 and K a 2 for H2S are 10

–7

2–

and and 1.3 × 10–13

respectively.

Solution:

[H2S] = 0.1 ; [HCl] = 0.3M ∴ [H+] = 0.3 M H2S H+ + HS[H + ][HS- ] 10 −7 × 0.1 = 3.3 × 10-8 Ka = 10-7 = ⇒ [HS-] = [H 2 S] 0.3 − + 2− HS H + S  H +   S 2 −  −13 K a2 = 1.3 × 10 =  HS −  1 1.3 × 10 −13 × × 10 −7 3 ⇒  S 2 −  = = 1.44 × 10-20 0.3

Problem 5:

The dissociation constant of a weak acid HA is 4.9 × 10-8. After making the necessary approximations, calculate (i) percentage ionization, (ii) pH and (iii) OH− concentration in a decimolar solution of the acid. Water has a pH of 7.

Solution:

(i) If α is the degree of dissociation of the weak of the weak acid HA, we will have HA H+ + A− (0.1 M) (1 - α) (0.1 M)α (0.1 M) α + 2  H   A  (0.1M )α Ka =     = 1−α [ HA ] Assuming α S2 > S3 > S4 (b) S1 > S2 = S3 > S4 (c) S1 > S3 > S2 > S4 (d) S4 > S2 > S3 > S1

15.

The pH at which Mg(OH)2 begins to precipitate from a solution containing 0.10M Mg2+ ions [Kp of Mg(OH)2 = 1 × 10–11] is (a) 5 (b) 9 (c) 4 (d) 10

16.

The concentration of hydroxyl ion in solution left after mixing 100 mL of 0.1 M MgCl2 and 100 mL of 0.2 M NaOH (Ksp of Mg(OH)2 = 1.2 × 10–11] is (a) 2.8 × 10–3 (b) 2.8 × 10–2 –4 (c) 2.8 × 10 (d) 2.8 × 10–5.

17.

The solubility of Fe(OH)3 would be maximum in (a) 0.1 M NaOH (b) 0.1 M HCl (c) 0.1 M KOH (d) 0.1 M H2SO4.

18.

pKa values of three acids A, B and C are 4.5, 3.5 and 6.5 respectively. Which of the following represents the correct order of acid strength ? (a) A > B > C (b) C > A > B

[Type text]

[Type text] (c)

B>A>C

(d)

C > B > A.

19.

The solubility of M(OH)3 is x mol L–1. Its Ksp would be (a) 9x3 (d) 3x4 4 (c) 27x (d) 9x4

20.

A solution of HCl is diluted so that its pH changes by 0.3. How does concentration of H+ ion change? (a) 0.5 times of initial value (b) 0.3 times of initial value –3 (b) 10 times increases (d) None

LEVEL - II 1.

A solution is a mixture of 0.05 M NaCl and 0.05 M Nal. The concentration of iodide ion in the solution when AgCl just starts precipitating is equal to : (Ksp AgCl = 1 × 10–10 M2; Ksp AgI = 4 × 10–16 M2) (a) 4 × 10–6 M (b) 2 × 10–8 M (c) 2 × 10–7 M (d) 8 × 10–15 M

2.

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 (a) un ionized in the small intestine and in the stomach (b) completely ionized in the small intestine and in the stomach (c) ionized in the stomach and almost un ionised in the small intestine (d) ionized in the small intestine and almost un ionised in the stomach

3.

If pKb for fluoride ion at 25°C is 10.83, the ionization const. of HF in water at the temperature is (a) 1.74 × 10–5 (b) 3.52 × 10–3 (c) 6.75 × 10–4 (d) 5.38 × 10–2

4.

The pH of 0.1M CH3COOH is 2.873. What is pH of 0.1M NH4OH. Ka(CH3COOH) = 1.8 × 10–5 and Kb(NH4OH) = 1.8 × 10–5 (a) 11.127 (b) 2.873 (b) 7 (d) 9.53

5.

To a 50 ml of 0.1 M HCl solution, 10 ml of 0.1 M NaOH is added and the resulting solution is diluted to 100 ml. What is change in pH of the HCl solution? (a) 4.398 (b) .398 (b) 0.1M (d) None.

6.

The pOH of 10–8 M HCl is (a) 8 (b) Between 6 and 7

[Type text]

(b) 6 (d) Between 7 and 8.

[Type text]

7.

Given that the dissociation constant for H2O, Kw = 1 × 10–14 mol2 litre–2, what is the pH of a 0.001 M KOH solution ? (a) 10–11 (b) 10–3 (c) 3 (d) 11.

8.

An acid solution of pH 6 is diluted thousand times. The pH of solution becomes approx. (a) 6.69 (b) 6 (c) 4 (d) 9

9.

The pH of an aqueous solution of a 0.1 M solution of a weak monoprotic acid, which is 1%, ionized is (a) 1 (b) 2 (c) 3 (d) 11

10.

In a mixture of weak of acid and its salt with strong acid, the ratio of concentration of salt to acid is increased ten times the pH of the solution (a) Increases by 10 (b) Decreases by 10 (c) Decreases by 1 (d) Increases by 1.

LEVEL - II 1.

Which of the following is the correct. Expression of degree of dissociation for weak acid. (Assume α is negligible w.r.t to unity). Ka Kw × c (a) (b) c Ka (c)

Kw 1 × K b of conjugatebase c

(d)

K b of conjugate base × c

2.

When equal volumes of the following solutions are mixed, precipitation of AgCl (Ksp = 1.8 × 10–10) will occur only with (b) 10–3 M[Ag+] and 10–3 M [Cl–] (a) 10–4 M [Ag+] and 10–4M[HCl] (c) 10–6M [Ag+] and 10–6 M [Cl-–] (d) 10–2 M [Ag+] and 10–2M [Cl–].

3.

Correct options for the following reaction is :NH3 + H2O N+H4 + OH +  (a) K a  N H 4  + Kb (NH3) = 10-14 (b) K a (NH + ) × K b ( NH 3 ) = 1/ K w 4   (c) K a (NH 4+ ) × K b ( NH3 ) = K w (d) pK a ( NH + ) + pK b ( NH3 ) = 14 4

[Type text]

[Type text] 4.

pH of 0.03 M NaH2PO4 (a) Increases with dilution (c) does not depend upon concentration

(b) decreases with dilutions (d) changes with change in temperature

5.

Which of the following statements is are right (a) The pH of 1 × 10-8 M solution of HCl is 8 (b) The conjugate also of H2 PO −4 is HPO24 − (c) auto protolysis constant of water increases with temperature (d) when a solution of weak monoprtic acid is titrated against a strong base, at half 1 neutralization point pH = pK a 2

6.

When a 2 M aqueous solution of CH3COOH is diluted with water. The change which may occure is (a) The concentration of H+ increases (b) the degree of dissociation of Acid increases (c) the value of ka increases (d) the number of H+ ion increases

7.

Which of the following statement are true for a solution saturated with AgCl and AgBr. If their solubilities in mol lit-1 in separate solutions are x and y respectively. (a) [Ag+] = [Br−] + [Cl−] (b) [Cl−] > [Br−] − (c) [Br ] > y (d) [Ag+] = x + y

8.

Pure NH3 is placed in a vessel at a temperature where its dissociation constant (K) is appreciable, At equilibrium (a) K does not change with pressure (b) K does not change with [NH3] (c) concentration of NH3 does not change with pressure (d) concentration of Hydrogen in less than that of nitrogen.

9.

Which of the following solution will have no effect on pH with dilution (a) 0.1 M NaHS (b) 5 M H2CO3 + 5M NaHCO3 (c) 1 M CH3COONH4 (d) 1 M NH4Cl

10.

Which of the following solution will have pH =13 assume complete dissociations (a) 2g of NaOH in 500 ml solution (b) 100 ml of solution of 0.05 M Ca(OH)2 (c) 100 ml of solution of 0.1 N Ca(OH)2 (d) 4g of NaOH in 500 ml solution.

SECTION – III

Comprehension Type Questions

Write-up I

If a sparingly soluble salt is placed in water, after some time an equilibrium is established when the rate of dissolution of ions from the solid equals the rate of precipitation of ions from the saturated solution at a particular temperature. Thus, a dynamic equilibrium exists between the undissociated solid species and the dissolved ionic species in a saturated solution at a particular temperature. For example, in AgCl, we have the following equilibrium : [Type text]

[Type text] AgCl (s) Ag+ (aq) + Cl− (aq.) The equilibrium constant  Ag +  Cl −  K eq =  [ AgCl ] Q [AgCl] is constant ∴ Keq × [AgCl] = [Ag+] [Cl-] ⇒ Ksp(AgCl) = [Ag+] [Cl-] If there would not have been a saturated solution, then from equation (1), Keq. [AgCl] ≠ Ksp, but Keq. [AgCl] = QAgCl, where Q is ionic product. It implies that for a saturated solution, Q = Ksp Ksp is temperature dependent. When Q < Ksp, then the solution is unsaturated and then will be no precipitate. When Q = Ksp, then solution will be saturated, no ppt. when Q = Ksp, the solution will be supersaturated and there will be formation of precipitate. 1.

pH of a saturated solution of Ba(OH)2 is 12. Hence Ksp of Ba(OH)2 is: (a) 5 × 10-7 M3 (b) 5 × 10-4 M2 (c) 1 × 10-6 M3 (d) 4 × 10-6 M3

2.

A solution is a mixture of 0.05 M NaCl and 0.05 M NaI. The concentration of iodide ion in the solution when AgCl just starts precipitating is equal to: (Ksp AgCl = 1 × 10-10 M2; Ksp AgI = 4 × 10-16 M2) (a) 4 × 10-6 M (b) 2 × 10-8 M (c) 2 × 10-7 M (d) 8 × 10-15 M

3.

Silver iodide is used in cloud seeding to produce rain AgI Ag+(aq) + I− (aq); Ksp = 8.5 × 10-7. AgNO3 and KI are mixed to give + [Ag ] = 0.010 M; [I−] = 0.015 M. Will AgI precipitate ? (a) yes (b) no + (c) can’t say (d) this depends on  NO3−  and [K ]

4.

Slaked lime, Ca(OH)2 is used extensively in sewage treatment. What is the maximum pH that can be established in Ca(OH)2(aq.) Ca(OH)2(s) (a) 1.66 (c) 7

Ca2+ (aq) + 2OH− (aq) ; (b) 12.34 (d) 14

Ksp = 5.5 × 10-6

Write-up II

Acidity or alkalinity of a solution depend upon the concentration of hydrogen ion relative to that of hydroxyl ions. The product of hydrogen ion & hydroxyl ion concentration is given by Kw = [H+] [OH–]

[Type text]

[Type text] the value of which depends only on the temperature & not on the individual ionic concentration. If the concentration of hydrogen ions exceeds that of the hydroxyl ions, the solution is said to be acidic; whereas, if concentration of hydroxyl ion exceeds that of the hydrogen ions, the solution is said to be alkaline. The pH corresponding to the acidic and alkaline solutions at 25ºC will be less than and greater than seven, respectively. To confirm the above facts 0.5 M CH3COOH is taken for the experiments. [Given : Ka of acetic acid = 1.8 ×10–5] 1.

2.

Degree of dissociation of acetic acid is (a) 66 × 10–2 (c) 3 × 10–3

(b) 6 × 10–3 (d) 5 × 10–3

pH of the solution will be (a) 2.52 (c) 5

(b) 2.22 (d) 3.92

3.

If pH of the solution is doubled, what will be the concentration of acetic acid (a) 1.8 × 10–5 M (b) 1.0 M –5 (c) 1.374 × 10 M (d) 1.25 × 10–3 M

4.

To what volume at 25º C must 1 dm3 of this solution be diluted in order to double the pH(a) 3.37 × 104 dm3 (b) 2.34 × 102 dm3 (d) 3.18 × 103 dm3 (c) 1.68 × 104 dm3

5.

Now to increase the hydrogen ion concentration 100 dm3 of 0.1M HCl solution is added to100 dm3 of 0.5 M acetic acid solution, then what will be the pH of the final solution (a) 6 (b) 1.3 (c) 3 (d) 1 Write-up III

In qualitative analysis, cations of group II as well as group IV both precipitated in the form of sulphides due to low value of Ksp of group II sulphides, group reagent is H2S in presence of dil. HCl and due to high value of Ksp of group IV sulphides, group reagent is H2S in presence of NH4OH and NH4Cl. In a 0.1M H2S solution, Sn2+, Cd2+ and Ni2+ ions are present in equimolar concentration (0.1 M). Given: K a1 Ka2(H2S) = 10-7, K a 2 (H2S) = 10–14, Ksp (SnS) = 8 × 10–29 Ksp (CdS) = 10–28, Ksp (NiS) =3 × 10–21 6.

If HCl solution is passed slowly then which sulphide will precipitate first (a) SnS (b) CdS (c) NiS (d) none of these

7.

At what pH precipitate of NiS will form -

[Type text]

[Type text] (a) 12.76 (c) 1.24

(b) 7 (d) 4

8.

Which of the following sulphide is more soluble in pure water(a) CdS (b) NiS (c) SnS (d) all have equal solubility

9.

If 0.1 M HCl is mixed in the solution containing only 0.1 M Cd2+ and saturated with H2S then concentration of Cd2+ remains in the solution after CdS has precipitated (a) 8.2 × 10-8 (b) 8.2 × 10-9 (c) 5.6 ×10-7 (d) 5.6 ×10-8

Answers to Assignments SECTION - I LEVEL - I 1.

(a)

2.

(d)

3

(b)

4

(c)

5.

(c)

6

(c)

7

(a)

8

(b)

9

(b)

10

(d)

11

(a)

12.

(a)

13.

(b)

14.

(c)

15.

(b)

16

(c)

17.

(d)

18.

(c)

19.

(c)

20.

(a)

LEVEL - II 1.

(a)

2.

(d)

3

(c)

4

(a)

5.

(b)

6

(d)

7

(d)

8

(a)

9

(c)

10

(d)

SECTION - II 1.

(a, c)

2.

(a, b, d)

3

(c, d)

4

(c, d)

5.

(b, c)

6

(b, d)

7

(a, d)

8

(a, b, c)

9

(a, b, c)

10

(a, b, c)

4. 9.

(b) (b)

SECTION - III 1. 6. 11.

(a) (a) (c)

[Type text]

2. 7. 12.

(c) (c) (b)

3. 8. 13.

(a) (a) (a)

5. 10.

(b) (a)

[Type text]

[Type text]