Ionic Equilibrium IIT JEE Questions

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[Arvind Kumar]

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Solved Examples on Equilibrium
Question 1:

Calculate the pH of the solution when 0.1 M CH3 COOH (50 ml) and 01. M NaOH (50 ml) are mixed, [Ka (CH3COOH)=10-5]
Solution:

CH3 COOH \rightleftharpoons CH3 COO_ + H+ …(I)

NaOH → Na+ + OH-

H+ + OH_ \rightleftharpoonsH2O …(II)

(I) + (II)

CH3COOH + OH- CH3COO- + H2O . (III)

0.05-X 0.05-x x

Keq of eq. (III) = Ka/Kw

conc. of H2O remain constant

109 = x/(0.05-x)2

because value of eq. Const.is very high

here for x» 0.05

let 0.05-x=a

109=0.05/a2

a = 7.07\times10-6

pOH= 6-log 7.07

pOH= 6 – 0.85

pH= 14-6+0.85 = 8.85

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Question 2:

Calculate the pH at the equivalence point of the titration between 0.1M CH3*COOH ( 25 ml) with 0.05 M NaOH. Ka (CH3COOH) = 1.8 \times 10–5.
Solution:

We have already seen that even though when CH3COOH is titrated with NaOH the reaction does not go to completion but instead reaches equilibrium. We can assume that the reaction is complete and then salt gets hydrolysed because, this assumption will help us to do the problem easily and it does not effect our answer.

[H^+]=\sqrt{\frac{K_wK_a}{C}}

First of all we would calculate the concentration of the salt, CH3COONa. For reaching equivalence point,

N1V1 = N2V2

0.1 ´ 25 = 0.05 ´ V2

\Rightarrow V2 = 50 ml

Therefore [CH3COONa] = (0.1\times25)/75 =0.1/3

[H+] = \sqrt{\frac{10^-^14\times 1.8\times 10^-^5}{0.1/3}}

\Rightarrow[H+] = 2.32 ´ 10–5

\Rightarrow pH = – log 2.32 ´ 10–5 = 8.63

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Question 3:

Given the solubility product of Pb3 (PO4)2 is 1.5 x 10-32.Determine the solubility in gms/litre.
Solution:

Solubility product of Pb3 (PO4)2 = 1.5 \times 10–32

Pb3 (PO4)2 \rightleftharpoons 3Pb2+ + 2PO43-

If x is the solubility of Pb3 (PO4)2

Then Ksp = (3x)3 (2x)2 = 108 x5

x = 1.692 \times 10–7 moles/lit

Molecular mass of Pb3(PO4)2 = 811

x = 1.692 ´ 10–7 ´ 811 g/lit = 1.37 \times 10–4 g/lit

Solubility product is

Ksp(SrC2O4) = [Sr2+] [C2O42–] = (5.4 \times 10–4)2- = 2.92 \times 10–7

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Question 4:

What is pH of 1M CH3COOH solution? To what volume must one litre of this solution be diluted so that the pH of resulting solution will be twice the original value. Given : Ka = 1.8 \times 10–5
Solution:



H3CCOOH + H2O \rightleftharpoons H3CCOO– + H3O+

t = 0 1M 0 0

-xM xM xM

__________________________________________

t = teq (1-x)M x x

pH = – log [H3O+] = – log {4.2 \times 10–3} = 3 – log 4.2 = 2.37

Now, let 1L of 1M ACOH solution be diluted to VL to double the pH and the conc. of diluted solution be C.

H3CCOOH + H2O \rightleftharpoons H3CCOO– + H3O+

t = 0 C 0 0

– 1.8 \times 10–5 1.8 \times 10–5 1.8 \times 10–5

__________________________________________________ __

t= teq C – 1.8 \times 10–5 1.8 \times 10–5 1.8 \times 10–5

New pH = 2 \times old pH = 2 \times 2.37 = 4.74

pH = – log [H3O+] = 4.74

[H3O+] = 1.8 \times 10–5

C = 3.6 \times 10–5 L

on dilution

M1V1 = M2V2

1M \times 1L = 3.6 \times 10–5 L \times V2

V*2 = 2.78 \times 104 L

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Question 5 :

Find the concentration of H+, HCO3- and CO32-, in a0.01M solution of carbonic acid if the pH of this is 4.18.

Ka1(H2CO3) = 4.45 \times 10–7 and Ka2 = 4.69 \times 10–11
Solution:

pH = – log[H+]

4.18 = – log [H+]

[H+] = 6.61 \times 10–5

H2CO3 \rightleftharpoons H+ + HCO3-

again, HCO3- \rightleftharpoons H+ + CO32-

[CO32-] = 4.8 \times 10–11

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Question 6:

Calculate the molar solubility of Mg(OH)2 in 1MNH4Cl

KspMg(OH)2 = 1.8 \times 10–11

Kb(NH3) = 1.8 \times 10–5
Solution:

Mg(OH)2(s) \leftrightharpoons Mg++ + 2OH– K1 = Ksp

2NH4+ + 2OH- \leftrightharpoons 2NH4OH K2 = 1/K2b

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Question 7:

An aqueous solution of metal bromide MBr2 (0.05M) in saturated with H2S. What is the minimum pH at which MS will ppt.?

Ksp =(MS) = 6 \times 10–21

Concentration of standard H2S = 0.1

Ka1(H2S) = 1 \times 10–7

Ka2(H2S) = 1.3 \times 10–13
Solution:

In saturated solution of MS

MS(s) \leftrightharpoons M++ + S2-

The precipitate of MS will form only if [S––] exceeds the concentration of 1.2 \times 10–19

H2S \leftrightharpoons H+ + HS– Ka1

H2S– \leftrightharpoons H+ + S-- Ka2

——————————————

H2S \leftrightharpoons 2H+ + S2– K = 1.3 \times 10–20

[H+] = 0.109

pH = 0.96

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Question 8 :

How much AgBr could dissolve in 1.0 L of 0.4 M NH3? Assume that [Ag (NH3)2]+ is the only complex formed given, Kf [Ag(NH3)2+]=1.0\times108, Ksp (AgBr)= 5.0\times10-13
Solution:

AgBr \rightleftharpoons Ag+ + Br-

Ag+ + 2NH3 \rightleftharpoons Ag (NH2)2+

Let x= solubility ,

Then x= [Br-]=[Ag+]+[Ag(NH3)2+]