Verified Solution • 2025–26 Session moderate

QJEEAdv20Chac5c: JEE Advanced 2020 Chemistry - Class 11 JEEAdv20Chac5c Solution: An acidified solution of 0.05 M Zn 2+ is saturated with 0.1 M H …

An acidified solution of 0.05 M Zn2+ is saturated with 0.1 M H2S. What is the minimum molar concentration (M) of H+ required to prevent the precipitation of ZnS?

Use Ksp(ZnS) = 1.25 ×\times 10 −- 22 and overall dissociation constant of

H2S, Knet = K1K2 = 1 ×\times 10-21.

Step-by-Step Verified Solution

ZnS(s)⟶Zn2+(aq)(0.05M)+S2−(aq)

Ksp(ZnS)=[Zn2+][S2−]=1.25×10−22{K_{sp}}(ZnS) = [Z{n^{2 + }}][{S^{2 - }}] = 1.25 \times {10^{ - 22}}

0.05×[S2−]=1.25×10−220.05 \times [{S^{2 - }}] = 1.25 \times {10^{ - 22}}

⇒[S2−]=1.25×10−220.05⇒25×10−22M\Rightarrow [{S^{2 - }}] = {{1.25 \times {{10}^{ - 22}}} \over {0.05}} \Rightarrow 25 \times {10^{ - 22}}M

For H2S, H2S(0.1M)⟶2H++S−2(25×10−22M)

Knet=1×10−21=[H+]2[S2−][H2S]{K_{net}} = 1 \times {10^{ - 21}} = {{{{[{H^ + }]}^2}[{S^{2 - }}]} \over {[{H_2}S]}}

1×10−21=[H+]2×25×10−22[0.1]1 \times {10^{ - 21}} = {{{{[{H^ + }]}^2} \times 25 \times {{10}^{ - 22}}} \over {[0.1]}}

[H+]2=125{[{H^ + }]^2} = {1 \over {25}}

[H+]=15=0.2M[{H^ + }] = {1 \over 5} = 0.2M

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