JEE Advance - Chemistry (2020 - Paper 2 Offline - No. 18)
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.
Use Ksp(ZnS) = 1.25 $$ \times $$ 10$$-$$22 and overall dissociation constant of
H2S, Knet = K1K2 = 1 $$ \times $$ 10-21.
Answer
0.2
0.2M
- OR
Explanation
$$ZnS(s)\buildrel {} \over
\longrightarrow \mathop {Z{n^{2 + }}(aq)}\limits_{(0.05\,M)} + {S^{2 - }}(aq)$$
$${K_{sp}}(ZnS) = [Z{n^{2 + }}][{S^{2 - }}] = 1.25 \times {10^{ - 22}}$$
$$0.05 \times [{S^{2 - }}] = 1.25 \times {10^{ - 22}}$$
$$ \Rightarrow [{S^{2 - }}] = {{1.25 \times {{10}^{ - 22}}} \over {0.05}} \Rightarrow 25 \times {10^{ - 22}}M$$
For H2S, $$\mathop {{H_2}S}\limits_{(0.1M)} \buildrel {} \over \longrightarrow 2{H^ + } + \mathop {{S^{ - 2}}}\limits_{(25 \times {{10}^{ - 22}}M)} $$
$${K_{net}} = 1 \times {10^{ - 21}} = {{{{[{H^ + }]}^2}[{S^{2 - }}]} \over {[{H_2}S]}}$$
$$1 \times {10^{ - 21}} = {{{{[{H^ + }]}^2} \times 25 \times {{10}^{ - 22}}} \over {[0.1]}}$$
$${[{H^ + }]^2} = {1 \over {25}}$$
$$[{H^ + }] = {1 \over 5} = 0.2M$$
$${K_{sp}}(ZnS) = [Z{n^{2 + }}][{S^{2 - }}] = 1.25 \times {10^{ - 22}}$$
$$0.05 \times [{S^{2 - }}] = 1.25 \times {10^{ - 22}}$$
$$ \Rightarrow [{S^{2 - }}] = {{1.25 \times {{10}^{ - 22}}} \over {0.05}} \Rightarrow 25 \times {10^{ - 22}}M$$
For H2S, $$\mathop {{H_2}S}\limits_{(0.1M)} \buildrel {} \over \longrightarrow 2{H^ + } + \mathop {{S^{ - 2}}}\limits_{(25 \times {{10}^{ - 22}}M)} $$
$${K_{net}} = 1 \times {10^{ - 21}} = {{{{[{H^ + }]}^2}[{S^{2 - }}]} \over {[{H_2}S]}}$$
$$1 \times {10^{ - 21}} = {{{{[{H^ + }]}^2} \times 25 \times {{10}^{ - 22}}} \over {[0.1]}}$$
$${[{H^ + }]^2} = {1 \over {25}}$$
$$[{H^ + }] = {1 \over 5} = 0.2M$$
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