Question

An acidified solution of $0.05MZ{n}^{2+}$ is saturated with $0.1M{H}_{2}S$. What is the minimum molar concentration $(M)$ of $H^{+}$required to prevent the precipitation of $ZnS$ ?
Use $K_{sp}(ZnS)=1.25\times 10^{-22}$ and overall dissociation constant of $H_{2}S.K_{NET}=K_1{K}_2=1\times 10^{-21}$

Answer 1

Answer: 0.20

$ZnS(s)\rightleftharpoons Z{n}^{2+}(aq)+S^2(aq)$
$\left[S^{2-}\right]=\frac{K_{sp}}{\left[Z{n}^{2+}\right]}=\frac{1.25\times 10^{-22}}{0.05}=25\times 10^{-22}$
$H_{2}S\rightleftharpoons 2H^{+}+S^{2-}$
$10^{-21}=\frac{{\left[H^{+}\right]}^2\times 25\times 10^{-22}}{0.1}$
${\left[H^{+}\right]}^2=\frac{10^{-22}\times 10^{22}}{25}=\frac{1}{25}$
$\left[H^{+}\right]=\frac{1}{5}=0.20M$