Question

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

Answer 1

For $ZnS\rightleftharpoons {Zn}^{2+}+S^{2-}$

$K_{sp}\left(ZnS\right)=\left[Z{n}^{+2}\right]\left[S^{2-}\right]$
To stop precipitation of $ZnS$,
$\left[Z{n}^{+2}\right]\left[S^{2-}\right]\le K_{sp}\left(ZnS\right)$
$\left[S^{2-}\right]\le \frac{1.25\times {10}^{-22}}{0.05}$
For $H_{2}S\rightleftharpoons 2H^{+}+S^{2-}$

$\left[S^{2-}\right]=\frac{K_{NET}\times \left[H_{2}S\right]}{{\left[H^{+}\right]}^2}$

$\frac{K_{NET}\times \left[H_{2}S\right]}{[H^{+}{]}^2}\le \frac{1.25\times {10}^{-22}}{{10}^{-2}\times 5}$
$[H^{+}{]}^2\ge \frac{{10}^{-21}\times {10}^{-1}\times 4}{{10}^{-20}}$
$\left[H^{+}\right]\ge 2\times {10}^{-1}=0.2$
Alternate:
$\left[Z{n}^{+2}\right]\left[S^{2-}\right]\le K$
$\left[S^{2-}\right)\le \frac{1.25\times {10}^{-22}}{0.05}=\frac{1}{4}\times {10}^{-20}$
$H_{2}S\rightleftharpoons 2H^{+}+S^{2-}$
${\left[H^{+}\right]}^2=\frac{K_1{K}_2\times \left[H_{2}S\right]}{\left[S^{2-}\right]}=\frac{{10}^{-21}\times 0.1}{\left[S^{2-}\right]}$
$\left[S^{2-}\right]=\frac{{10}^{-22}}{[H^{+}{]}^2}\le \frac{1}{4}\times {10}^{-20}$
$\left[H^{+}\right]\ge 0.2$