A solution is 0.1 M in each of ${\mathrm{H}}^{+}, {\mathrm{Zn}}^{2+}$ and ${\mathrm{Fe}}^{2+}$. If ${\mathrm{K}}_{\mathrm{s}}\left(\mathrm{ZnS}\right) = 2.5 \mathrm{x} {10}^{- 22} {\mathrm{M}}^2, {\mathrm{K}}_{\mathrm{s}}\left(\mathrm{FeS}\right) = 4 \mathrm{x} {10}^{-19}$${\mathrm{M}}^2$ and ${\mathrm{K}}_{\mathrm{a}1} {\mathrm{K}}_{\mathrm{a}2}$ of ${\mathrm{H}}_2\mathrm{S}$ is $1.0 \mathrm{x} {10}^{- 21} {\mathrm{M}}^2.$ Predict, which of the following statements is correct when the concentrations of ${\mathrm{H}}_2\mathrm{S}$ in the solution is $0.1 \mathrm{M}$?

${\mathrm{H}}_2\mathrm{S}\rightleftharpoons 2{\mathrm{H}}^{+}+{\mathrm{S}}^{2-};K_{\mathrm{a}}{K}_{\mathrm{a}2}=1.0\times {10}^{-21}{\mathrm{M}}^2$

Since $K_{\mathrm{a}1}{K}_{\mathrm{a}2}={\left[{\mathrm{H}}^{+}\right]}^2\left[{\mathrm{S}}^{2-}\right]/\left[{\mathrm{H}}_2\mathrm{S}\right],$ we have 

$\left[{\mathrm{S}}^{2-}\right]=\frac{K_{\mathrm{a}1}{K}_{\mathrm{a}2}\left[{\mathrm{H}}_2\mathrm{S}\right]}{{\left[{\mathrm{H}}^{+}\right]}^2}=\frac{\left(1.0\times {10}^{-21}{\mathrm{M}}^2\right)\left(0.1\mathrm{M}\right)}{\left(0.1{\mathrm{M}}^2\right.}=1.0\times {10}^{-20}\mathrm{M}$

Ionic products of given ions are

For $\mathrm{ZnS}$             $Q_c=\left[{\mathrm{Zn}}^{2+}\right]\left[{\mathrm{S}}^{2-}\right]=\left(0.1\mathrm{M}\right)\left({10}^{-20}\mathrm{M}\right)={10}^{-21}{\mathrm{M}}^2$

For $\mathrm{FeS}$            $Q_c=\left[{\mathrm{Fe}}^{2+}\right]\left[{\mathrm{S}}^{2-}\right]=\left(0.1\mathrm{M}\right)\left({10}^{-20}\mathrm{M}\right)={10}^{-21}{\mathrm{M}}^2$       

Since only for $Q_{\mathrm{c}}\left(\mathrm{ZnS}\right)>K_{\mathrm{s}}\left(\mathrm{ZnS}\right)$ only $\mathrm{ZnS}$ is precipitated.