Chemistry - Ionic Equilibrium Question with Solution | TestHub

The solubility of metal sulphides in saturated solution of ${\mathrm{H}}_2\mathrm{S}\left\{\left[{\mathbf{H}}_2\mathbf{S}\right]=0.1\mathbf{M}\right\}$ can be represented by

$$\begin{gathered} \text{displaylines}MS+2H^{+}\to M^{2+}+H_{2}S; \\ {K}_{eq}=\frac{\left[{\mathrm{M}}^{2+}\right]\left[{\mathrm{H}}_2\mathrm{S}\right]}{{\left[{\mathrm{H}}^{+}\right]}^2} \end{gathered}$$

The value of ${\mathrm{K}}_{\mathrm{eq}}$ is given for few metal sulphide. If conc. of each metal ion in solution is $0.01\mathbf{M}$, which metal sulphide(s) are selectively precipitated at total $\left[{\mathbf{H}}^{+}\right]=1M$ in a saturated ${\mathrm{H}}_2\mathrm{S}$ solution.