The solubility of metal sulphide in saturated solution of $\mathrm{H}_{2} \mathrm{~S}$ (concentration $=0.1 \mathrm{M}$ ) can be represented as:
$\mathrm{MS}(\mathrm{s})+2 \mathrm{H}^{+}(\mathrm{aq}) \rightleftharpoons \mathrm{M}^{2+}(\mathrm{aq})+\mathrm{H}_{2} \mathrm{~S}(\mathrm{aq})$
$K_{\mathrm{eq}}=\frac{\left[\mathrm{M}^{2+}\right]\left[\mathrm{H}_{2} \mathrm{~S}\right]}{\left[\mathrm{H}^{+}\right]^{2}}$
The values of $K_{\mathrm{eq}}$ for the metal sulphides, $\mathrm{MnS}, \mathrm{ZnS}, \mathrm{CoS}$ and $\mathrm{PbS}$ are $3 \times 10^{10}$
$3 \times 10^{-2}, 3$ and $3 \times 10^{-7}$, respectively. If the concentration of each metal ion in a saturated solution of $\mathrm{H}_{2} \mathrm{~S}$ is $0.01 \mathrm{M}$, which metal sulphide(s) will precipitate at $\left[\mathrm{H}^{+}\right]=1.0 \mathrm{M} ?$
(a) $\mathrm{MnS}, \mathrm{ZnS}, \mathrm{CoS}$
(b) $\mathrm{PbS}$
(c) $\mathrm{PbS}, \mathrm{ZnS}, \mathrm{CoS}$
(d) $\mathrm{PbS}, \mathrm{ZnS}$