An aqueous solution of metal chloride MCl₂ (0.05 M) is saturated with H₂S (0.1 M). The minimum pH at which metal sulphide will be precipitated is ____.

$\left[{\mathrm{K}}_{\mathrm{sp}}\mathrm{MS}=5\times {10}^{-21},{\mathrm{K}}_1\left({\mathrm{H}}_2\mathrm{S}\right)={10}^{-7},{\mathrm{K}}_2\left({\mathrm{H}}_2\mathrm{S}\right)={10}^{-14}\right.$

${\mathrm{K}}_{\mathrm{sp}}\mathrm{MS}=\left[{\mathrm{M}}^{2+}\right]\left[{\mathrm{S}}^{2-}\right]$

$5\times {10}^{-21}=\left(0.05\right)\left[{\mathrm{S}}^{2-}\right]$

For precipitation $Q_{\mathrm{sp}}=K_{\mathrm{sp}}$

$\therefore \left[{\mathrm{S}}^{2-}\right]=\frac{5\times {10}^{-21}}{5\times {10}^{-2}}={10}^{-19}$

$\text{ For }{\mathrm{H}}_2\mathrm{S}:$

$\begin{array}{l}{\mathrm{H}}_2\mathrm{S}\rightleftharpoons {\mathrm{H}}^{\oplus }+{\mathrm{HS}}^{\ominus }\left({\mathrm{K}}_1\right)\\ {\mathrm{HS}}^{\ominus }\rightleftharpoons {\mathrm{H}}^{\oplus }+{\mathrm{S}}^{2-}\left({\mathrm{K}}_2\right)\\ {\mathrm{K}}_1\times {\mathrm{K}}_2=\frac{{\left[{\mathrm{H}}^{\oplus }\right]}^2\left[{\mathrm{S}}^{2-}\right]}{{\mathrm{H}}_2\mathrm{S}}\\ {10}^{-7}\times {10}^{-14}=\frac{{\left[{\mathrm{H}}^{\oplus }\right]}^2\left[{10}^{-19}\right]}{0.1}\\ {\left[{\mathrm{H}}^{\oplus }\right]}^2={10}^{-3},\end{array}$

$\Rightarrow \text{ [Taking negative logarithm both sides] }$

$\begin{array}{l}-2\log \left[{\mathrm{H}}^{\oplus }\right]=3\\ \mathrm{pH}=\frac{3}{2}=1.5\end{array}$