A solution which is 10^-2 M each in ${\mathrm{Mn}}^{2+};{\mathrm{Fe}}^{+2},{\mathrm{Zn}}^{2+};{\mathrm{Hg}}^{2+}$ is treated with ${10}^{-20}$ sulphide ion. If $K_{sp}$ $\mathrm{MnS};\mathrm{FeS};\mathrm{ZnS};\mathrm{HgS}$ are ${10}^{-13},{10}^{-18},{10}^{-24},{10}^{-53}$    respectively. Which one will precipitate first?

When ionic product predominates solubility product; then precipitation takes place.

Ionic product of $HgS=\left[H{g}^{2+}\right]\left[S^{2-}\right]$

$\begin{array}{l}=\left[{10}^{-2}\right]\left[{10}^{-20}\right]={10}^{-22} < {\mathrm{K}}_{\mathrm{sp}} \left({10}^{-53}\right)\\ \end{array}$

The metal sulphide having the lowest solubility will precipitate first provided the ionic product is higher than the K_sp​. Here, all salts are of the same valence type. So, the sulphide having the lowest K_sp​ value will precipitate first provided