The solubility products of $\mathrm{NiS}, \mathrm{ZnS}, \mathrm{CdS}$ and $\mathrm{HgS}$ are $4.7 \times…
The solubility products of $\mathrm{NiS}, \mathrm{ZnS}, \mathrm{CdS}$ and $\mathrm{HgS}$ are $4.7 \times 10^{-5}, 1.6 \times 10^{-24}, 8 \times 10^{-27}$ and $4 \times 10^{-53}$ respectively. An aqueous solution contains $\mathrm{Ni}^{2+}, \mathrm{Zn}^{2+}, \mathrm{Cd}^{2+}$ and $\mathrm{Hg}^{2+}$ of equal concentration. $\mathrm{H}_2 \mathrm{~S}$ gas was passed into this solution very slowly. The first and the last ions that precipitate as sulphides are respectively.
$\mathrm{Ni}^{2+}, \mathrm{Hg}^{2+}$
$\mathrm{Hg}^{2+}, \mathrm{Cd}^{2+}$
$\mathrm{Zn}^{2+}, \mathrm{Hg}^{2+}$
$\mathrm{Hg}^{2+}, \mathrm{Ni}^{2+}$
Solution
$
\begin{aligned}
& K_{\mathrm{sp}} \text { of } \mathrm{NiS}=4.7 \times 10^{-5} \\
& K_{\mathrm{sp}} \text { of } \mathrm{ZnS}=1.6 \times 10^{-24} \\
& K_{\mathrm{sp}} \text { of } \mathrm{CdS}=8 \times 10^{-27} \\
& K_{\mathrm{sp}} \text { of } \mathrm{HgS}=4 \times 10^{-53}
\end{aligned}
$
Since, the solubility products of NiS is highest and that of $\mathrm{HgS}$ is lowest, it implies that $\mathrm{Hg}^{2+}$ will start precipitating quickly than any other salt on introduction of $\mathrm{H}_2 \mathrm{~S}$ (sulphide ion source). On the same account $\mathrm{Ni}^{2+}$ will precipitate last