A solution containing $\mathrm{As}^{3+}, \mathrm{Cd}^{2+}, \mathrm{Ni}^{2+}$ and $\mathrm{Zn}^{2+}$ is made…
A solution containing $\mathrm{As}^{3+}, \mathrm{Cd}^{2+}, \mathrm{Ni}^{2+}$ and $\mathrm{Zn}^{2+}$ is made alkaline with dilute $\mathrm{NH}_{4} \mathrm{OH}$ and treated with $\mathrm{H}_{2} \mathrm{~S}$. The precipitate obtained will consist of
$\mathrm{As}_{2} \mathrm{~S}_{3}$ and $\mathrm{CdS}$
CdS, NiS and ZnS
NiS and $\mathrm{ZnS}$
Sulphide of all ions
Solution
$\mathrm{As}^{3+}$ and $\mathrm{Cd}^{2+}$ are the radicals of group II, whereas $\mathrm{Ni}^{2+} \& \mathrm{Zn}^{2+}$ are the radicals of group IV. The solubility product of group IV radicals is higher as compared to group II. $\mathrm{NH}_{4} \mathrm{OH}$ increases the ionisation of $\mathrm{H}_{2} \mathrm{~S}$ by removing $\mathrm{H}^{+}$ of $\mathrm{H}_{2} \mathrm{~S}$ as unionisable water. $\mathrm{H}_{2} \mathrm{~S} ightleftharpoons 2 \mathrm{H}^{+}+\mathrm{S}^{2-}$
$$
\mathrm{H}^{+}+\mathrm{OH}^{-} \longrightarrow \mathrm{H}_{2} \mathrm{O}
$$
Thus excess of sulphide ions are present which leads to the precipitation of all the four ions. Note : HCl decreases ionisation of $\mathrm{H}_{2} \mathrm{~S}$ whereas $\mathrm{NH}_{4} \mathrm{OH}$ increases the ionisation of $\mathrm{H}_{2} \mathrm{~S} .$
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