Question

A solution which is ${10}^{-3}M$ each in $M{n}^{2+},F{e}^{2+}$ and $H{g}^{2+}$ is treated with ${10}^{-16}M$ sulphide ion, If $K_{sp}$ of $MnS,FeS,ZnSandHgS$ are ${10}^{-15},{10}^{-25},{10}^{-20}$ and ${10}^{-54}$ respectively, which one will precipitate first?

• A. $FeS$
• B. $MnS$
• C. $HgS$
• D. $ZnS$

Answer 1

The correct option is C $HgS$

Ionic product of the solution $={10}^{-3}\times {10}^{-16}={10}^{-19}.$
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 $K_{sp}{10}^{-19}$.
$HgS$ has the lowest $K_{sp}$ value $\left({10}^{-54}\right)$ so it will precipitate first.