Question

A solution which is ${10}^{-3}$ M each in $M{n}^{2+},F{e}^{2+},Z{n}^{2+}$ and $H{g}^{2+}$ is treated with ${10}^{-16}$ M sulphide ion. If $K_{sp}$ of $MnS,FeS,ZnS$ and $HgS$ 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

Answer: (C)

HgS

Ionic product in 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.