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)

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 $(10^{-54)}$, so it will precipitate first.