A solution which is 10^–3 M each in Mn²⁺, Fe²⁺, Zn²⁺ and Hg²⁺ is treated with 10^–16M sulphide ion. If K_sp of MnS, FeS, ZnS and HgS are 10^–15, 10^–23,10^–20 and 10^–54 respectively. Which of the following will be precipitated first?

 

Given data: It is given that 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 ${10}^{–15}, {10}^{–23},{10}^{–20}$ and ${10}^{–54}$ respectively,  the following will be precipitated first

Explanation: 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   value$({10}^{-5\left(D\right)}$, so it will precipitate first.

Hence, option (C) is correct.