1. Tardigrade
  2. Question
  3. Chemistry
  4. A solution which is 10-3 M each in Mn2+, Fe2+, Zn2+ and Hg2+ is treated with 10-16 M sulphide ion. If Ksp of MnS, FeS, ZnS and HgS are 10-15, 10-25, 10-20 and 10-54 respectively, which one will precipitate first?

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Q. A solution which is $10^{-3}\, M$ each in $Mn^{2+}$, $Fe^{2+}$, $Zn^{2+}$ and $Hg^{2+}$ 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?

Equilibrium

Solution:

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.