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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, which one will precipitate first
$FeS$
$MgS$
$HgS$
$ZnS$
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 $K_{\text {sp. }}$. Here, all salts are of the same valence type. So, the sulphide having the lowest $K_{s p}$ value will precipitate first provided $K_{s p}\,<\,10^{-19}$. HgS has the lowest $K _{\text {sp }}$ value $\left(10^{-54}\right)$, so it will precipitate first.
