A solution which is $10^{- 3} M$ each in $M n^{2 +}, F e^{2 +}$ and $H g^{2 +}$ is treated with $10^{- 16} M$ sulphide ion, If $K_{s p}$ of $M n S, F e S, Z n S a n d H g S$ are $10^{- 15}, 10^{- 25}, 10^{- 20}$ and $10^{- 54}$ respectively, which one will precipitate first?

F e S

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M n S

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H g S

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Z n S

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Solution

The correct option is C $H g S$
Ionic product of 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 $K_{s p}$ .
Here, all salts are of the same valence type. So, the sulphide having the lowest $K_{s p}$ value will precipitate first provided $K_{s p} 10^{- 19}$ .
$H g S$ has the lowest $K_{s p}$ value $(10^{- 54})$ so it will precipitate first.

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A solution which has $10^{- 3}$ M each of $M n^{2 +}, F e^{2 +}, Z n^{2 +}$ and $H g^{2 +}$ is treated with $10^{- 16}$ M sulphide ion. If $K_{s p}$ of $M n S$ , $F e S$ , $Z n S$ and $H g S$ are $10^{- 15}$ , $10^{- 23}$ , $10^{- 20}$ and $10^{- 54}$ respectively, which one will precipitate first?