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A solution is $10^{- 3}$ each in ${Mn}^{2 +}, {Fe}^{2 +}, {Zn}^{2 +}$ and ${Hg}^{2 +}$ is treated with $10^{- 16} M$ sulphide ion. If $K_{s p}$ values of $MnS, FeS, ZnS$ and $HgS$ are $10^{- 15}, 10^{- 23}, 10^{- 20}$ and $10^{- 54}$ respectively, which one will precipitate first?

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$FeS$

$ZnS$

$HgS$

$MgS$

answer is C.

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Detailed Solution

$K_{i p}$ for all $= 10^{- 3} \times 10^{- 16} = 10^{- 19}$

$≫ K_{sp}$ of $HgS$ (lowest $K_{sp}$ )

So, $HgS$ will ppt. first

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