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(C) Precipitation occurs when the ionic product $(Q_{sp})$ exceeds the solubility product $(K_{sp})$.Given concentration of metal ions $[M^{2+}] = 10^

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$HgS$

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(C) Precipitation occurs when the ionic product $(Q_{sp})$ exceeds the solubility product $(K_{sp})$.Given concentration of metal ions $[M^{2+}] = 10^{-3} \ M$ and $[S^{2-}] = 10^{-16} \ M$.The ionic product $Q_{sp} = [M^{2+}][S^{2-}] = 10^{-3} 	imes 10^{-16} = 10^{-19}$.Comparing $Q_{sp}$ with $K_{sp}$ values:For $MnS$: $K_{sp} = 10^{-15} > 10^{-19}$ (No precipitation).For $FeS$: $K_{sp} = 10^{-23} For $ZnS$: $K_{sp} = 10^{-20} For $HgS$: $K_{sp} = 10^{-54} The substance that precipitates first is the one with the lowest $K_{sp}$ value,as it requires the least concentration of sulphide ions to exceed its solubility product.Since $10^{-54}$ is the smallest $K_{sp}$,$HgS$ will precipitate first.

$A$ 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 the $K_{sp}$ 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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