A solution which is 10^{-3} M each in Mn^{2+ | vedclass

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(C) The ionic product for each metal sulphide is calculated as: $Q = [M^{2+}][S^{2-}] = 10^{-3} 	imes 10^{-16} = 10^{-19}$.Precipitation occurs when

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$HgS$

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(C) The ionic product for each metal sulphide is calculated as: $Q = [M^{2+}][S^{2-}] = 10^{-3} 	imes 10^{-16} = 10^{-19}$.Precipitation occurs when the ionic product exceeds the solubility product $(K_{sp})$.Comparing the $K_{sp}$ values: $MnS (10^{-15}), FeS (10^{-23}), ZnS (10^{-20}), HgS (10^{-54})$.Since $HgS$ has the lowest $K_{sp}$ $(10^{-54})$,it will be the first to satisfy the condition $Q > K_{sp}$ as the sulphide ion concentration increases,and it 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 $K_{sp}$ 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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