If [S^{2-}] = 0.6 times 10^{-2} mol m^{-3} | vedclass

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Question

(A) Precipitation occurs when the ionic product $(Q_{sp})$ exceeds the solubility product $(K_{sp})$.For a salt $MS$,$Q_{sp} = [M^{2+}][S^{2-}]$.Given

Vedclass · Accepted Answer

$HgS$ $(K_{sp} = 10^{-54})$

Vedclass · Answer

(A) Precipitation occurs when the ionic product $(Q_{sp})$ exceeds the solubility product $(K_{sp})$.For a salt $MS$,$Q_{sp} = [M^{2+}][S^{2-}]$.Given $[S^{2-}] = 0.6 	imes 10^{-2} \ mol \ m^{-3} = 0.6 	imes 10^{-5} \ mol \ L^{-1}$ and $[M^{2+}] = 1 	imes 10^{-16} \ mol \ m^{-3} = 1 	imes 10^{-19} \ mol \ L^{-1}$.$Q_{sp} = (1 	imes 10^{-19}) 	imes (0.6 	imes 10^{-5}) = 0.6 	imes 10^{-24}$.$A$ substance precipitates first if its $K_{sp}$ value is the smallest relative to the ionic product.Comparing the given $K_{sp}$ values:$HgS: 10^{-54}$$MnS: 10^{-11}$$FeS: 10^{-21}$$ZnS: 10^{-25}$Since $HgS$ has the lowest $K_{sp}$ $(10^{-54})$,it will reach the precipitation condition $(Q_{sp} > K_{sp})$ first.

If $[S^{2-}] = 0.6 \times 10^{-2} \ mol \ m^{-3}$ and $[Hg^{2+}] = [Mn^{2+}] = [Fe^{2+}] = [Zn^{2+}] = 1 \times 10^{-16} \ mol \ m^{-3}$,which of the following will precipitate first in an aqueous solution?

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