Which of the following is most soluble?

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$MnS (K_{sp}^{\circ} = 7 \times 10^{- 16})$

$CuS (K_{sp}^{\circ} = 8 \times 10^{- 37})$

${Ag}_{2} S (K_{sp}^{\circ} = 6 \times 10^{- 51})$

${Bi}_{2} S_{3} (K_{sp}^{\circ} = 1 \times 10^{- 70})$

answer is D.

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Detailed Solution

The following is most soluble is $MnS (K_{sp}^{\circ} = 7 \times 10^{- 16})$ .

An electrolyte becomes more soluble as the ionic product, or solubility product, rises. Now,

$K_{sp} (MnS) > K_{sp} (CuS) > K_{sp} ({Ag}_{2} S) > K_{sp} ({Bi}_{2} S_{3})$

MnS will be the most soluble as a result.

Hence the option D is correct.

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