Discussion
If "S" is the solubility of X3Y2 in pure water, assuming that neither kind of ion reacts with water, then, the solubility product, Ksp is
Answer: B
\(K_{sp}\) is \(108\,S^{5}\).
Each formula unit dissolving gives 3 X ions and 2 Y ions, so a solubility S gives [X] = 3S and [Y] = 2S:
\[ K_{sp} = (3S)^{3}(2S)^{2} = 27S^{3} \times 4S^{2} = 108S^{5} \]
The options with \(S^{2}\), \(S^{4}\) or \(S^{6}\) use the wrong total number of ions (the power of S is 3 + 2 = 5), and 27 drops the factor 4 from \((2S)^{2}\).
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