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The solubility product, Ksp of a sparingly soluble salt, AX2, is 3.2 × 10-14 mol3lit-3. Its solubility is

  • A.4 × \(10^{-3}\)mol \(lit^{-1}\)
  • B.2 × \(10^{-2}\)mol \(lit^{-1}\)
  • C.2 × \(10^{-5}\)mol \(lit^{-1}\)
  • D.8 × \(10^{-4}\)mol \(lit^{-1}\)
  • E.2 × \(10^{-3}\)mol \(lit^{-1}\)

Answer: C

The solubility is \(2 \times 10^{-5}\) mol lit-1.

For AX2 dissolving to give A2+ and 2X-, a solubility \(s\) gives \([A^{2+}] = s\) and \([X^{-}] = 2s\):

\[ \begin{aligned} K_{sp} &= s(2s)^{2} = 4s^{3} \\ s^{3} &= \frac{3.2 \times 10^{-14}}{4} = 8 \times 10^{-15} \\ s &= 2 \times 10^{-5} \text{ mol lit}^{-1} \end{aligned} \]

Forgetting the 4 (using \(K_{sp} = s^{3}\)) gives about \(3.2 \times 10^{-5}\), and the \(10^{-3}\) and \(10^{-2}\) options misplace the cube root of the power of ten.

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