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The solubility product of sparingly soluble salt $$\text{Ag}_2\text{CrO}_4$$ is $$4 \times 10^{-12}$$. The molar solubility of the salt (in mol per litre) is _____________
Correct Answer: 0.0001
$$\text{Ag}_2\text{CrO}_4 \rightleftharpoons 2\text{Ag}^+ + \text{CrO}_4^{2-}$$, so with solubility $$s$$, $$K_{sp} = (2s)^2 s = 4s^3$$. Then $$4s^3 = 4 \times 10^{-12}$$ gives $$s = 10^{-4}\text{ mol L}^{-1}$$.
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