Question

If the solubility product $K_{sp}$ of a sparingly soluble salt $M{X}_2$ at ${25}^{o}C$ is $1.0\times {10}^{-11}$, then the solubility of the salt in mole $litr{e}^{-1}$ at this temperature will be:

• A. $2.46\times {10}^{14}$
• B. $1.35\times {10}^{-4}$
• C. $2.60\times {10}^{-7}$
• D. $1.20\times {10}^{-10}$

Answer 1

The correct option is B $1.35\times {10}^{-4}$

$M{X}_2\rightleftharpoons {\underset{\left(S\right)}{M}}^{2+}+{\underset{{\left(2S\right)}^2}{2X}}^{-};K_{sp}=4S^3$

$S=\sqrt[3]{3\frac{K_{sp}}{4}}=\sqrt[3]{3\frac{1\times {10}^{-11}}{4}}=1.35\times {10}^{-4}$