Question

If the solubility product $K_{sp}$ of a sparingly soluble salt $M{X}_2$ at ${25}^{\circ }C$ is

$1.0\times {10}^{-11}$, the solubility of the salt in mole/litre at this temperature will

be

• A. $2.46\times {10}^{14}$
• B. $1.36\times {10}^{-4}$
• C. $2.60\times {10}^{-7}$
• D. $1.20\times {10}^{-10}$

Answer 1

The correct option is B

$1.36\times {10}^{-4}$

$M{X}_2\leftrightharpoons M^{+}+2X^{-}$;
At equilibrium, $\left[M^{+}\right]=s,\left[X^{-}\right]=2s$
$K_{sp}=4s^3$

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