Question

At ${25}^{\circ }C$, the solubility product of $Mg(OH{)}_2$ is $1.0\times {10}^{-11}$. At which $pH$, will $M{g}^{2+}$ ions start precipitating in the form of $Mg(OH{)}_2$ from a solution of $0.001MM{g}^{2+}$ ions ?

• A. 10
• B. 15
• C. 1
• D. 20

Answer 1

The Correct Option is A

Explanation:
We know,$M{g}^{2+}+2O{H}^{-}\rightleftharpoons Mg(OH{)}_2$$K_{sp}=\left[M{g}^{2+}\right]{\left[O{H}^{-}\right]}^2$ (Equation for equilibrium)$\left[O{H}^{-}\right]=\sqrt{\frac{K_{sp}}{\left[M{g}^{2+}\right]}}={10}^{-4}$$\therefore pOH=4$ and $pH=10$Hence, the correct answer is '10'.