Question

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

• A. 9
• B. 10
• C. 11
• D. 8

Answer 1

Answer: (B)

10

Answer 2

$Mg ^{2+} + 2OH^{-} {\rightleftharpoons} Mg\left(OH\right)_{2}$ $K_{sp} = \left[Mg^{2+}\right]\left[OH^{-}\right]^{2}$ $\left[OH^{-}\right] = \sqrt{\frac{K_{sp}}{\left[Mg^{2+}\right]}} = 10^{-4}$ $\therefore p^{OH} = 4$ and $p^{H} = 10$