Question

The pH at which Mg(OH)$_2$ $[K_{sp} = 1 \times 10^{-11}]$ begins to precipitate from a solution containing 0.10 M Mg$^{2+}$ ions is \\_\\_\\_\\_.

Answer 1

Precipitation occurs when $Q_p = K_{sp}$. The solubility product expression is:

$[\text{Mg}^{2+}][\text{OH}^-]^2 = K_{sp}$

Given:

$0.1 \times [\text{OH}^-]^2 = 10^{-11}$

Solving for $[\text{OH}^-]$:

$[\text{OH}^-] = 10^{-5}$

Then,

$\text{pOH} = 5 \Rightarrow \text{pH} = 9$