Question

The $pH$ of a saturated solution of a metal hydroxide of formula $X(OH{)}_2$ is $12.0$ at $298K$. What is the solubility product of a metal hydroxide at $298K$ (in $mo{l}^3{L}^{-3})$?

• A. $2\times {10}^{-6}$
• B. $1\times {10}^{-7}$
• C. $5\times {10}^{-5}$
• D. $2\times {10}^{-5}$
• E. $5\times {10}^{-7}$

Answer 1

Answer: (E)

$5\times {10}^{-7}$

Given, $pH=12$

$\therefore \left[H^{+}\right]=1\times {10}^{-pH}=1\times {10}^{-12}$

We know that
$\left[H^{+}\right]\left[O{H}^{-}\right]=K_w=1\times {10}^{-14}$

$\therefore \left[O{H}^{-}\right]=\frac{1\times {10}^{-14}}{\left[H^{+}\right]}$

$=\frac{1\times {10}^{-14}}{1\times {10}^{-12}}=1\times {10}^{-2}M$

$X(OH{)}_2$ dissolves as

$X(OH{)}_2\to \underset{X}{X^{2+}}+\underset{2X}{2O{H}^{-}}$

$\therefore X^{2+}=\frac{\left[O{H}^{-}\right]}{2}=\frac{1\times {10}^{-2}}{2}=5\times {10}^{-3}$

$K_{sp}=\left[X^{2+}\right][2O{H}^{-}{]}^2$

$=(5\times {10}^{-3})(1\times {10}^{-2}{)}^2$

$=(5\times {10}^{-3})(1\times {10}^{-4})$

$=5\times {10}^{-7}mo{l}^3{L}^{-3}$.