Question

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

• 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)

Given, $pH =12$

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

We know that

$\left[ H ^{+}\right]\left[ OH ^{-}\right] =K_{w}=1 \times 10^{-14}$

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

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

$X( OH )_{2}$ dissolves as

$X ( OH )_{2} \longrightarrow \underset{X}{X ^{2+}}+ \underset{1\times 10^-2}{2OH ^{-}}$

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

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

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

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

$=5 \times 10^{-7}\, mol ^{3}\, L ^{-3}$