Question

pH of a saturated solution of $Ba(OH{)}_2$ is 12. The value of solubility product $\left(K_{sp}\right)ofBa(OH{)}_2$ is

• A. $3.3\times {10}^{-7}$
• B. $5.0\times {10}^{-7}$
• C. $4.0\times {10}^{-6}$
• D. $5.0\times {10}^{-6}$

Answer 1

The correct option is B $5.0\times {10}^{-7}$

pH of solution = 12
$\left[H^{+}\right]={10}^{-12}$
$\left[O{H}^{-}\right]=\frac{{10}^{-14}}{{10}^{-12}}={10}^{-2}$
$Ba\left(O{H}_2\right)\rightleftharpoons \underset{s}{B{a}^{2+}}+\underset{2s}{2O{H}^{-}}$
$2s={10}^{-2}\Rightarrow s=\frac{{10}^{-2}}{2}$
$K_{sp}=\left(s\right)(2s{)}^2=4s^3$
$=4\times {\left(\frac{{10}^{-2}}{2}\right)}^3=\frac{4}{8}\times {10}^{-6}=5\times {10}^{-7}$