Question

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

Answer: (B)

$5.0\times {10}^{-7}$

$Ba(OH{)}_2\rightleftharpoons \underset{s}{B{a}^{2+}}+\underset{2s}{2O{H}^{-}}$
$[OH{]}^{-}={10}^{-2}$
$2s={10}^{-2}$
$s=\frac{{10}^{-2}}{2}$
$K_{sp}=4s^3$
$=4\times {\left(\frac{{10}^{-2}}{2}\right)}^3$
$=5\times {10}^{-7}=5\times {10}^{-7}$