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

Answer 1

Answer: (D)

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

$Ba(OH{)}_2\rightleftarrows B{a}^{2+}+2O{H}^{-}$

$pH=12\Rightarrow p\left(OH\right)=14-pH$

$\to p\left(OH\right)=14-12=2$

$\left[O{H}^{-}\right]={10}^{-POH}={10}^{-2}or1\times {10}^{-2}$

as conc. of $B{a}^{2+}$ is half of $O{H}^{-}$

$\to B{a}^{2+}=0.5\times {10}^{-2}$

$K_{sp}=(0.5\times {10}^{-2})(1\times {10}^{-2}{)}^2$

$K_{sp}=0.5\times {10}^{-6}$

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