Question

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

$Ba(OH{)}_2\rightleftarrows B{a}^{2+}+2O{H}^{-}$
$pH=12\Rightarrow p(OH)=14-pH$
$\to p(OH)=14-12=2$
$\left[O{H}^{-}\right]=10^{-POH}=10^{-2}$ or $1\times 10^{-2}$
as conc. of $B{a}^{2+}$ is half of $O{H}^{-}$
$\to B{a}^{2+}=0.5\times 10^{-2}$
$K_{sp}=\left(0.5\times 10^{-2}\right){\left(1\times 10^{-2}\right)}^2$
$K_{sp}=0.5\times 10^{-6}$
$=5\times 10^{-7}$