Question

$pH$ of $Ba{\left(OH\right)}_2$ solution is 12. Its solubility product is :

• A. $5.0\times {10}^{-7}$
• B. $0.6\times {10}^{-12}$
• C. $4.0\times {10}^{-8}$
• D. $5.0\times {10}^{-9}$

Answer 1

Answer: (A)

Given: $pH=12$

Hence, $pOH=14-pH=14-12=2$ or $\left[O{H}^{-}\right]={10}^{-2}$.

$\left[B{a}^{2+}\right]=0.5\times \left[O{H}^{-}\right]=0.5\times {10}^{-2}$.

The expression for the solubility product of barium hydroxide is,

$K_{sp}=\left[B{a}^{2+}\right][O{H}^{-}{]}^2=(0.5\times {10}^{-2}\left)\right(1\times {10}^{-2}{)}^2=5\times {10}^{-7}$.