Question

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

• A.
• B.
• C.
• D.

Answer 1

The correct option is D

$SincepH=12$
$\therefore pOH=14-12=2$
$\therefore {\left[OH\right]}^{-}={10}^{-2}M$
We know Ba ${\left(OH\right)}_2\rightleftharpoons B{a}^{2+}+2O{H}^{-}$
$\therefore \left[B{a}^{2+}\right]=\frac{{10}^{-2}}{2}M$
$\therefore K_{sp}=\left[B{a}^{2+}\right]{\left[O{H}^{-}\right]}^2=\left(\frac{{10}^{-2}}{2}\right)\times ({10}^{-2}{)}^2=5\times {10}^{-7}$