Question

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

• A. 10-6
• B. 4 × 10-6
• C. 5 × 10-7
• D. None of these

Answer 1

The correct option is C

5 × 10-7

$pH$ = $12$ so $\left[O{H}^{-}\right]$ = ${10}^{-2}M$

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

$5\times {10}^{-3}M{10}^{-2}M$

So $K_{sp}$ = $\left[B{a}^{2+}\right][O{H}^{-}{]}^2$

= $(5\times {10}^{-3})({10}^{-2}{)}^2$

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