Question

$pH$ of a saturated solution of $Ba(OH{)}_2$ is 12. Hence, Ksp of $Ba(OH{)}_2$ is:

• A. $5\times {10}^{-7}{M}^3$
• B. $5\times {10}^{-4}{M}^2$
• C. $1\times {10}^{-6}{M}^3$
• D. $4\times {10}^{-6}{M}^3$

Answer 1

The correct option is A $5\times {10}^{-7}{M}^3$

pH of solution = 12

$\left[H^{+}\right]={10}^{-12}$

$\left[O{H}^{-}\right]=\frac{{10}^{-14}}{{10}^{-12}}={10}^{-2}$

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

$23={10}^{-2}\Rightarrow 3=\frac{{10}^{-2}}{2}$

$K_{sp}=\left(s\right)(2s{)}^2=4s^3$

$=4\times {\left(\frac{{10}^{-2}}{2}\right)}^3$

$=\frac{4}{8}\times {10}^{-6}$

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