Chemistry - Ionic Equilibrium Question with Solution | TestHub

As$\text{pH}=\text{12}$
$\text{pOH}=\text{14}-\text{12}=2$
$\left[\text{OH}\right]=10^{-2}\mathrm{M}=0.0\text{1M}/\mathrm{L}$
In case of$\underset{\mathrm{s}}{\text{Ba}(\text{OH}{)}_2}\rightleftharpoons \underset{\mathrm{s}}{\text{B}{\text{a}}^{+2}}+\underset{2\mathrm{s}}{\text{2O}{\text{H}}^{-2}}$
As$\text{2s}=0.0\text{1}\mathrm{m}/\mathrm{L}$
$\mathrm{s}=\frac{0.01}{2}=5\times 10^{-3}\mathrm{}\mathrm{m}/\mathrm{L}$
Ksp for$\text{Ba}{\left(\text{OH}\right)}_2=\text{4}{\text{s}}^3=4{\left(5\times 10^{-3}\right)}^3=5\times 10^{-7}{\mathrm{M}}^3$
Hence$\mathrm{‘}n\mathrm{’}$= 5