Chemistry - Ionic Equilibrium Question with Solution | TestHub

Given that,

$\mathrm{pH}$ of $\mathrm{Ba}{\left(\mathrm{OH}\right)}_2=12$

$\because \mathrm{pH}+\mathrm{pOH}=14$

$\therefore \mathrm{pOH}=14-\mathrm{pH}=14-12=2$

Now, $\mathrm{pOH}=-\log \left[{\mathrm{OH}}^{-}\right]$

$\Rightarrow 2=-\log \left[\mathrm{OH}\right]$

$\left[{\mathrm{OH}}^{-}\right]={10}^{-2} \mathrm{M}$

$\underset{\mathrm{S} \mathrm{mol} {\mathrm{L}}^{-1}}{\mathrm{Ba}{\left(\mathrm{OH}\right)}_2\left(\mathrm{s}\right)}\rightleftharpoons \underset{\mathrm{S}}{{\mathrm{Ba}}^{2+}}+\underset{2\mathrm{S}}{2{\mathrm{OH}}^{-}}$

$\left[{\mathrm{OH}}^{-}\right]=2\mathrm{S}={10}^{-2} \mathrm{M}$

$\therefore \mathrm{S}=0.5\times {10}^{-2}=5\times {10}^{-3} \mathrm{M}$

${\mathrm{K}}_{\mathrm{sp}}=\left[{\mathrm{Ba}}^{2+}\right]{\left[{\mathrm{OH}}^{-}\right]}^2$

$\Rightarrow (5\times {10}^{-3}){\left({10}^{-2}\right)}^2=5\times {10}^{-7}{\mathrm{M}}^3$

Therefore, the value of ${\mathrm{K}}_{\mathrm{sp}}$ is $5.0\times {10}^{-7} {\mathrm{M}}^3$.