Chemistry - Ionic Equilibrium Question with Solution | TestHub

$\mathrm{C}\mathrm{d}\mathrm{S}\rightleftharpoons \underset{0.1}{\mathrm{C}{\mathrm{d}}^{2+}}+\underset{\mathrm{s}}{{\mathrm{S}}^{2-}}$
For precipitation of CdS
$\left[{\mathrm{Cd}}^{2+}]\right[{\mathrm{S}}^{2-}]={\mathrm{K}}_{\mathrm{sp}(\mathrm{CdS})}$
$0.1\times [{\mathrm{S}}^{2-}]=8\times 10^{-27}$
$[{\mathrm{S}}^{2-}]=8\times 10^{-26}$
For precipitation of ZnS
$\left[{\mathrm{Zn}}^{2+}]\right[{\mathrm{S}}^{2-}]={\mathrm{K}}_{\mathrm{sp}(\mathrm{ZnS})}$
$0.1\times [{\mathrm{S}}^{2-}]=10^{-21}$
$[{\mathrm{S}}^{2-}]=10^{-20}$
CdS will start precipitate as the conc of$[{\mathrm{S}}^{2-}]=8\times 10^{-26}$and ZnS will start precipitate as the conc of$[{\mathrm{S}}^{2-}]=10^{-20}$so we can raise the concentration of$[{\mathrm{S}}^{2-}]$upto$10^{-20}$
Now
$\underset{\underset{\approx 0.1}{0.1(1-\mathrm{\alpha })}}{{\mathrm{H}}_2\mathrm{S}}\rightleftharpoons \underset{\underset{=\frac{\mathrm{x}}{20}\mathrm{M}}{2\times 0.1\mathrm{\alpha }}}{2{\mathrm{H}}^{+}}+\underset{0.1\mathrm{\alpha }=10^{-20}}{{\mathrm{S}}^{2-}}$
$\frac{{\left(\frac{\mathrm{x}}{20}\right)}^2\times 10^{-20}}{0.1}=10^{-21}$
${\left(\frac{\mathrm{x}}{20}\right)}^2=10^{-2}$
$\frac{\mathrm{x}}{20}=10^{-1}$
$\mathrm{x}=2$