Chemistry - Ionic Equilibrium Question with Solution | TestHub

An electrolyte ${\mathrm{MX}}_2$ undergoes dissociation as follows :- ${\mathrm{MX}}_2\rightleftharpoons {\mathrm{M}}^{+2}+2{\mathrm{X}}^{-}$ $\begin{array}{cccc}\text{Concentration }& {\mathrm{MX}}_2& {\mathrm{M}}^{+2}& {\mathrm{X}}^{-}\\ \text{Initial concentration }& 1& 0& 0\\ \text{Concentration at Equilibrium }& 1-\mathrm{s}& \mathrm{s}& 2\mathrm{s}\end{array}$ Thus from the above condition we can say that, ${\mathrm{K}}_{\mathrm{sp}}=\mathrm{s}\times (2\mathrm{s}{)}^2=4\times (\mathrm{s}{)}^3$ Here, s (the solubility) is $0.5\times 10^{-4}\mathrm{mole}/\mathrm{lit}$. $\begin{array}{rl}& \therefore {\mathrm{K}}_{\mathrm{sp}}=4\times {\left(0.5\times 10^{-4}\right)}^3\\ & \therefore {\mathrm{K}}_{\mathrm{sp}}=5\times 10^{-13}\end{array}$