NEET UG 2002 — Chemistry Physical Chemistry

An electrolyte $\mathrm{MX}_2$MX2​ undergoes dissociation as follows :- $\mathrm{MX}_2 \rightleftharpoons \mathrm{M}^{+2}+2 \mathrm{X}^{-}$MX2​⇌M+2+2X− $\begin{array}{|c|c|c|c|} \hline \text {Concentration } & \mathrm{MX}_2 & \mathrm{M}^{+2} & \mathrm{X}^{-} \\ \hline \text {Initial concentration } & 1 & 0 & 0 \\ \hline \text {Concentration at Equilibrium } & 1-\mathrm{s} & \mathrm{s} & 2 \mathrm{~s} \\ \hline \end{array}$Concentration Initial concentration Concentration at Equilibrium ​MX2​11−s​M+20s​X−02 s​​ Thus from the above condition we can say that, $\mathrm{K}_{\mathrm{sp}}=\mathrm{s} \times(2 \mathrm{~s})^2=4 \times(\mathrm{s})^3$Ksp​=s×(2 s)2=4×(s)3 Here, s (the solubility) is $0.5 \times 10^{-4} \mathrm{~mole} / \mathrm{lit}$0.5×10−4 mole/lit. $\begin{aligned} & \therefore \mathrm{K}_{\mathrm{sp}}=4 \times\left(0.5 \times 10^{-4}\right)^3 \\ & \therefore \mathrm{K}_{\mathrm{sp}}=5 \times 10^{-13} \end{aligned}$​∴Ksp​=4×(0.5×10−4)3∴Ksp​=5×10−13​