Chemistry - Ionic Equilibrium Question with Solution | TestHub

$$\begin{array}{rl}& {\mathrm{K}}_{sp}=3.2\times 10^{-11}\\ & {\mathrm{AX}}_2\rightleftharpoons {\mathrm{A}}^{2+}+2{\mathrm{X}}^{-}\\ & \mathrm{S}2\mathrm{S}\\ & {\mathrm{K}}_{sp}=\mathrm{S}\times (2\mathrm{S}{)}^2=4{\mathrm{S}}^3;\\ & =3.2\times 10^{-11}=4{\mathrm{S}}^3\end{array}$$ or, ${\mathrm{S}}^3=0.8\times 10^{-11}=8\times 10^{-12}$ $$\therefore \mathrm{S}=2\times 10^{-4}\text{. }$$ Related Theory The solubility of a substance in a solvent is the total amount of the solute that can be dissolved in the solvent at equilibrium. On the other hand, the solubility product constant is an equilibrium constant that provides insight into the equilibrium between the solid solute and its constituent ions that are dissociated across the solution.