Chemistry - Ionic Equilibrium Question with Solution | TestHub

$$\begin{array}{rl}& {\mathrm{Cr}}_2{\left({\mathrm{SO}}_4\right)}_3\leftrightharpoons 2{\mathrm{Cr}}^{+++}+3{\mathrm{SO}}_4^{--}\\ & {\mathrm{K}}_{\mathrm{sp}}=(2s{)}^2(3s{)}^3=4{\mathrm{s}}^2\times 27s^3=108s^5\\ & \mathrm{s}={\left(\frac{{\mathrm{K}}_{sp}}{108}\right)}^{1/5}\\ & {\mathrm{Hg}}_2{\mathrm{Cl}}_2\leftrightharpoons 2{\mathrm{Hg}}^{2+}+2{\mathrm{Cl}}^{-}\\ & {\mathrm{K}}_{\mathrm{sp}}=(2s{)}^2\times (2s{)}^2=16s^4\\ & \mathrm{s}={\left(\frac{{\mathrm{K}}_{sp}}{16}\right)}^{1/4}\\ & {\mathrm{BaSO}}_4\leftrightharpoons {\mathrm{Ba}}^{++}+{\mathrm{SO}}_4^{-}\\ & {\mathrm{K}}_{\mathrm{sp}}=s^2\\ & \mathrm{s}=\sqrt{{\mathrm{K}}_{\mathrm{sp}}}\\ & {\mathrm{CrCl}}_3\leftrightharpoons {\mathrm{Cr}}^{3+}+3{\mathrm{Cl}}^{-}\\ & s\end{array}$$ $$\begin{array}{rl}& {\mathrm{K}}_{\mathrm{sp}}=s\times (3s{)}^3=27s^4\\ & s={\left(\frac{{\mathrm{K}}_{sp}}{27}\right)}^{1/4}\end{array}$$ Hence the correct order of solubilities of salts is $$\sqrt{{\mathrm{K}}_{\mathrm{sp}}}>{\left(\frac{{\mathrm{K}}_{\mathrm{sp}}}{16}\right)}^{1/4}>{\left(\frac{{\mathrm{K}}_{\mathrm{sp}}}{27}\right)}^{1/4}>{\left(\frac{{\mathrm{K}}_{\mathrm{sp}}}{108}\right)}^{1/5}$$