JEE Main 2013 — Chemistry Physical Chemistry

$$\text { } \begin{aligned} & \mathrm{Cr}_2\left(\mathrm{SO}_4\right)_3 \leftrightharpoons 2 \mathrm{Cr}^{+++}+3 \mathrm{SO}_4^{--} \\ & \mathrm{K}_{\mathrm{sp}}=(2 s)^2(3 s)^3=4 \mathrm{~s}^2 \times 27 s^3=108 s^5 \\ & \mathrm{~s}=\left(\frac{\mathrm{K}_{s p}}{108}\right)^{1 / 5} \\ & \mathrm{Hg}_2 \mathrm{Cl}_2 \leftrightharpoons 2 \mathrm{Hg}^{2+}+2 \mathrm{Cl}^{-} \\ & \mathrm{K}_{\mathrm{sp}}=(2 s)^2 \times(2 s)^2=16 s^4 \\ & \mathrm{~s}=\left(\frac{\mathrm{K}_{s p}}{16}\right)^{1 / 4} \\ & \mathrm{BaSO}_4 \leftrightharpoons \mathrm{Ba}^{++}+\mathrm{SO}_4^{-} \\ & \mathrm{K}_{\mathrm{sp}}=s^2 \\ & \mathrm{~s}=\sqrt{\mathrm{K}_{\mathrm{sp}}} \\ & \mathrm{CrCl}_3 \leftrightharpoons \mathrm{Cr}^{3+}+3 \mathrm{Cl}^{-} \\ & s \end{aligned}$$ ​Cr2​(SO4​)3​⇋2Cr++++3SO4−−​Ksp​=(2s)2(3s)3=4 s2×27s3=108s5 s=(108Ksp​​)1/5Hg2​Cl2​⇋2Hg2++2Cl−Ksp​=(2s)2×(2s)2=16s4 s=(16Ksp​​)1/4BaSO4​⇋Ba+++SO4−​Ksp​=s2 s=Ksp​​CrCl3​⇋Cr3++3Cl−s​ $$\begin{aligned} & \mathrm{K}_{\mathrm{sp}}=s \times(3 s)^3=27 s^4 \\ & s=\left(\frac{\mathrm{K}_{s p}}{27}\right)^{1 / 4} \end{aligned}$$​Ksp​=s×(3s)3=27s4s=(27Ksp​​)1/4​ Hence the correct order of solubilities of salts is $$\sqrt{\mathrm{K}_{\mathrm{sp}}}>\left(\frac{\mathrm{K}_{\mathrm{sp}}}{16}\right)^{1 / 4}>\left(\frac{\mathrm{K}_{\mathrm{sp}}}{27}\right)^{1 / 4}>\left(\frac{\mathrm{K}_{\mathrm{sp}}}{108}\right)^{1 / 5}$$Ksp​​>(16Ksp​​)1/4>(27Ksp​​)1/4>(108Ksp​​)1/5