Which one of the following arrangements represents the correct order of solubilities of sparingly soluble salts $\mathrm{Hg}_{2} \mathrm{Cl}_{2}, \mathrm{Cr}_{2}\left(\mathrm{SO}_{4}ight)_{3}, \mathrm{BaSO}_{4}$
and $\mathrm{CrCl}_{3}$ respectively?

Question

Aicharya · Accepted Answer

$\mathrm{Cr}_{2}\left(\mathrm{SO}_{4}ight)_{3} \leftrightharpoons \underset{2 s}{2 \mathrm{Cr}^{3+}}+\underset{3 s}{3} \mathrm{SO}_{4}^{2-}$
$\mathrm{K}_{\mathrm{sp}}=(2 s)^{2}(3 s)^{3}=4 \mathrm{~s}^{2} \times 27 s^{3}=108 s^{5}$
$\mathrm{s}=\left(\frac{\mathrm{K}_{s p}}{108}ight)^{1 / 5}$
$\mathrm{Hg}_{2} \mathrm{Cl}_{2} \leftrightharpoons \underset{2 s}{2 \mathrm{Hg}^{2+}}+\underset{2 s}{2 \mathrm{Cl}^{-}}$
$\mathrm{K}_{\mathrm{sp}}=(2 s)^{2} \times(2 s)^{2}=16 s^{4}$
$\mathrm{s}=\left(\frac{\mathrm{K}_{s p}}{16}ight)^{1 / 4}$
$\mathrm{BaSO}_{4} \leftrightharpoons \mathrm{Ba}^{2+}+\mathrm{SO}_{4}^{2-}$
$\mathrm{K}_{\mathrm{sp}}=s^{2}$
$\mathrm{s}=\sqrt{\mathrm{K}_{\mathrm{sp}}}$
$\mathrm{CrCl}_{3} \leftrightharpoons \mathrm{Cr}_{s}^{3+}+3 \mathrm{Cl}^{-}$
$\mathrm{K}_{\mathrm{sp}}=s \times(3 s)^{3}=27 s^{4}$
$s=\left(\frac{\mathrm{K}_{s p}}{27}ight)^{1 / 4}$
Hence the correct order of solubilities of salts is
$\sqrt{\mathrm{K}_{\mathrm{sp}}}>\left(\frac{\mathrm{K}_{\mathrm{sp}}}{16}ight)^{1 / 4}>\left(\frac{\mathrm{K}_{\mathrm{sp}}}{27}ight)^{1 / 4}>\left(\frac{\mathrm{K}_{\mathrm{sp}}}{108}ight)^{1 / 5}$ undefined

Which one of the following arrangements represents the correct order of solubilities of sparingly soluble…

Solution