Question

Which one of the following arrangements represents the correct order of solubilities of sparingly soluble salts $Hg_2Cl_2, Cr_2(SO_4)_3, BaSO_4$ and $CrCl_3$ respectively ?

• A. $BaSO_4 > Hg_2Cl_2 > Cr_2(SO_4)_3 > CrCl_3$
• B. $BaSO_4 > Hg_2Cl_2 > CrCl_3 > Cr_2(SO_4)_3$
• C. $BaSO_4 > CrCl_3 > Hg_2Cl_2 > Cr_2 (SO_4 )_3$
• D. $Hg_2Cl_2 > BaSO_4 > CrCl_3 > Cr_2 (SO_4 )_3$

Answer 1

Answer: (B)

$ Cr _2\left( SO _4\right)_3 \rightleftharpoons 2 \underset{2s}{Cr} ^{+++}+3 \underset{3s}{So} _4^{--}$
$K_{s p}(2 s)^2(3 s)^3=4 s^2 \times 27 s^3=108 s^5 $
$s=\left(\frac{K_{s p}}{108}\right)^{1 / 5}$
$Hg _2 Cl _2 \rightleftharpoons 2 \underset{2s}{Hg} ^{2+}+2 \underset{2s}{Cl} ^{-} $
$K_{s p}=(2 s)^2 \times(2 s)^2=16 s^4$
$s=\left(\frac{K_{s p}}{16}\right)^{1 / 4} $
$ BaSO _4 \rightleftharpoons \underset{s}{Ba} ^{++} \underset{s}{SO} _4^{--}$
$K_{s p}=s^2$
$s=\sqrt{K_{s p}}$
$ CrCl _2 \rightleftharpoons \underset{s}{Cr} ^{3+}+3 \underset{3s}{Cl} ^{-} $
$K_{s p}=s \times(3 s)^3=27 s^4$
$s=\left(\frac{K_{s p}}{27}\right)^{1 / 4}$
Hence the correct order of solubilities of salts is
$\sqrt{K_{s p}}>\left(\frac{K_{s p}}{16}\right)^{1 / 4}>\left(\frac{K_{s p}}{27}\right)^{1 / 4}>\left(\frac{K_{s p}}{108}\right)^{1 / 5}$