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(B) The solubility $(s)$ of a sparingly soluble salt depends on its $K_{sp}$ and its stoichiometric dissociation.For $BaSO_4$ ($1:1$ type): $K_{sp} =

Vedclass · Accepted Answer

$BaSO_4 > Hg_2Cl_2 > CrCl_3 > Cr_2(SO_4)_3$

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(B) The solubility $(s)$ of a sparingly soluble salt depends on its $K_{sp}$ and its stoichiometric dissociation.For $BaSO_4$ ($1:1$ type): $K_{sp} = s^2 \implies s = (K_{sp})^{1/2}$.For $Hg_2Cl_2$ ($1:2$ type): $K_{sp} = (s)(2s)^2 = 4s^3 \implies s = (K_{sp}/4)^{1/3}$.For $CrCl_3$ ($1:3$ type): $K_{sp} = (s)(3s)^3 = 27s^4 \implies s = (K_{sp}/27)^{1/4}$.For $Cr_2(SO_4)_3$ ($2:3$ type): $K_{sp} = (2s)^2(3s)^3 = 108s^5 \implies s = (K_{sp}/108)^{1/5}$.Comparing the powers of solubility,the order of solubility for these salts is $BaSO_4 > Hg_2Cl_2 > CrCl_3 > Cr_2(SO_4)_3$.

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?

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