Question
Which one of the following arrangements represents the correct order of solubilities of sparingly soluble salts $$H{g_2}C{l_2},C{r_2}{\left( {S{O_4}} \right)_3},BaS{O_4}$$ and $$CrC{l_3}$$ respectively ?
A.
$$BaS{O_4} > H{g_2}C{l_2} > C{r_2}{\left( {S{O_4}} \right)_3} > CrC{l_3}$$
B.
$$BaS{O_4} > H{g_2}C{l_2} > CrC{l_3} > C{r_2}{\left( {S{O_4}} \right)_3}$$
C.
$$BaS{O_4} > CrC{l_3} > H{g_2}C{l_2} > C{r_2}{\left( {S{O_4}} \right)_3}$$
D.
$$H{g_2}C{l_2} > BaS{O_4} > CrC{l_3} > C{r_2}{\left( {S{O_4}} \right)_3}$$
Answer :
$$BaS{O_4} > H{g_2}C{l_2} > CrC{l_3} > C{r_2}{\left( {S{O_4}} \right)_3}$$
Solution :
$$\eqalign{
& C{r_2}{\left( {S{O_4}} \right)_3} \rightleftharpoons \mathop {2C}\limits_{2s} {r^{3 + }} + \mathop {3S}\limits_{3s} O_4^{2 - } \cr
& {K_{sp}} = {\left( {2s} \right)^2}{\left( {3s} \right)^3} = 4{s^2} \times 27{s^3} = 108{s^5} \cr
& s = {\left( {\frac{{{K_{sp}}}}{{108}}} \right)^{\frac{1}{5}}} \cr
& H{g_2}C{l_2} \rightleftharpoons 2\mathop {Hg_{}^{2 + }}\limits_{2s} + 2\mathop {C{l^ - }}\limits_{2s} \cr
& {K_{sp}} = {\left( {2s} \right)^2} \times {\left( {2s} \right)^2} = 16{s^4} \cr
& s = {\left( {\frac{{{K_{sp}}}}{{16}}} \right)^{\frac{1}{4}}} \cr
& BaS{O_4} \rightleftharpoons \mathop {Ba_{}^{2 + }}\limits_s + \mathop {SO}\limits_s \,_4^{2 - } \cr
& {K_{sp}} = {s^2} \cr
& s = \sqrt {{K_{sp}}} \cr
& CrC{l_3} \rightleftharpoons C\mathop {r_{}^{3 + }}\limits_s + 3\mathop {C{l^ - }}\limits_{3s} \cr
& {K_{sp}} = s \times {\left( {3s} \right)^3} = 27{s^4} \cr
& s = {\left( {\frac{{{K_{sp}}}}{{27}}} \right)^{\frac{1}{4}}} \cr} $$
Hence the correct order of solubilities of salts is
$$\sqrt {{K_{sp}}} > {\left( {\frac{{{K_{sp}}}}{{16}}} \right)^{\frac{1}{4}}} > {\left( {\frac{{{K_{sp}}}}{{27}}} \right)^{\frac{1}{4}}} > {\left( {\frac{{{K_{sp}}}}{{108}}} \right)^{\frac{1}{5}}}$$