Equilibrium Question 845

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-

Options:

A) $ BaSO_4>Hg_2Cl_2>Cr{{(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} $

Show Answer

Answer:

Correct Answer: B

Solution:

$ Cr_2{{(SO_4)}_3}\rightarrow \underset{2s}{\mathop{2C{r^{3+}}}},+\underset{3s}{\mathop{3SO_4^{2-}}}K _{sp}={{(2s)}^{2}}{{(3s)}^{3}}=4s^{2}\times 27s^{3}=108s^{5} $

$ s={{( \frac{K _{sp}}{108} )}^{1/5}} $

$ Hg_2Cl_2\rightarrow \underset{2s}{\mathop{2H{g^{2+}}}},+\underset{2s}{\mathop{2C{l^{-}}}}, $

$ K _{sp}={{(2s)}^{2}}\times {{(2s)}^{2}}=16s^{4} $

$ s={{( \frac{K _{sp}}{16} )}^{1/4}} $

$ BaSO_4\rightarrow \underset{s}{\mathop{B{a^{2+}}}},+\underset{s}{\mathop{SO_4^{2-}}}, $

$ K _{sp}=s^{2} $

$ s=\sqrt{K _{sp}} $

$ CrCl_3\rightarrow \underset{s}{\mathop{C{r^{3+}}}},+\underset{3s}{\mathop{3C{l^{-}}}},~ $

$ K _{sp}=s\times {{(3s)}^{3}}=27s^{4} $

$ s={{( \frac{K _{sp}}{27} )}^{1/4}} $ Hence the correct order of solubilities of salts is $ \sqrt{K _{sp}}>{{( \frac{K _{sp}}{16} )}^{1/4}}>{{( \frac{K _{sp}}{27} )}^{1/4}}>{{( \frac{K _{sp}}{108} )}^{1/5}} $



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