Equilibrium Question 811
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}} $
BETA
