Equilibrium Question 234
Question: The solubility product of $ PbCl_2 $ at $ 20^{o}C $ is $ 1.5\times {10^{-4}}. $ Calculate the solubility [Bihar CEE 1995; BHU 2002]
Options:
A) $ 3.75\times {10^{-4}} $
B) $ 3.34\times {10^{-2}} $
C) $ 3.34\times 10^{2} $
D) None of these
Show Answer
Answer:
Correct Answer: B
Solution:
$ \underset{S}{\mathop{PbCl_2}}, $ ⇌ $ \underset{S,}{\mathop{P{b^{++}}}},+\underset{{{(2S)}^{2}}}{\mathop{2C{l^{-}}}}, $
$ K _{sp}=S\times {{(2S)}^{2}}=4S^{3} $
$ S=\sqrt[3]{\frac{K _{sp}}{4}}=\sqrt[3]{\frac{1.5\times {10^{-4}}}{4}}=3.34\times {10^{-2}} $ .
