Equilibrium Question 205
Question: At 298 K, the solubility of $ PbCl_2 $ is $ 2\times {10^{-2}}mol/lit $ , then $ K _{sp}= $ [RPMT 2002]
Options:
A) $ 1\times {10^{-7}} $
B) $ 3.2\times {10^{-7}} $
C) $ 1\times {10^{-5}} $
D) $ 3.2\times {10^{-5}} $
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Answer:
Correct Answer: D
Solution:
$ PbCl_2 $ ⇌ $ \underset{(S)}{\mathop{P{b^{2+}}}},+\underset{{{(2S)}^{2}}}{\mathop{2C{l^{-}}}}, $
$ K _{sp}=4S^{3}=4\times {{(2\times {10^{-2}})}^{3}}=3.2\times {10^{-5}} $