Equilibrium Question 208

Question: The solubility of $ PbCl_2 $ at $ 25^{o}C $ is $ 6.3\times {10^{-3}} $ mole/litre. Its solubility product at that temperature is [NCERT 1979; CPMT 1985]

Options:

A) $ (6.3\times {10^{-3}})\times (6.3\times {10^{-3}}) $

B) $ (6.3\times {10^{-3}})\times (12.6\times {10^{-3}}) $

C) $ (6.3\times {10^{-3}})\times {{(12.6\times {10^{-3}})}^{2}} $

D) $ (12.6\times {10^{-3}})\times (12.6\times {10^{-3}}) $

Show Answer

Answer:

Correct Answer: C

Solution:

$ PbCl_2\to \underset{S}{\mathop{P{b^{++}}}},+\underset{2S}{\mathop{2C{l^{-}}}}, $

$ K _{sp}=S\times {{(2S)}^{2}} $

$ =[6.3\times {10^{-3}}]\times {{[12.6\times {10^{-3}}]}^{2}} $ .



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