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}} $ .
