Equilibrium Question 153
Question: The solubility in water of a sparingly soluble salt $ AB_2 $ is $ 1.0\times {10^{-5}}mol,{l^{-1}} $ . Its solubility product number will be [AIEEE 2003]
Options:
A) $ 4\times {10^{-15}} $
B) $ 4\times {10^{-10}} $
C) $ 1\times {10^{-15}} $
D) $ 1\times {10^{-10}} $
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Answer:
Correct Answer: A
Solution:
$ AB_2 $ ⇌ $ \underset{1\times {10^{-5}}}{\mathop{{A^{+}},}},+,\underset{2\times {10^{-5}}}{\mathop{2{B^{-}}}}, $
$ K _{sp}=[1\times {10^{-5}}]{{[2\times {10^{-5}}]}^{2}}=4\times {10^{-15}} $
