JEE Main 12 Jan 2019 Evening Question 27
Question: If $ K _{sp} $ of $ Ag _2CO _3 $ is $ 8\times {10^{-12}}, $ the molar solubility of $ Ag _2CO _3 $ in $ 0.1,M,AgNO _3 $ is
[JEE Main Online Paper Held On 12-Jan-2019 Evening]
Options:
A) $ 8\times {10^{-11}}M $
B) $ 8\times {10^{-10}}M $
C) $ 8\times {10^{-13}}M $
D) $ 8\times {10^{-12}}M $
Show Answer
Answer:
Correct Answer: B
Solution:
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Let the solubility of $ Ag _2CO _3 $ be s.
Then, $ Ag _2CO _3\underset{2s}{\mathop{2A{g^{+}}}},+\underset{s}{\mathop{CO_3^{2-}}}, $ $ AgNO _3\underset{0.1}{\mathop{A{g^{{}}}}},+\underset{0.1}{\mathop{NO_3^{-}}}, $ $ [A{g^{+}}]=(2s+0.1) $ $ [CO_3^{2-}]=s $ $ K _{sP}={{[A{g^{+}}]}^{2}}[CO_3^{2-}]={{(2s+0.1)}^{2}}(s) $ $ =s\times (4s^{2}+0.01+0.4s)=4s^{3}+0.01s+0.4s^{2} $
Neglecting $ s^{3} $ and $ s^{2}, $ we get $ 8\times {10^{-12}}=0.01s $ $ s=\frac{8\times {10^{-12}}}{0.01}=8\times {10^{-10}}M $
