Equilibrium Question 188

Question: The solubility of $ AgCl $ in $ 0.2MNaCl $ solution $ (K _{sp} $ for $ AgCl=1.20\times {10^{-10}}) $ is [MP PET 1996]

Options:

A) $ 0.2M $

B) $ 1.2\times {10^{-10}},M $

C) $ 0.2\times {10^{-10}},M $

D) $ 6\times {10^{-10}},M $

Show Answer

Answer:

Correct Answer: D

Solution:

$ \underset{a}{\mathop{AgCl}}, $ ⇌ $ \underset{a}{\mathop{A{g^{+}}}},+\underset{a}{\mathop{C{l^{-}}}}, $

$ \underset{0.02}{\mathop{NaCl}}, $ ⇌ $ \underset{0.02,}{\mathop{N{a^{+}}}},+\underset{0.02}{\mathop{C{l^{-}}}}, $

$ K _{sp}AgCl=1.20\times {10^{-10}} $

$ K _{sp}AgCl=[A{g^{+}}][C{l^{-}}] $

$ =a\times [a+0.2] $

$ =a^{2}+0.2a $

$ a^{2} $ is a very small so it is a neglected. $ K _{sp}AgCl=0.2a $

$ 1.20\times {10^{-10}}=0.2a $

$ a=\frac{1.20\times {10^{-10}}}{0.20}=6\times {10^{-10}} $ mole



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