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
