Equilibrium Question 849

Question: The solubility (in $ mol{L^{-1}} $ ) of $ AgCl $

$ (K _{sp}=1.0\times {10^{-10}}) $ in a 0.1 M $ KCl $ solution will be

Options:

A) $ 1.0\times {10^{-9}} $

B) $ 1.0\times {10^{-10}} $

C) $ 1.0\times {10^{-5}} $

D) $ 1.0\times {10^{-11}} $

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Answer:

Correct Answer: A

Solution:

Let solubility of $ AgCl=x,mole/L $

$ AgCl\rightarrow A{g^{+}}+C{l^{-}} $ i.e. $ {K_{sp(Agcl)}}=x\times x $

$ KCl\xrightarrow{{}}{K^{+}}+\underset{0.1}{\mathop{C{l^{-}}}}, $

$ [ C{l^{-}} ] $ from $ KCl=0.1m $ Total $ [ C{l^{-}} ] $ in solution $ =x+0.1 $

$ K _{sp}(AgCl)=[A{g^{+}}][C{l^{-}}]=x(x+0.1) $

$ 1.0\times {10^{-10}}=x(x+0.1) $

$ 1.0\times {10^{-10}}=x^{2}+0.1x $

$ 1.0\times {10^{-10}}=0.1x( as,x^{2}«1 ) $

$ x=1.0\times {10^{-9}}mol/L $



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