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}} $
Show Answer
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 $
