A) \[1.0\times {{10}^{-9}}\]
B) \[1.0\times {{10}^{-10}}\]
C) \[1.0\times {{10}^{-5}}\]
D) \[1.0\times {{10}^{-11}}\]
Correct Answer: A
Solution :
[a] Let solubility of \[AgCl=x\,mole/L\] \[AgCl\rightleftharpoons A{{g}^{+}}+C{{l}^{-}}\] i.e. \[{{K}_{sp(Agcl)}}=x\times x\] \[KCl\xrightarrow{{}}{{K}^{+}}+\underset{0.1}{\mathop{C{{l}^{-}}}}\,\] \[\left[ C{{l}^{-}} \right]\] from \[KCl=0.1m\] Total \[\left[ C{{l}^{-}} \right]\] in solution \[=x+0.1\] \[{{K}_{sp}}(AgCl)=[A{{g}^{+}}][C{{l}^{-}}]=x(x+0.1)\] \[1.0\times {{10}^{-10}}=x(x+0.1)\] \[1.0\times {{10}^{-10}}={{x}^{2}}+0.1x\] \[1.0\times {{10}^{-10}}=0.1x\left( as\,{{x}^{2}}<<1 \right)\] \[x=1.0\times {{10}^{-9}}mol/L\]You need to login to perform this action.
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