A) \[{{10}^{-4}}M(A{{g}^{+}})\,and\,{{10}^{-4}}M(C{{l}^{-}})\]
B) \[{{10}^{-5}}M(A{{g}^{+}})\,and\,{{10}^{-5}}M(C{{l}^{-}})\]
C) \[{{10}^{-16}}M(A{{g}^{+}})\,and\,{{10}^{-6}}M(C{{l}^{-}})\]
D) \[{{10}^{-10}}M(A{{g}^{+}})\,and\,{{10}^{-10}}M(C{{l}^{-}})\]
Correct Answer: A
Solution :
[a] \[AgCl\rightleftharpoons [A{{g}^{+}}]+[C{{l}^{-}}]\] \[=\left( \frac{{{10}^{-4}}\times V}{2V} \right)\left( \frac{{{10}^{-4}}\times V}{2V} \right)=2.5\times {{10}^{-9}}\] This value is greater than \[{{K}_{sp}}=1.8\times {{10}^{-10}}\] (2) IP=\[\left( \frac{{{10}^{-5}}\times V}{2V} \right)\left( \frac{{{10}^{-5}}\times V}{2V} \right)=2.5\times {{10}^{-11}}<{{K}_{sp}}\] (3) IP=\[\left( \frac{{{10}^{-16}}\times V}{2V} \right)\left( \frac{{{10}^{-6}}\times V}{2V} \right)=2.5\times {{10}^{-23}}<{{K}_{sp}}\] (4) IP=\[\left( \frac{{{10}^{-10}}\times V}{2V} \right)\left( \frac{{{10}^{-10}}\times V}{2V} \right)=2.5\times {{10}^{-21}}<{{K}_{sp}}\]
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