question_answer

For the reaction:\[\operatorname{Ag}\left( CN \right)_{2}^{-}A{{g}^{+}}+2C{{N}^{-}},\]the \[{{\operatorname{K}}_{c}}\] at \[25{}^\circ C\] is \[4\times {{10}^{-14}}.\] What will be the \[[A{{g}^{+}}]\] in solution which was originally 0.1 M in KCN and 0.03M in \[{{\operatorname{AgNO}}_{3}}.\]

A) \[75\times {{10}^{-13}}\]

B) \[75\times {{10}^{-12}}\]

C) \[7.5\times {{10}^{-12}}\]

D) \[7.5\times {{10}^{-13}}\]

Correct Answer: D

Solution :

\[\therefore \]\[{{[Ag{{(CN)}_{2}}]}^{-}}=0.03M\]

Since,\[{{\operatorname{K}}_{c}}\] is too small and dissociation of

\[\operatorname{Ag}\left( CN \right)_{2}^{-}\]is very less and thus,

\[\therefore \]\[0.04+2a\approx 0.04\text{ }\operatorname{and}\text{ }0.03-a\approx 0.03\]

\[\therefore \]\[{{[Ag{{(CN)}_{2}}]}^{-}}=0.03;[A{{g}^{+}}]=a,;[C{{N}^{-}}]=0.04\]

Now \[{{K}_{C}}=\frac{[A{{g}^{+}}]{{[C{{N}^{-}}]}^{2}}}{[Ag(CN)_{2}^{-}]}=\frac{a\times {{(0.04)}^{2}}}{0.03}\]

\[\therefore \]      \[a=7.5\times {{10}^{-13}}\]