• # 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) $7.5\times {{10}^{-13}}$

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 ${{K}_{C}}=\frac{[A{{g}^{+}}]{{[C{{N}^{-}}]}^{2}}}{[Ag(CN)_{2}^{-}]}=\frac{a\times {{(0.04)}^{2}}}{0.03}$ $\therefore$      $a=7.5\times {{10}^{-13}}$

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