• # question_answer The dissociation constants for acetic acid and HCN at $25{}^\circ C$ are$1.5\times {{10}^{-5}}$and $4.5\times {{10}^{-10}},$ respectively. The equilibrium constant for the equilibrium. $C{{N}^{-}}+C{{H}_{3}}COOH\rightleftharpoons HCN+C{{H}_{3}}CO{{O}^{-}}$would be A) $3.0\times {{10}^{5}}$B) $3.0\times {{10}^{-5}}$C) $3.0\times {{10}^{-4}}$D) $3.0\times {{10}^{4}}$

Given, $C{{H}_{3}}COOH\rightleftharpoons C{{H}_{3}}O{{O}^{-}}+{{H}^{+}};$ ${{K}_{a1}}=1.5\times {{10}^{-5}}$ (i) ${{K}_{a2}}=4.5\times {{10}^{-10}}$ (ii) $HCN\rightleftharpoons {{H}^{+}}+C{{N}^{-}};$ $C{{N}^{-}}+C{{H}_{3}}COOH\rightleftharpoons$ $HCN+C{{H}_{3}}CO{{O}^{-}}$ $K=?$ On subtracting Eq. (ii) from Eq. (i), we get $C{{H}_{3}}COOH+C{{N}^{-}}\rightleftharpoons HCN+C{{H}_{3}}CO{{O}^{-}}$ $K=\frac{{{K}_{a1}}}{{{K}_{a2}}}=\frac{1.5\times {{10}^{-5}}}{4.5\times {{10}^{-10}}}=\frac{{{10}^{5}}}{3}=3.33\times {{10}^{4}}$