$k_{a}ofHCNandC{H}_{3}COOH$ are respectively $4.5\times {10}^{-10}$ and $1.5\times {10}^{-5}$ at 298k.Thus, equilibrium constant for $C{N}^{-}+C{H}_{3}COOH\stackrel{ }{\rightleftharpoons }C{H}_{3}CO{O}^{-}+HCN$ would be

$\begin{array}{l}HCN\underset{ }{\rightleftharpoons }{H}^{+}+C{N}^{-}{k}_a=4.5\times {10}^{-10}\\ C{H}_{3}COOH\underset{ }{\rightleftharpoons }{H}^{+}+C{H}_{3}CO{O}^{-},k_a^1=1.5\times {10}^{-5}\\ C{N}^{-}+H^{+}\underset{ }{\rightleftharpoons }HCN{k}_a^2=\frac{1}{k_a}\\ C{H}_{3}COOH+C{N}^{-}\stackrel{ }{\rightleftarrows }HCN+C{H}_{3}CO{O}^{-},k_{eq}=k_a^{\text{'}}{k}_a^{"}\\ =1.5\times {10}^{-5}\times \frac{1}{4.5\times {10}^{-10}}=3.33\times {10}^{+4}\end{array}$