The dissociation constants for acetic acid and HCN at 25^ C are 1.5...

Given, $\mathrm{CH}_3 \mathrm{COOH} \rightleftharpoons \mathrm{CH}_3 \mathrm{COO}^{-}+\mathrm{H}^{+}$ \begin{array}{r}\mathrm{K}_{\mathrm{a}}=1.5 \times10^{-5} \ldots \text{ (i) }\\ \mathrm{HCN} \rightleftharpoons \mathrm{H}^{+}+\mathrm{CN}^{-}; \quad \mathrm{K}_{\mathrm{a}}=4.5 \times10^{-10} \ldots \text{ (ii) }\\ \mathrm{CN}^{-}+\mathrm{CH}}_{3} \mathrm{COOH} \rightleftharpoons \mathrm{HCN}+\mathrm{CH}}_{3} \mathrm{COO}^{-}\\ \mathrm{K}=\text{ ? } \end{array}On substracting Eq. (ii) from Eq. (i), we get \begin{aligned}& \mathrm{CH}}_{3} \mathrm{COOH}+\mathrm{CN}^{-} \rightleftharpoons \mathrm{HCN}+\mathrm{CH}}_{3} \mathrm{COO}^{-}; \\ & \mathrm{K}=\frac{\mathrm{K}_{\mathrm{a}}}{\mathrm{K}_{\mathrm{a}_1}}=\frac{1.5 \times 10^{-5}}{4.5 \times 10^{-10}}=\frac{10^5}{3}=3.33 \times10^4 \end{aligned}While adding two equations, dissociation constants are multiplied and when subtracting the equations, dissociation constants are divided. Alternative \begin{aligned}& \mathrm{CH}}_{3} \mathrm{COOH} \rightleftharpoons \mathrm{CH}}_{3} \mathrm{COO}^{-}+\mathrm{H}^{+}; \mathrm{K}_{\mathrm{a}}=1.5 \times10^{-5} \\ & \mathrm{~K}_{\mathrm{a}}=\frac{\left[\mathrm{CH}_3 \mathrm{COO}^{-}\right]\left[\mathrm{H}^{+}\right]}{\left[\mathrm{CH}_3 \mathrm{COOH}\right]}=1.5 \times10^{-5} \\ & \mathrm{HCN} \rightleftharpoons \mathrm{H}^{+}+\mathrm{CN}^{-}; \mathrm{K}_{\mathrm{a}}=4.5 \times10^{-10} \\ & \mathrm{~K}_{\mathrm{a}}=\frac{\left[\mathrm{H}^{+}\right]\left[\mathrm{CN}^{-}\right]}{[\mathrm{HCN}]}=4.5 \times10^{-10} \\ & \mathrm{CN}^{-}+\mathrm{CH}}_{3} \mathrm{COOH} \rightleftharpoons \mathrm{HCN}+\mathrm{CH}}_{3} \mathrm{COO}^{-}\\ & \mathrm{K}_{\mathrm{c}}=\frac{[\mathrm{HCN}]\left[\mathrm{CH}_3 \mathrm{COO}^{-}\right]}{\left[\mathrm{CN}^{-}\right]\left[\mathrm{CH}_3 \mathrm{COOH}\right]}\\ & \end{aligned}From Eqs. (i), (ii) and (iii), \mathrm{K}_c=$\frac{1.5 \times 10^{-5}}{4.5 \times 10^{-10}}=$3.33 $\times$10^4