Chemistry - Chemical Equilibrium Question with Solution | TestHub

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