The dissociation constants for acetic acid and HCN at 25° C are 1.5 × 10^-5 and 4.5 × 10^-10, respectively. The equilibrium constant for the…

Given, $\mathrm{CH}_3 \mathrm{COOH} \rightleftharpoons \mathrm{CH}_3 \mathrm{COO}^{-}+\mathrm{H}^{+}$CH3​COOH⇌CH3​COO−+H+ $$\begin{array}{r} \mathrm{K}_{\mathrm{a}}=1.5 \times 10^{-5} \ldots \text { (i) } \\ \mathrm{HCN} \rightleftharpoons \mathrm{H}^{+}+\mathrm{CN}^{-} ; \quad \mathrm{K}_{\mathrm{a}}=4.5 \times 10^{-10} \ldots \text { (ii) } \\ \mathrm{CN}^{-}+\mathrm{CH}_3 \mathrm{COOH} \rightleftharpoons \mathrm{HCN}+\mathrm{CH}_3 \mathrm{COO}^{-} \\ \mathrm{K}=\text { ? } \end{array}$$Ka​=1.5×10−5… (i) HCN⇌H++CN−;Ka​=4.5×10−10… (ii) CN−+CH3​COOH⇌HCN+CH3​COO−K= ? ​ On substracting Eq. (ii) from Eq. (i), we get $$\begin{aligned} & \mathrm{CH}_3 \mathrm{COOH}+\mathrm{CN}^{-} \rightleftharpoons \mathrm{HCN}+\mathrm{CH}_3 \mathrm{COO}^{-} ; \\ & \mathrm{K}=\frac{\mathrm{K}_{\mathrm{a}}}{\mathrm{K}_{\mathrm{a}_1}}=\frac{1.5 \times 10^{-5}}{4.5 \times 10^{-10}}=\frac{10^5}{3}=3.33 \times 10^4 \end{aligned}$$​CH3​COOH+CN−⇌HCN+CH3​COO−;K=Ka1​​Ka​​=4.5×10−101.5×10−5​=3105​=3.33×104​ While adding two equations, dissociation constants are multiplied and when subtracting the equations, dissociation constants are divided. Alternative $$\begin{aligned} & \mathrm{CH}_3 \mathrm{COOH} \rightleftharpoons \mathrm{CH}_3 \mathrm{COO}^{-}+\mathrm{H}^{+} ; \mathrm{K}_{\mathrm{a}}=1.5 \times 10^{-5} \\ & \mathrm{~K}_{\mathrm{a}}=\frac{\left[\mathrm{CH}_3 \mathrm{COO}^{-}\right]\left[\mathrm{H}^{+}\right]}{\left[\mathrm{CH}_3 \mathrm{COOH}\right]}=1.5 \times 10^{-5} \\ & \mathrm{HCN} \rightleftharpoons \mathrm{H}^{+}+\mathrm{CN}^{-} ; \mathrm{K}_{\mathrm{a}}=4.5 \times 10^{-10} \\ & \mathrm{~K}_{\mathrm{a}}=\frac{\left[\mathrm{H}^{+}\right]\left[\mathrm{CN}^{-}\right]}{[\mathrm{HCN}]}=4.5 \times 10^{-10} \\ & \mathrm{CN}^{-}+\mathrm{CH}_3 \mathrm{COOH} \rightleftharpoons \mathrm{HCN}+\mathrm{CH}_3 \mathrm{COO}^{-} \\ & \mathrm{K}_{\mathrm{c}}=\frac{[\mathrm{HCN}]\left[\mathrm{CH}_3 \mathrm{COO}^{-}\right]}{\left[\mathrm{CN}^{-}\right]\left[\mathrm{CH}_3 \mathrm{COOH}\right]} \\ & \end{aligned}$$​CH3​COOH⇌CH3​COO−+H+;Ka​=1.5×10−5 Ka​=[CH3​COOH][CH3​COO−][H+]​=1.5×10−5HCN⇌H++CN−;Ka​=4.5×10−10 Ka​=[HCN][H+][CN−]​=4.5×10−10CN−+CH3​COOH⇌HCN+CH3​COO−Kc​=[CN−][CH3​COOH][HCN][CH3​COO−]​​ From Eqs. (i), (ii) and (iii), $$\mathrm{K}_c=\frac{1.5 \times 10^{-5}}{4.5 \times 10^{-10}}=3.33 \times 10^4$$Kc​=4.5×10−101.5×10−5​=3.33×104