Question

The dissociation constants for acetic acid and $HCN$ at ${25}^{o}C$ are $1.5\times {10}^{-5}$ and $4.5\times {10}^{-10}$ respectively. The equilibrium constant for the following equilibrium would be:

${CN}^{-}+{CH}_{3}COOH\rightleftharpoons HCN+{CH}_{3}CO{O}^{-}$

• A. $3\times {10}^4$
• B. $3\times {10}^5$
• C. $3\times {10}^{-5}$
• D. $3\times {10}^{-4}$

Answer 1

The correct option is A $3\times {10}^4$

The dissociation constants for acetic acid and $HCN$ at ${25}^{o}C$ are $1.5\times {10}^{-5}$ and $4.5\times {10}^{-10}$ respectively. The equilibrium constant for the equilibrium ${CN}^{-}+{CH}_{3}COOH\rightleftharpoons HCN+{CH}_{3}CO{O}^{-}$ would be $3\times {10}^4$.
${CH}_{3}COOH\rightleftharpoons {CH}_{3}CO{O}^{-}+H^{+};K_1=1.5\times {10}^{-5}...\left(i\right)$

$HCN\rightleftharpoons H^{+}+{CN}^{-};K_2=4.5\times {10}^{-10}...\left(ii\right)$

Reverse equation (ii)
$H^{+}+{CN}^{-}\rightleftharpoons HCN;K_3=\frac{1}{K_2}=\frac{1}{4.5\times {10}^{-10}}...\left(iii\right)$

Add equation (i) and (iii) to get $K$
${CN}^{-}+{CH}_{3}COOH\rightleftharpoons HCN+{CH}_{3}CO{O}^{-}$

$K=K_1\times K_3=1.5\times {10}^{-5}\times \frac{1}{4.5\times {10}^{-10}}=3\times {10}^4$.