The dissociation constants for acetic acid and $\mathrm{HCN}$ at $25^{\circ} \mathrm{C}$ are $1.5 \times…
The dissociation constants for acetic acid and $\mathrm{HCN}$ at $25^{\circ} \mathrm{C}$ are $1.5 \times 10^{-5}$ and $4.5 \times 10^{-10}$, respectively. The equilibrium constant for the equilibrium,
$\mathrm{CN}^{-}+\mathrm{CH}_3 \mathrm{COOH} \quad \mathrm{HCN}+\mathrm{CH}_3 \mathrm{COO}^{-}$
would be
$3.0 \times 10^5$
$3.0 \times 10^{-5}$
$3.0 \times 10^{-4}$
$3.0 \times 10^4$
Solution
Given, $\mathrm{CH}_3 \mathrm{COOH} \rightleftharpoons \mathrm{CH}_3 \mathrm{COO}^{-}+\mathrm{H}^{+}$
$\begin{array}{r}
\mathrm{K}_{\mathrm{a}}=1.5 \times 10^{-5} \ldots \text { (i) } \\
\mathrm{HCN} \rightleftharpoons \mathrm{H}^{+}+\mathrm{CN}^{-} ; \quad \mathrm{K}_{\mathrm{a}}=4.5 \times 10^{-10} \ldots \text { (ii) } \\
\mathrm{CN}^{-}+\mathrm{CH}_3 \mathrm{COOH} \rightleftharpoons \mathrm{HCN}+\mathrm{CH}_3 \mathrm{COO}^{-} \\
\mathrm{K}=\text { ? }
\end{array}$
On substracting Eq. (ii) from Eq. (i), we get
$\begin{aligned}
& \mathrm{CH}_3 \mathrm{COOH}+\mathrm{CN}^{-} \rightleftharpoons \mathrm{HCN}+\mathrm{CH}_3 \mathrm{COO}^{-} ; \\
& \mathrm{K}=\frac{\mathrm{K}_{\mathrm{a}}}{\mathrm{K}_{\mathrm{a}_1}}=\frac{1.5 \times 10^{-5}}{4.5 \times 10^{-10}}=\frac{10^5}{3}=3.33 \times 10^4
\end{aligned}$
While adding two equations, dissociation constants are multiplied and when subtracting the equations, dissociation constants are divided.
Alternative
$\begin{aligned}
& \mathrm{CH}_3 \mathrm{COOH} \rightleftharpoons \mathrm{CH}_3 \mathrm{COO}^{-}+\mathrm{H}^{+} ; \mathrm{K}_{\mathrm{a}}=1.5 \times 10^{-5} \\
& \mathrm{~K}_{\mathrm{a}}=\frac{\left[\mathrm{CH}_3 \mathrm{COO}^{-}\right]\left[\mathrm{H}^{+}\right]}{\left[\mathrm{CH}_3 \mathrm{COOH}\right]}=1.5 \times 10^{-5} \\
& \mathrm{HCN} \rightleftharpoons \mathrm{H}^{+}+\mathrm{CN}^{-} ; \mathrm{K}_{\mathrm{a}}=4.5 \times 10^{-10} \\
& \mathrm{~K}_{\mathrm{a}}=\frac{\left[\mathrm{H}^{+}\right]\left[\mathrm{CN}^{-}\right]}{[\mathrm{HCN}]}=4.5 \times 10^{-10} \\
& \mathrm{CN}^{-}+\mathrm{CH}_3 \mathrm{COOH} \rightleftharpoons \mathrm{HCN}+\mathrm{CH}_3 \mathrm{COO}^{-} \\
& \mathrm{K}_{\mathrm{c}}=\frac{[\mathrm{HCN}]\left[\mathrm{CH}_3 \mathrm{COO}^{-}\right]}{\left[\mathrm{CN}^{-}\right]\left[\mathrm{CH}_3 \mathrm{COOH}\right]} \quad \ldots (iii)
\end{aligned}$
From Eqs. (i), (ii) and (iii),
$\mathrm{K}_c=\frac{1.5 \times 10^{-5}}{4.5 \times 10^{-10}}=3.33 \times 10^4$