The dissociation constants for acetic acid and HCN at $25^\circ ext{C}$ are $1.5 imes 10^{-5}$ and $4.5

Question

The dissociation constants for acetic acid and HCN at $25^\circ 	ext{C}$ are $1.5 	imes 10^{-5}$ and $4.5 	imes 10^{-10}$, respectively. The equilibrium constant for the equilibrium, $CN^- + CH_3COOH ightleftharpoons HCN + CH_3COO^-$ would be:

Accepted Answer

For the given reaction: $CN^- + CH_3COOH ightleftharpoons HCN + CH_3COO^-$
The equilibrium constant $K$ is given by:
$K = \frac{[HCN][CH_3COO^-]}{[CN^-][CH_3COOH]}$
Multiplying the numerator and denominator by $[H^+]$, we get:
$K = \frac{[CH_3COO^-][H^+]}{[CH_3COOH]} 	imes \frac{[HCN]}{[CN^-][H^+]}$
$K = \frac{K_a(CH_3COOH)}{K_a(HCN)}$
Given $K_a(CH_3COOH) = 1.5 	imes 10^{-5}$ and $K_a(HCN) = 4.5 	imes 10^{-10}$,
$K = \frac{1.5 	imes 10^{-5}}{4.5 	imes 10^{-10}} = \frac{1}{3} 	imes 10^5 = 0.333 	imes 10^5 = 3.33 	imes 10^4$
The closest value among the given options is $3.0 	imes 10^4$.

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