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The dissociation constants for acetic acid and HCN at $25^\circ C$ are $1.5\times10^{-5}$ and $4.5\times10^{-10}$, respectively. The equilibrium constant for the equilibrium $CN^- + CH_3COOH \rightleftharpoons HCN + CH_3COO^-$ would be
A
$3.0\times10^4$
B
$3.0\times10^5$
C
$3.0\times10^{-5}$
D
$3.0\times10^{-4}$
Detailed Solution
(i) $CH_3COOH \rightleftharpoons CH_3COO^- + H^+$; $K_1 = 1.5\times10^{-5}$
(ii) $HCN \rightleftharpoons H^+ + CN^-$; $K_2 = 4.5\times10^{-10}$
Reverse (ii): $H^+ + CN^- \rightleftharpoons HCN$; $K = \frac{1}{K_2}$
Adding (i) and the reversed (ii): $CN^- + CH_3COOH \rightleftharpoons HCN + CH_3COO^-$; $K = \frac{K_1}{K_2}$
$K = \frac{1.5\times10^{-5}}{4.5\times10^{-10}} = 0.333\times10^5$
$K = 3.3\times10^4 \approx 3.0\times10^4$
(ii) $HCN \rightleftharpoons H^+ + CN^-$; $K_2 = 4.5\times10^{-10}$
Reverse (ii): $H^+ + CN^- \rightleftharpoons HCN$; $K = \frac{1}{K_2}$
Adding (i) and the reversed (ii): $CN^- + CH_3COOH \rightleftharpoons HCN + CH_3COO^-$; $K = \frac{K_1}{K_2}$
$K = \frac{1.5\times10^{-5}}{4.5\times10^{-10}} = 0.333\times10^5$
$K = 3.3\times10^4 \approx 3.0\times10^4$