At 25^{circ}C,the dissociation constants of a | vedclass

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(A) The given reaction is: $CN^- + CH_3COOH \rightleftharpoons HCN + CH_3COO^-$This reaction can be expressed as the difference between the dissociati

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$3.0 	imes 10^{4}$

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(A) The given reaction is: $CN^- + CH_3COOH ightleftharpoons HCN + CH_3COO^-$This reaction can be expressed as the difference between the dissociation of $CH_3COOH$ and $HCN$:$CH_3COOH ightleftharpoons CH_3COO^- + H^+$; $K_a(CH_3COOH) = 1.5 	imes 10^{-5}$$HCN ightleftharpoons H^+ + CN^-$; $K_a(HCN) = 4.5 	imes 10^{-10}$To get the target reaction,we add the first equation to the reverse of the second equation:$CH_3COOH + CN^- ightleftharpoons CH_3COO^- + H^+ + H^+ + CN^- ightleftharpoons HCN + CH_3COO^-$The equilibrium constant $K$ for the reaction is given by:$K = \frac{K_a(CH_3COOH)}{K_a(HCN)}$$K = \frac{1.5 	imes 10^{-5}}{4.5 	imes 10^{-10}}$$K = \frac{1.5}{4.5} 	imes 10^{5} = \frac{1}{3} 	imes 10^{5} = 0.333 	imes 10^{5} = 3.33 	imes 10^{4}$Given the options,the closest value is $3.0 	imes 10^{4}$.

At $25^{\circ}C$,the dissociation constants of acetic acid and $HCN$ are $1.5 \times 10^{-5}$ and $4.5 \times 10^{-10}$ respectively. What is the equilibrium constant for the following reaction? $(CN^- + CH_3COOH \rightleftharpoons HCN + CH_3COO^-)$

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