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(C) The dissociation of acetic acid is given by: $CH_3COOH ightleftharpoons CH_3COO^{-} + H^{+}$,$K_{a1} = 1.5 	imes 10^{-5}$ The dissociation of $

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$3.0 	imes 10^4$

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(C) The dissociation of acetic acid is given by: $CH_3COOH ightleftharpoons CH_3COO^{-} + H^{+}$,$K_{a1} = 1.5 	imes 10^{-5}$ The dissociation of $HCN$ is given by: $HCN ightleftharpoons H^{+} + CN^{-}$,$K_{a2} = 4.5 	imes 10^{-10}$ For the reaction $CN^{-} + CH_3COOH ightleftharpoons HCN + CH_3COO^{-}$,the equilibrium constant $K$ is calculated as: $K = \frac{K_{a1}}{K_{a2}}$ Substituting the values: $K = \frac{1.5 	imes 10^{-5}}{4.5 	imes 10^{-10}} = \frac{1.5}{4.5} 	imes 10^{5} = \frac{1}{3} 	imes 10^{5} = 0.333 	imes 10^{5} = 3.33 	imes 10^{4}$ Rounding to the nearest provided option,$K \approx 3.0 	imes 10^{4}$.

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

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