If the equilibrium constant of CH_3COOH + H_2 | vedclass

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(B) The given reaction is the neutralization of a weak acid by a strong base.$CH_3COOH + H_2O \rightleftharpoons CH_3COO^{-} + H_3O^{+}$ $(K_a = 1.8 \

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$1.8 	imes 10^{9}$

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(B) The given reaction is the neutralization of a weak acid by a strong base.$CH_3COOH + H_2O ightleftharpoons CH_3COO^{-} + H_3O^{+}$ $(K_a = 1.8 	imes 10^{-5})$$H_3O^{+} + OH^{-} ightleftharpoons 2H_2O$ $(K = 1/K_w = 1/10^{-14} = 10^{14})$Adding these two reactions gives:$CH_3COOH + OH^{-} ightleftharpoons CH_3COO^{-} + H_2O$The equilibrium constant $K$ for this reaction is $K = K_a 	imes (1/K_w) = (1.8 	imes 10^{-5}) 	imes 10^{14} = 1.8 	imes 10^{9}$.

If the equilibrium constant of $CH_3COOH + H_2O \rightleftharpoons CH_3COO^{-} + H_3O^{+}$ is $1.8 \times 10^{-5}$,what is the equilibrium constant for the reaction $CH_3COOH + OH^{-} \rightleftharpoons CH_3COO^{-} + H_2O$?

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