The dissociation constant of acetic acid is x | vedclass

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(C) The mixture of $CH_3COOH$ and $CH_3COONa$ forms an acidic buffer solution.Using the Henderson-Hasselbalch equation:$pH = pK_a + \log \frac{[Salt]}

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$10$

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(C) The mixture of $CH_3COOH$ and $CH_3COONa$ forms an acidic buffer solution.Using the Henderson-Hasselbalch equation:$pH = pK_a + \log \frac{[Salt]}{[Acid]}$Calculate the concentrations in the final $50 \ mL$ mixture:$[CH_3COONa] = \frac{25 \ mL 	imes 0.2 \ M}{50 \ mL} = 0.1 \ M$$[CH_3COOH] = \frac{25 \ mL 	imes 0.02 \ M}{50 \ mL} = 0.01 \ M$Substitute the values into the equation:$5 = pK_a + \log \frac{0.1}{0.01}$$5 = pK_a + \log(10)$$5 = pK_a + 1$$pK_a = 4$Since $pK_a = -\log(K_a)$:$K_a = 10^{-4} = 10 	imes 10^{-5}$Comparing this with $x 	imes 10^{-5}$,we get $x = 10$.

The dissociation constant of acetic acid is $x \times 10^{-5}$. When $25 \ mL$ of $0.2 \ M \ CH_3COONa$ solution is mixed with $25 \ mL$ of $0.02 \ M \ CH_3COOH$ solution,the $pH$ of the resultant solution is found to be equal to $5$. The value of $x$ is $..........$.

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