20 mL of 0.1 (N) acetic acid is mixed with | vedclass

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(B) The reaction is: $CH_3COOH + NaOH ightarrow CH_3COONa + H_2O$Initial millimoles of $CH_3COOH = 20 \ mL 	imes 0.1 \ N = 2 \ mmol$.Initial millim

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(B) The reaction is: $CH_3COOH + NaOH ightarrow CH_3COONa + H_2O$Initial millimoles of $CH_3COOH = 20 \ mL 	imes 0.1 \ N = 2 \ mmol$.Initial millimoles of $NaOH = 10 \ mL 	imes 0.1 \ N = 1 \ mmol$.After the reaction,$1 \ mmol$ of $CH_3COOH$ remains and $1 \ mmol$ of $CH_3COONa$ is formed.This forms an acidic buffer solution.Using Henderson-Hasselbalch equation:$pH = pK_a + \log \frac{[Salt]}{[Acid]}$$pH = 4.74 + \log \frac{1 \ mmol}{1 \ mmol}$$pH = 4.74 + \log(1) = 4.74 + 0 = 4.74$.

$20 \ mL$ of $0.1 \ (N)$ acetic acid is mixed with $10 \ mL$ of $0.1 \ (N)$ solution of $NaOH$. The $pH$ of the resulting solution is ($pK_a$ of acetic acid is $4.74$). (in $.74$)

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