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(D) The reaction is: $CH_3COOH + NaOH ightarrow CH_3COONa + H_2O$.Initial moles of $CH_3COOH = 50 \ mL 	imes 0.1 \ M = 5 \ mmol$.Initial moles of $

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$458$

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(D) The reaction is: $CH_3COOH + NaOH ightarrow CH_3COONa + H_2O$.Initial moles of $CH_3COOH = 50 \ mL 	imes 0.1 \ M = 5 \ mmol$.Initial moles of $NaOH = 20 \ mL 	imes 0.1 \ M = 2 \ mmol$.After reaction,moles of $CH_3COONa$ (salt) $= 2 \ mmol$ and remaining moles of $CH_3COOH$ (acid) $= 5 - 2 = 3 \ mmol$.Using the Henderson-Hasselbalch equation: $pH = pKa + \log_{10} (\frac{[salt]}{[acid]})$$pH = 4.76 + \log_{10} (\frac{2}{3})$$pH = 4.76 + (\log 2 - \log 3) = 4.76 + (0.30 - 0.48) = 4.76 - 0.18 = 4.58$.Thus,$pH = 4.58 = 458 	imes 10^{-2}$.

$20 \ mL$ of $0.1 \ M$ $NaOH$ is added to $50 \ mL$ of $0.1 \ M$ acetic acid solution. The $pH$ of the resulting solution is $....... \times 10^{-2}$ (Nearest integer).
Given: $pKa$ $(CH_3COOH)$ $= 4.76$,$\log 2 = 0.30$,$\log 3 = 0.48$.

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$20 \ mL$ of $0.1 \ M$ $NaOH$ is added to $50 \ mL$ of $0.1 \ M$ acetic acid solution. The $pH$ of the resulting solution is $....... \times 10^{-2}$ (Nearest integer).Given: $pKa$ $(CH_3COOH)$ $= 4.76$,$\log 2 = 0.30$,$\log 3 = 0.48$.