If 20 mL of 0.1 M NaOH is added to 30 mL | vedclass

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(A) The reaction is: $NaOH_{(aq)} + CH_3COOH_{(aq)} 	o CH_3COONa_{(aq)} + H_2O_{(l)}$Initial moles of $NaOH = 20 \ mL 	imes 0.1 \ M = 2 \ mmol$.Init

Vedclass · Accepted Answer

$4.44$

Vedclass · Answer

(A) The reaction is: $NaOH_{(aq)} + CH_3COOH_{(aq)} 	o CH_3COONa_{(aq)} + H_2O_{(l)}$Initial moles of $NaOH = 20 \ mL 	imes 0.1 \ M = 2 \ mmol$.Initial moles of $CH_3COOH = 30 \ mL 	imes 0.2 \ M = 6 \ mmol$.After the reaction,$2 \ mmol$ of $CH_3COONa$ is formed,and $4 \ mmol$ of $CH_3COOH$ remains.This forms an acidic buffer solution.Using the Henderson-Hasselbalch equation: $pH = pK_a + \log \frac{[Salt]}{[Acid]}$$pH = 4.74 + \log \frac{2}{4}$$pH = 4.74 + \log(0.5)$$pH = 4.74 - 0.3010 = 4.439 \approx 4.44$.

If $20 \ mL$ of $0.1 \ M \ NaOH$ is added to $30 \ mL$ of $0.2 \ M \ CH_3COOH$ $(pK_a = 4.74)$,find the $pH$ of the resulting solution.

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If $20 \ mL$ of $0.1 \ M \ NaOH$ is added to $30 \ mL$ of $0.2 \ M \ CH_3COOH$ $(pK_a = 4.74)$,the $pH$ of the resulting solution is

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