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

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(A) The reaction is: $CH_3COOH + NaOH 	o CH_3COONa + H_2O$Initial millimoles of $CH_3COOH = 30 \ mL 	imes 0.2 \ M = 6 \ mmol$Initial millimoles of $

Vedclass · Accepted Answer

$4.44$

Vedclass · Answer

(A) The reaction is: $CH_3COOH + NaOH 	o CH_3COONa + H_2O$Initial millimoles of $CH_3COOH = 30 \ mL 	imes 0.2 \ M = 6 \ mmol$Initial millimoles of $NaOH = 20 \ mL 	imes 0.1 \ M = 2 \ mmol$After the reaction,$2 \ mmol$ of $NaOH$ reacts with $2 \ mmol$ of $CH_3COOH$ to form $2 \ mmol$ of $CH_3COONa$.Remaining $CH_3COOH = 6 - 2 = 4 \ mmol$Formed $CH_3COO^- = 2 \ mmol$Using the Henderson-Hasselbalch equation: $pH = pK_a + \log \frac{[CH_3COO^-]}{[CH_3COOH]}$$pH = 4.74 + \log \frac{2}{4} = 4.74 + \log(0.5)$$pH = 4.74 - 0.30 = 4.44$

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

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