50 mL of 0.02 M NaOH solution is mixed with | vedclass

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(D) The reaction between $NaOH$ and acetic acid $(CH_3COOH)$ is: $CH_3COOH + NaOH \rightarrow CH_3COONa + H_2O$. Initial moles of $CH_3COOH = 0.06 \ M

Vedclass · Accepted Answer

$4.46$

Vedclass · Answer

(D) The reaction between $NaOH$ and acetic acid $(CH_3COOH)$ is: $CH_3COOH + NaOH ightarrow CH_3COONa + H_2O$. Initial moles of $CH_3COOH = 0.06 \ M 	imes 0.050 \ L = 0.003 \ mol$. Initial moles of $NaOH = 0.02 \ M 	imes 0.050 \ L = 0.001 \ mol$. After the reaction,$0.001 \ mol$ of $CH_3COONa$ is formed and $0.002 \ mol$ of $CH_3COOH$ remains. Since a buffer solution is formed,we use the Henderson-Hasselbalch equation: $pH = pK_a + \log \frac{[Salt]}{[Acid]}$. $pH = 4.76 + \log \frac{0.001}{0.002} = 4.76 + \log (0.5)$. $pH = 4.76 - 0.30 = 4.46$.

$50 \ mL$ of $0.02 \ M$ $NaOH$ solution is mixed with $50 \ mL$ of $0.06 \ M$ acetic acid solution. The $pH$ of the resulting solution is $......$ ($pK_a$ of acetic acid is $4.76$,$\log 2 = 0.30$).

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