3 ; g of acetic acid is added to 250 ; mL of | vedclass

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(D) $1$. Calculate the milliequivalents $(meq)$ in the $500 \; mL$ solution: $meq$ of $HCl = 250 \; mL 	imes 0.1 \; M = 25 \; meq$. $meq$ of $CH_3COO

Vedclass · Accepted Answer

$5.23$

Vedclass · Answer

(D) $1$. Calculate the milliequivalents $(meq)$ in the $500 \; mL$ solution: $meq$ of $HCl = 250 \; mL 	imes 0.1 \; M = 25 \; meq$. $meq$ of $CH_3COOH = \frac{3 \; g}{60 \; g/mol} = 0.05 \; mol = 50 \; meq$. $2$. Calculate the $meq$ in $20 \; mL$ of the solution: $meq$ of $HCl = \frac{25 \; meq}{500 \; mL} 	imes 20 \; mL = 1 \; meq$. $meq$ of $CH_3COOH = \frac{50 \; meq}{500 \; mL} 	imes 20 \; mL = 2 \; meq$. $3$. Calculate the $meq$ of $NaOH$ added: $meq$ of $NaOH = 5 \; M 	imes 0.5 \; mL = 2.5 \; meq$. $4$. Reaction: $HCl$ reacts first: $1 \; meq$ of $NaOH$ neutralizes $1 \; meq$ of $HCl$. Remaining $NaOH = 2.5 - 1 = 1.5 \; meq$. This $1.5 \; meq$ of $NaOH$ reacts with $CH_3COOH$: $CH_3COOH + NaOH ightarrow CH_3COONa + H_2O$. Remaining $CH_3COOH = 2 - 1.5 = 0.5 \; meq$. Formed $CH_3COONa = 1.5 \; meq$. $5$. Calculate $pH$ using the Henderson-Hasselbalch equation: $pH = pK_a + \log \frac{[Salt]}{[Acid]} = 4.75 + \log \frac{1.5}{0.5} = 4.75 + \log 3 = 4.75 + 0.4771 = 5.2271 \approx 5.23$.

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