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(B) The initial moles of acetic acid and sodium acetate in $1 \ 	ext{L}$ of solution are $0.01 \ 	ext{mole}$ each. When $1 	imes 10^{-3} \ 	ext{mo

Vedclass · Accepted Answer

$4.66$

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(B) The initial moles of acetic acid and sodium acetate in $1 \ 	ext{L}$ of solution are $0.01 \ 	ext{mole}$ each. When $1 	imes 10^{-3} \ 	ext{mole}$ of $HCl$ is added, it reacts with the salt (sodium acetate) to form acetic acid: $CH_{3}COO^{-} + H^{+} \longrightarrow CH_{3}COOH$ New moles of salt = $0.01 - 0.001 = 0.009 \ 	ext{mole}$. New moles of acid = $0.01 + 0.001 = 0.011 \ 	ext{mole}$. Using the Henderson-Hasselbalch equation: $pH = pK_{a} + \log \frac{[	ext{salt}]}{[	ext{acid}]} = 4.75 + \log \frac{0.009}{0.011} = 4.75 + \log(0.818) \approx 4.75 - 0.087 = 4.663$ Thus, the final $pH$ is approximately $4.66$.

$1 \times 10^{-3} \ \text{mole}$ of $HCl$ is added to a buffer solution made up of $0.01 \ \text{M}$ acetic acid and $0.01 \ \text{M}$ sodium acetate. The final $pH$ of the buffer will be (given, $pK_{a}$ of acetic acid is $4.75$ at $25^{\circ} \text{C}$)

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