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Calculate $\mathrm{pH}$ of a buffer prepared by adding 10 $\mathrm{mL}$ of $0.10 \mathrm{M}$ acetic acid to $20 \mathrm{~mL}$ of $0.1 \mathrm{M}$ sodium acetate $\left[\mathrm{p} K_{a}\left(\mathrm{CH}_{3} \mathrm{COOH}\right)=4.74\right]$
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The correct answer is:
$5.04$
$\left[\mathrm{CH}_{3} \mathrm{COOH}\right]=\text { millimoles of } \mathrm{CH}_{3} \mathrm{COOH}$
$=0.1 \times 10=1.0$
$\begin{array}{l}
{\left[\mathrm{CH}_{3} \mathrm{COONa}\right]=\text { millimoles of } \mathrm{CH}_{3} \mathrm{COONa}} \\
=0.1 \times 20=2.0
\end{array}$
From, Henderson Hasselbalch equation,
$\begin{array}{l}
\mathrm{pH}=\mathrm{pK}_{\mathrm{a}}+\log \frac{\text { [conjugate base] }}{[\text { acid }]} \\
=4.74+\log \frac{2}{1}=4.74+0.30=5.04
\end{array}$
$=0.1 \times 10=1.0$
$\begin{array}{l}
{\left[\mathrm{CH}_{3} \mathrm{COONa}\right]=\text { millimoles of } \mathrm{CH}_{3} \mathrm{COONa}} \\
=0.1 \times 20=2.0
\end{array}$
From, Henderson Hasselbalch equation,
$\begin{array}{l}
\mathrm{pH}=\mathrm{pK}_{\mathrm{a}}+\log \frac{\text { [conjugate base] }}{[\text { acid }]} \\
=4.74+\log \frac{2}{1}=4.74+0.30=5.04
\end{array}$
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