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A buffer solution is prepared by mixing equimolar acetic acid and sodium acetate. If ' $\mathrm{K}_{\mathrm{a}}$ ' of acetic acid is $1.78 \times 10^{-5}$, find the $\mathrm{pH}$ of buffer solution.
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Verified Answer
The correct answer is:
4.75
For acidic buffer,
$$
\begin{aligned}
\mathrm{pH} & =\mathrm{pK}_{\mathrm{a}}+\log _{10} \frac{[\text { Salt }]}{[\text { Acid }]}=\mathrm{pK}_{\mathrm{a}}+\log _{10} \frac{1}{1}=\mathrm{pK}_{\mathrm{a}} \\
\mathrm{pK}_{\mathrm{a}} & =-\log _{10} \mathrm{~K}_{\mathrm{a}}=-\log _{10}\left(1.78 \times 10^{-5}\right) \\
& =-\left(\log _{10} 1.78+\log _{10} 10^{-5}\right) \\
& =-(0.25-5)=4.75
\end{aligned}
$$
Therefore, $\mathrm{pH}$ of buffer solution $=4.75$
$$
\begin{aligned}
\mathrm{pH} & =\mathrm{pK}_{\mathrm{a}}+\log _{10} \frac{[\text { Salt }]}{[\text { Acid }]}=\mathrm{pK}_{\mathrm{a}}+\log _{10} \frac{1}{1}=\mathrm{pK}_{\mathrm{a}} \\
\mathrm{pK}_{\mathrm{a}} & =-\log _{10} \mathrm{~K}_{\mathrm{a}}=-\log _{10}\left(1.78 \times 10^{-5}\right) \\
& =-\left(\log _{10} 1.78+\log _{10} 10^{-5}\right) \\
& =-(0.25-5)=4.75
\end{aligned}
$$
Therefore, $\mathrm{pH}$ of buffer solution $=4.75$
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