KCET · Chemistry · Ionic Equilibrium
A buffer solution contains \(0.1\) mole of sodium acetate in \(1000 \mathrm{~cm}^{3}\) of \(0.1 \mathrm{M}\) acetic acid. To the above buffer solution, \(0.1\) mole of sodium acetate is further added and dissolved. The \(\mathrm{pH}\) of the resulting buffer is equal to
- A \(\mathrm{pK}_{\mathrm{a}}-\log 2\)
- B \(\mathrm{pK}_{\mathrm{a}}\)
- C \(\mathrm{pK}_{\mathrm{a}}+2\)
- D \(\mathrm{pK}_{\mathrm{a}}+\log 2\)
Answer & Solution
Correct Answer
(D) \(\mathrm{pK}_{\mathrm{a}}+\log 2\)
Step-by-step Solution
Detailed explanation
Initially \(\left[\mathrm{CH}_{3} \mathrm{COONa}\right]=0.1 \mathrm{~mol}\)
\(\left[\mathrm{CH}_{2} \mathrm{COOH}\right]=0.1 \mathrm{~mol}\)
when \(0.1 \mathrm{~mol} \mathrm{CH}_{3} \mathrm{COONa}\) is added.
\[
\begin{aligned}
{\left[\mathrm{CH}_{3} \mathrm{COONa}\right] } &=0.1+0.1=0.2 \mathrm{~mol} \\
\mathrm{pH} &=\mathrm{pK}_{\mathrm{a}}+\log \frac{\text { [salt }]}{\text { [acid] }} \\
&=\mathrm{pK}_{\mathrm{a}}+\log \frac{0.2}{0.1} \\
&=\mathrm{pK}_{\mathrm{a}}+\log 2
\end{aligned}
\]
\(\left[\mathrm{CH}_{2} \mathrm{COOH}\right]=0.1 \mathrm{~mol}\)
when \(0.1 \mathrm{~mol} \mathrm{CH}_{3} \mathrm{COONa}\) is added.
\[
\begin{aligned}
{\left[\mathrm{CH}_{3} \mathrm{COONa}\right] } &=0.1+0.1=0.2 \mathrm{~mol} \\
\mathrm{pH} &=\mathrm{pK}_{\mathrm{a}}+\log \frac{\text { [salt }]}{\text { [acid] }} \\
&=\mathrm{pK}_{\mathrm{a}}+\log \frac{0.2}{0.1} \\
&=\mathrm{pK}_{\mathrm{a}}+\log 2
\end{aligned}
\]
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