Q39 / 160 · WBJEE 2010 · Chemistry
A weak acid of dissociation constant $10^{-5}$ is being titrated with aqueous $\mathrm{NaOH}$ solution. The $\mathrm{pH}$ at the point of one-third neutralisation of the acid will be
- A$5+\log 2-\log 3$
- Correct answerB$5-\log 2$
- C$5-\log 3$
- D$5-\log 6$
Solution
Hints: $\mathrm{K}_{\mathrm{a}}=10^{-5} \Rightarrow \mathrm{pK}_{\mathrm{a}}=-\log \mathrm{K}_{\mathrm{a}}=-\log 10^{-5}=5$
$$
\begin{array}{llcc}
& \mathrm{HA}+\mathrm{NaOH} \longrightarrow \mathrm{NaA}+\mathrm{H}_2 \mathrm{O} \\
\text { Initial } & 1 \mathrm{~mole} 0 & 0 & 0 \\
\text { Final } & (1-1 / 3) \mathrm{mole} & 1 / 3 \mathrm{~mole} & 1 / 3 \mathrm{~mole} \\
& =2 / 3 \mathrm{~mole} & 1 / 3 \mathrm{~mole}
\end{array}
$$
(Assumed weak acid to be monoprotic, since only one dissociation constant value is provided)
Final solution acts as an acidic buffer.
$$
\Rightarrow \mathrm{pH}=\mathrm{pK}_{\mathrm{a}}+\log \frac{[\text { salt }]}{[\text { Acid }]} \Rightarrow \mathrm{pH}=5+\log \frac{\frac{1}{3}}{\frac{2}{3}}=5+\log \frac{1}{2} \Rightarrow \mathrm{pH}=5-\log 2
$$
Hence, the correct answer is option: 'B'.