$20 \mathrm{~mL}$ of $0.2 \mathrm{M} \mathrm{NaOH}$ are added to $50 \mathrm{mLof}$ $0.2 \mathrm{M}…
$0.2 \mathrm{M} \mathrm{CH}_{3} \mathrm{COOH}\left(\mathrm{K}_{\mathrm{a}}=1.8 \times 10^{-5}ight)$ the $\mathrm{pH}$ of the solution is
- $4.56$
- $4.73$
- $9.45$
- $6.78$
Solution
\mathrm{NaOH} & + & \mathrm{CH}_3 \mathrm{COOH} & ightleftharpoons & \mathrm{CH}_3 \mathrm{COONa} & + & \mathrm{H}_2 \mathrm{O} \\
20 \times 0.2 & & 50 \times 0.2 & & 0 & & 0 \\
4 & & 10 & & 0 & & 0 \\
0 & & 6 & & 4 & & 4 \\
\end{array}\)
Conc. $=\frac{\text { Millimoles }}{\text { Total Volume }}$
$\left[\mathrm{CH}_{3} \mathrm{COOH}ight]=\frac{6}{70} ;\left[\mathrm{CH}_{3} \mathrm{COONa}ight]=\frac{4}{70}$
$\mathrm{pH}=-\log 1.8 \times 10^{-5}+\log \frac{4 / 70}{6 / 70}, \mathrm{pH}=4.56$ ^
Asked in: JEE-TOPICTESTS-CHEMISTRY