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$20 \mathrm{~mL}$ of $0.2 \mathrm{M} \mathrm{NaOH}$ is added to $50 \mathrm{~mL}$ of $0.2 \mathrm{M}$ acetic acid. The $\mathrm{pH}$ of this solution after mixing is $\left(\mathrm{K}_{\mathrm{a}}=1.8 \times 10^{-5}\right)$
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Verified Answer
The correct answer is:
$4.5$
$\mathrm{NaOH}+\mathrm{CH}_{3} \mathrm{COOH} \rightarrow \mathrm{CH}_{3} \mathrm{COONa}+\mathrm{H}_{2} \mathrm{O}$
$\begin{array}{lllll}\mathrm{m} \mathrm{mol} & 20 & \times 0.2 & 50 \times 0.2 & 0 & 0 \\ \text { added } & =4 & =10 & & \\ & & & \\ \text {-eáction } & 0 & 6 & 4 & 4\end{array}$
$\begin{array}{lcc}\mathrm{m} \text { mol after } & & \\ \text { reaction } 0 & 6 & 4 \\ {\left[\mathrm{CH}_{3} \mathrm{COOH}=\frac{6}{70}\right.} & : & {\left[\mathrm{CH}_{3} \mathrm{COONa}\right]}\end{array}=\frac{4}{70}$
Now since for a basic buffer
$$
\begin{array}{l}
\mathrm{pH}=-\operatorname{logk}_{\mathrm{a}}+\log \frac{[\mathrm{salt}]}{[\mathrm{base}]} \\
\begin{aligned}
\therefore \quad \mathrm{pH} &=-\log \left(1.8 \times 10^{-5}\right)+\log \frac{4 / 70}{6 / 70} \\
&=4.56
\end{aligned}
\end{array}
$$
$\begin{array}{lllll}\mathrm{m} \mathrm{mol} & 20 & \times 0.2 & 50 \times 0.2 & 0 & 0 \\ \text { added } & =4 & =10 & & \\ & & & \\ \text {-eáction } & 0 & 6 & 4 & 4\end{array}$
$\begin{array}{lcc}\mathrm{m} \text { mol after } & & \\ \text { reaction } 0 & 6 & 4 \\ {\left[\mathrm{CH}_{3} \mathrm{COOH}=\frac{6}{70}\right.} & : & {\left[\mathrm{CH}_{3} \mathrm{COONa}\right]}\end{array}=\frac{4}{70}$
Now since for a basic buffer
$$
\begin{array}{l}
\mathrm{pH}=-\operatorname{logk}_{\mathrm{a}}+\log \frac{[\mathrm{salt}]}{[\mathrm{base}]} \\
\begin{aligned}
\therefore \quad \mathrm{pH} &=-\log \left(1.8 \times 10^{-5}\right)+\log \frac{4 / 70}{6 / 70} \\
&=4.56
\end{aligned}
\end{array}
$$
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