Q27 / 120 · VITEEE 2011 · Chemistry

$20 \mathrm{~mL}$ 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 $\left(\mathrm{K}_{\mathrm{a}}=1.8 \times 10^{-5}\right)$

  • Correct answer
    A
    $4.5$
  • B
    $2.3$
  • C
    $3.8$
  • D
    4

Solution

$\mathrm{NaOH}+\mathrm{CH}_{3} \mathrm{COOH} \rightarrow \mathrm{CH}_{3} \mathrm{COONa}+\mathrm{H}_{2} \mathrm{O}$ $\begin{array}{lllll}\mathrm{m} \mathrm{mol} & 20 & \times 0.2 & 50 \times 0.2 & 0 & 0 \\ \text { added } & =4 & =10 & & \\ & & & \\ \text {-eáction } & 0 & 6 & 4 & 4\end{array}$ $\begin{array}{lcc}\mathrm{m} \text { mol after } & & \\ \text { reaction } 0 & 6 & 4 \\ {\left[\mathrm{CH}_{3} \mathrm{COOH}=\frac{6}{70}\right.} & : & {\left[\mathrm{CH}_{3} \mathrm{COONa}\right]}\end{array}=\frac{4}{70}$ Now since for a basic buffer $$ \begin{array}{l} \mathrm{pH}=-\operatorname{logk}_{\mathrm{a}}+\log \frac{[\mathrm{salt}]}{[\mathrm{base}]} \\ \begin{aligned} \therefore \quad \mathrm{pH} &=-\log \left(1.8 \times 10^{-5}\right)+\log \frac{4 / 70}{6 / 70} \\ &=4.56 \end{aligned} \end{array} $$

Hence, the correct answer is option: 'A'.

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