Question

3 grams of acetic acid is added to 250 mL of 0.1 M $HCl$ and the solution is made up to 500 mL. To 20 mL of this solution $\frac{1}{2}\text{mL}$ of 5 M $NaOH$ is added. The pH of this solution is .
(Given: $\text{log}3=0.4771,p{K}_a$ of acetic acid = 4.74, molar mass of acetic acid = 60 g/mole).

Answer 1

mmole of acetic acid in 20 mL = 2
mmole of $HCl$ in 20 mL = 1
mmole of $NaOH$ = 2.5

$\begin{array}{ccccccc}HCl& +& NaOH& \to & NaCl& +& H_{2}O\\ 1& & 2.5& & -& & -\\ -& & 1.5& & 1& & 1\end{array}$

$\begin{array}{ccccccc}C{H}_{3}COOH& +& NaOH\text{(remaining)}& \to & C{H}_{3}COONa& +& \text{water}\\ 2& & 1.5& & -& & -\\ 0.5& & 0& & 1.5& & -\end{array}$

$pH=p{K}_a+\text{log}\frac{1.5}{0.5}=4.74+\text{log}3=4.74+0.48=5.22$