If acetic acid is ionized to the extent of 2.3% in 0.02 M aqueous solution, the pH  of the solution is :

$$\begin{gathered} \mathrm{Normality}=0.02 \mathrm{N} \\ \mathrm{Molarity}=\mathrm{normality}\times \mathrm{n}-\mathrm{factor} \\ \mathrm{Molarity}=0.02\mathrm{N}\times 1=0.02 \mathrm{M} \end{gathered}$$
Degree of ionization, $\mathrm{\alpha }=\frac{2.3}{100}=0.023$

Concentration of hydrogen ion in the solution:

$$\begin{gathered} \left[{\mathrm{H}}^{+}\right] =0.02\mathrm{x} 0.023 \\ =0.00046 \mathrm{M} \end{gathered}$$

pH is calculated as:

$$\begin{gathered} \mathrm{pH} =-\log \left[{\mathrm{H}}^{+}\right] \\ =-\log (0.00046) \\ = 3.337\cong 3.34 \end{gathered}$$