20 mL of 0.1 M NaOH is added to 50 mL of 0.1 M acetic acid solution. The pH of the resulting solution is $\_\_\_\_$ $\times 10^{-2}$ (Nearest integer)
Given: $\mathrm{pKa}\left(\mathrm{CH}_3 \mathrm{COOH}\right)=4.76$
$$
\begin{aligned}
& \log 2=0.30 \\
& \log 3=0.48
\end{aligned}
$$
