50 mL of 0.1 M CH_3 COOH is titrated against 0.1 M NaOH solution. When 10 mL of NaOH is added, the pH of the solution becomes \ \ \ \ × 10^-1.…

$\displaystyle 50$ mL of $\displaystyle 0.1 \mathrm{M} \mathrm{CH}_3 \mathrm{COOH}$ is titrated against $\displaystyle 0.1$ M NaOH solution. When $\displaystyle 10$ mL of NaOH is added, the pH of the solution becomes \_\_\_\_ $\displaystyle \times 10^{-1}$. (nearest integer)Given: $\displaystyle \mathrm{pKa}\left(\mathrm{CH}_3 \mathrm{COOH}\right)=4.8, \log 2=0.3$