50 mL of 0.1 M CH_3 COOH is being titrated against 0.1 M NaOH. When 25 mL of NaOH has been added, the pH of the solution will be \ \ \ \ × 10^-2.…

$\displaystyle 50$ mL of $\displaystyle 0.1 \mathrm{M} \mathrm{CH}_3 \mathrm{COOH}$ is being titrated against $\displaystyle 0.1$ M NaOH . When $\displaystyle 25$ mL of NaOH has been added, the pH of the solution will be \_\_\_\_ $\displaystyle \times 10^{-2}$. (Nearest integer) (Given : $\displaystyle \mathrm{pK}_{\mathrm{a}}\left(\mathrm{CH}_3 \mathrm{COOH}\right)=4.76$ ) $$\begin{aligned} & \log 2=0.30 \\ & \log 3=0.48 \\ & \log 5=0.69 \\ & \log 7=0.84 \\ & \log 11=1.04 \end{aligned} $$