50   mL of 0 . 1   M   CH 3 COOH is being titrated against 0 . 1   M   NaOH . When…

${\mathrm{CH}}_3\mathrm{COOH}\left(\mathrm{aq}\right)+\mathrm{NaOH}\left(\mathrm{aq}.\right)\to {\mathrm{CH}}_3\mathrm{COONa}\left(\mathrm{aq}\right)+{\mathrm{H}}_2\mathrm{O}\left(\mathrm{l}\right)$

$$\begin{gathered} 5 \mathrm{mmole} 2.5 \mathrm{mmoles} \mathrm{initially} \\ 2.5 \mathrm{m} \mathrm{moles} 0 2.5 \mathrm{mmoles} \mathrm{after} \mathrm{reaction} \end{gathered}$$

Resultant solution is acidic buffer solution with same concentration of acid and salts.

$\mathrm{pH} = \mathrm{pKa} + \log \frac{\left[\mathrm{salt}\right]}{\left[\mathrm{acid}\right]}$

So, $\mathrm{pH}$ of solution $\mathrm{pH}=\mathrm{pKa}=4.76=476\times {10}^{-2}$