50mL of 0 . 1M CH _ 3 COOH is being titrated against 0 . 1MNaOH ....

${CH}_{3}COOH\left(aq\right) + NaOH\left(aq .\right) \to {CH}_{3}COONa\left(aq\right) + {H}_{2}O\left(l\right)$

$5mmole2 . 5mmolesinitially2 . 5mmoles02 . 5mmolesafterreaction$

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

$pH = pKa + log\dfrac{[ salt ]}{[ acid ]}$

So, $pH$ of solution $pH = pKa = 4 . 76 = 476 \times {10}^{- 2}$