20mL of 0 . 1MNaOH is added to 50mL of 0 . 1M acetic acid solution....

When a strong base is added to a weak acid solution, it results in the formation of a salt. Here, acid is present in a limiting reagent and base is present in excess amounts. So, by using the pH formula:

$\begin{cases}pH & = & pKa + log\dfrac{\left[salt\right]}{\left[acid\right]}\end{cases}$

${CH}_{3}COOH + NaOH ⟶ {CH}_{3}COONa + {H}_{2}O52 — — 302$

$pH = {pK}_{a} + \log \left(\dfrac{2}{3}\right)$

$pH = 4 . 76 + 0 . 30 - 0 . 48$

$= 4 . 76 – . 18$

$= 4 . 58$