Answer
$pH \approx 5.018$
Work Step by Step
Using the Henderson-Hasselbalch equation:
$pH = pKa +log\frac{[Base]}{[Acid]}$
1. Find the pKa:
$pKa = -log(Ka) = -log(1.8\times 10^{-5}) = 4.745$
2. Find [Base]:
$1CH_3COOH(aq) + 1KOH(aq) --> 1CH_3COO^-(aq) + K^+(aq) + H_2O(l)$
*Since the proportion is 1 to 1, 0.02 M of $CH_3COO^-$ were added.
$[Base] = Initial + 0.02 = 0.13 + 0.02 = 0.15M$
3. Find [Acid]:
*Since the proportion is 1 to 1, 0.02 M of $CH_3COOH$ reacted, so:
$[Acid] = Inital - 0.02 = 0.10 - 0.02 = 0.08M$
4. Now, find the pH:
$pH = 4.745 + log(\frac{0.15}{0.08})$
$pH = 4.745 + log(1.875)$
$pH \approx 4.745 + 0.273$
$pH \approx 5.018$