Answer
$pH \approx 4.86$
Work Step by Step
$Ka = \frac{[H^+][CH_3COO^-]}{[CH_3COOH]}$
$Ka = 1.8 \times 10^{-5}$
$[CH_3COO^-] = [CH_3COONa] = 0.13M$
$[CH_3COOH] = 0.10M$
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$1.8 \times 10^{-5} = \frac{[H^+][0.13]}{[0.10]}$
$1.8 \times 10^{-5} * 0.1 = {[H^+][0.13]}$
$1.8 \times 10^{-6} = {[H^+][0.13]}$
$[H^+] = \frac{1.8 \times 10^{-6}}{0.13}$
$[H^+] = 1.385 \times 10^{-5}$
$pH = -log[H^+]$
$pH = -log(1.385 \times 10^{-5}) \approx 4.859$
