The pH at the equivalence point, when a solution of 0.1 M acetic acid is titrated with a solution of 0.1 M sodium hydroxide is 

Conc. of sodium acetate formed $=\frac{0.1}{2}=0.05$ (equal volumes of acid and base are present)

Sodium acetate is a salt of weak acid and strong base

$pH=\frac{1}{2}p{K}_w+\frac{1}{2}p{K}_a+\frac{1}{2}logC$

$pH=\frac{1}{2}(-log1\times {10}^{-14})+\frac{1}{2}(-log(1.9\times {10}^{-5}))+\frac{1}{2}log\left(0.05\right)$

$=\frac{1}{2}\times 14+\frac{1}{2}\times 4.72+\frac{1}{2}(-1.30)$= 7 + 2.36 – 0.65 = 8.71