Answer
$pH=3.36$
$[H_3O^+]=4.40\dot{}10^{-4}$
Work Step by Step
Consumption of the $OH^-$:
$nOH^-=0.17\ M\dot{}5.0\ mL\dot{} 10^{-3} L/mL = 0.00085 mol$
$nHAc(initial) = 0.15\ M\dot{} 20\ mL \dot{} 10^{-3} L/mL =0.003\ mol$
$[HAc](final) = (0.003 - 0.00085)mol \div ((20+5)\ mL\times10^{-3}\ mL/L)=0.86\ M$
$[A^-] = 0.00085\ mol \div ((20+5)\ mL\times10^{-3}\ mL/L)=0.034\ M$
By the Henderson-Hasselbalch equation:
$pH = pKa + \log ([A^-]/[HA])$
$pKa$ of acetic acid $4.76$
$pH = 4.76 + \log(0.034/0.86)$
$pH = 4.76-1.40$
$pH=3.36$
$[H_3O^+]=10^{-pH}=4.40\dot{}10^{-4}$
