Answer
$pH=9.73$
$[H_3O^+]=1.87\dot{}10^{-10}$
Work Step by Step
Consumption of the $H_3O^+$:
$nH_3O^+=0.20\ M\dot{}25.0\ mL\dot{} 10^{-3} L/mL = 0.005 mol$
$nNH_3(before) = 0.40\ M\dot{} 50\ mL \dot{} 10^{-3} L/mL =0.02\ mol$
$[NH_3](after) = (0.02- 0.005)mol \div ((50+25)\ mL\times10^{-3}\ mL/L)=0.2\ M$
$[NH_4^+] = 0.005\ mol \div ((50+25)\ mL\times10^{-3}\ mL/L)=0.067\ M$
By the Henderson-Hasselbalch equation:
$pH = pKa + \log ([NH_3]/[NH_4^+])$
$pKb$ of ammonia $4.75$, $pKa = 14-pKb=9.25$
$pH = 9.25 + \log(0.2/0.067)$
$pH = 9.25+0.48$
$pH=9.73$
$[H_3O^+]=10^{-pH}=1.87\dot{}10^{-10}$