55.0 mL of 0.200 M aqueous NH3 (Kb=1.00×10^-5 at 298 K) is mixed with 50.0 mL of 0.200 M HCl and the resulting solution is diluted with water to 1.00…

Moles of $\mathrm{NH_3}=55.0\times0.200=11.0\ \mathrm{mmol}$ and moles of $\mathrm{HCl}=50.0\times0.200=10.0\ \mathrm{mmol}$. The strong acid converts $10.0\ \mathrm{mmol}$ of $\mathrm{NH_3}$ into $\mathrm{NH_4^+}$, leaving $1.0\ \mathrm{mmol}$ $\mathrm{NH_3}$ and $10.0\ \mathrm{mmol}$ $\mathrm{NH_4^+}$, a basic buffer. Dilution to $1.00\ \mathrm{L}$ changes both concentrations by the same factor ($[\mathrm{NH_3}]=1.0\times10^{-3}\ \mathrm{M}$, $[\mathrm{NH_4^+}]=1.0\times10^{-2}\ \mathrm{M}$), so the ratio is unaffected. Then $[\mathrm{OH^-}]=K_b\dfrac{[\mathrm{NH_3}]}{[\mathrm{NH_4^+}]}=10^{-5}\times\dfrac{1}{10}=1.0\times10^{-6}\ \mathrm{M}$, so $\mathrm{pOH}=6$ and $\mathrm{pH}=14-6=8$. A solution of this kind is mildly basic; $[\mathrm{OH^-}]=10^{-6}$ is still large enough compared with $10^{-7}$ from water that the approximation is acceptable for the stated answer. The pH is 8.