Calculate the $pH$ of a $0.10 \,M$ ammonia solution. Calculate the $pH$ after $50.0 \,mL$ of this solution is treated with $25.0 \,mL$ of $0.10 \,M$ $HCl$. The dissociation constant of ammonia,$K_{b} = 1.77 \times 10^{-5}$.

(N/A) $NH_{3} + H_{2}O \rightleftharpoons NH_{4}^{+} + OH^{-}$
$K_{b} = \frac{[NH_{4}^{+}][OH^{-}]}{[NH_{3}]} = 1.77 \times 10^{-5}$
Before neutralization:
$[NH_{4}^{+}] = [OH^{-}] = x$
$[NH_{3}] = 0.10 - x \approx 0.10 \,M$
$\frac{x^{2}}{0.10} = 1.77 \times 10^{-5} \implies x = \sqrt{1.77 \times 10^{-6}} = 1.33 \times 10^{-3} \,M = [OH^{-}]$
$[H^{+}] = \frac{K_{w}}{[OH^{-}]} = \frac{10^{-14}}{1.33 \times 10^{-3}} = 7.52 \times 10^{-12} \,M$
$pH = -\log(7.52 \times 10^{-12}) = 11.12$
After adding $25 \,mL$ of $0.1 \,M$ $HCl$ to $50 \,mL$ of $0.1 \,M$ $NH_{3}$:
Initial $mmol$ of $NH_{3} = 50 \times 0.1 = 5 \,mmol$
Initial $mmol$ of $HCl = 25 \times 0.1 = 2.5 \,mmol$
$NH_{3} + HCl \rightarrow NH_{4}^{+} + Cl^{-}$
Remaining $NH_{3} = 5 - 2.5 = 2.5 \,mmol$
Formed $NH_{4}^{+} = 2.5 \,mmol$
Total volume = $75 \,mL$
Using Henderson-Hasselbalch equation for a basic buffer:
$pOH = pK_{b} + \log\left(\frac{[Salt]}{[Base]}\right)$
$pK_{b} = -\log(1.77 \times 10^{-5}) = 4.75$
$pOH = 4.75 + \log\left(\frac{2.5/75}{2.5/75}\right) = 4.75 + \log(1) = 4.75$
$pH = 14 - pOH = 14 - 4.75 = 9.25$