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

$\begin{array}{rl}& {\mathrm{NH}}_3+{\mathrm{H}}_2\mathrm{O}\to {\mathrm{NH}}_4^{+}+{\mathrm{OH}}^{-}\\ & {\mathrm{K}}_{\mathrm{b}}=\left[{\mathrm{NH}}_4^{+}\right]\left[{\mathrm{OH}}^{-}\right]/\left[{\mathrm{NH}}_3\right]=1.77\times 10^{-5}\\ & \text{ Before neutralization, }\\ & \left[{\mathrm{NH}}_4^{+}\right]=\left[{\mathrm{OH}}^{-}\right]=\mathrm{x}\\ & \left[{\mathrm{NH}}_3\right]=0.10-\mathrm{x};0.10\\ & {\mathrm{x}}^2/0.10=1.77\times 10^{-5}\\ & \text{ Thus, }\mathrm{x}=1.33\times 10^{-3}=\left[{\mathrm{OH}}^{-}\right]\\ & \text{Therefore, }\left[{\mathrm{H}}^{+}\right]={\mathrm{K}}_{\mathrm{W}}/\left[{\mathrm{OH}}^{-}\right]=10^{-14}/\left(1.33\times 10^{-3}\right)=7.51\times 10^{-12}\\ & \text{ pH }=-\log \left(7.5\times 10^{-12}\right)=11.12\end{array}$
In the addition of 25 mL of 0.1 M HCl solution (i.e., 2.5 mmol of HCl) to 50 mL of 0.1M ammonia solution (i.e., 5 mmol of $N{H}_3$), 2.5 mmol of ammonia molecules are neutralized. The resulting 75 mL solution contains the remaining unneutralized 2.5 mmol of $N{H}_3$ molecules and 2.5 mmol of ${N{H}_4}^{+}$

The resulting 75 mL of solution contains 2.5 mmol of ${N{H}_4}^{+}$ ions (i.e., 0.033 M) and 2.5 mmol (i.e., O.033M) of unneutralised $N{H}_3$: molecules. This $N{H}_3$; exists in the equilibrium:

The final $75\mathrm{mL}$ solution after neutralisation already contains $2.5\mathrm{mmol}{\mathrm{NH}}_4^{+}$ions (i.e. $0.033\mathrm{M}$ ), thus total concentration of ${\mathrm{NH}}_4^{+}$ions is given as:
$\left[{\mathrm{NH}}_4^{+}\right]=0.033+\mathrm{y}$
As $y$ is small, $\left[{\mathrm{NH}}_4\mathrm{OH}\right];0.033\mathrm{M}$ and $\left[{\mathrm{NH}}_4^{+}\right]:0.033\mathrm{M}$.
We know,
$\begin{array}{rl}{\mathrm{K}}_{\mathrm{b}}& =\left[{\mathrm{NH}}_4^{+}\right]\left[{\mathrm{OH}}^{-}\right]/\left[{\mathrm{NH}}_4\mathrm{OH}\right]\\ & =\mathrm{y}(0.033)/(0.033)=1.77\times 10^{-5}\mathrm{M}\end{array}$
Thus, $\mathrm{y}=1.77\times 10^{-5}=\left[{\mathrm{OH}}^{-}\right]$
$\left[{\mathrm{H}}^{+}\right]=10^{-14}/1.77\times 10^{-5}=0.56\times 10^{-9}$
Hence, $\mathrm{pH}=9.24$