50 mL of 0.3 M HCI is mixed with 50 mL of 0.4 M NH₃ solution. If pK_a of ${\mathrm{NH}}_4^{+}$ is 9.26, pH of the mixture is

$$\begin{gathered} {\mathrm{pK}}_{\mathrm{a}}\left({\mathrm{NH}}_4^{+}\right)=9.26 \\ \therefore {\mathrm{pK}}_{\mathrm{b}}\left({\mathrm{NH}}_3\right)=14-9.26=4.74 \end{gathered}$$

$$\begin{gathered} {\mathrm{NH}}_3+\mathrm{HCl}\rightleftharpoons {\mathrm{NH}}_4\mathrm{Cl} \\ \mathrm{Initial} 20 15 0 \\ \mathrm{After} 5 0 15 \end{gathered}$$

Mixture is a buffer containing 5 millimoles of NH₃ (base) and 15 mitlimoles of ${\mathrm{NH}}_4^{+}$ (conjugare acid).

$\therefore \mathrm{pOH}={\mathrm{pK}}_{\mathrm{b}}+\log \frac{\left[{\mathrm{NH}}_4^{+}\right]}{\left[{\mathrm{NH}}_3\right]}=4.74+\log 3$

$$\begin{gathered} \therefore \mathrm{pH}=14-\mathrm{pOH}=9.26-\log 3=9.26-0.48 \\ = 8.78 \end{gathered}$$