40 ~mL of 0.1 M ammonia solution is mixed with 20 ~mL of 0.1 MHCl ....

$40 \mathrm{~mL}$ of $0.1 \mathrm{M}$ ammonia solution $=40 \times 0.1$ $=4$ milliequivalent ammonia solution $20 \mathrm{~mL}$ of $0.1 \mathrm{M} \mathrm{HCl}=20 \times 0.1$ $=2$ milli equivalent of $\mathrm{HCl}$ \begin{array}{lccc} & \mathrm{NH}_4 \mathrm{OH}+ & \mathrm{HCl} \rightarrow \mathrm{NH}_4 \mathrm{Cl}+\mathrm{H}_2 \mathrm{O} \\ \text{ Initial milli-eqv. } & 4 & 2 & 0 \\ \text{ Milli-eqv. after } & 4-2 & 0 & 2 \\ \text{ reaction } & =2 & & \end{array}$$\begin{gathered} \therefore \mathrm{pOH}=\mathrm{pK} K_b+\log \frac{\left[\mathrm{NH}_4 \mathrm{Cl}\right]}{\left[\mathrm{NH}_4 \mathrm{OH}\right]}=4.74+\log \frac{2}{2} \\ =4.74+\log 1=4.74 \\ \therefore \mathrm{pH}=14-4.74=9.26 \end{gathered}$$