Chemistry - Ionic Equilibrium Question with Solution | TestHub

$40\mathrm{mL}$ 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 $\mathrm{HCl}$ $$\begin{array}{lccc}& {\mathrm{NH}}_4\mathrm{OH}+& \mathrm{HCl}\to {\mathrm{NH}}_4\mathrm{Cl}+{\mathrm{H}}_2\mathrm{O}& \\ \text{ Initial milli-eqv. }& 4& 2& 0\\ \text{ Milli-eqv. after }& 4-2& 0& 2\\ \text{ reaction }& =2& & \end{array}$$ $$\begin{array}{c}\therefore \mathrm{pOH}=\mathrm{pK}{K}_b+\log \frac{\left[{\mathrm{NH}}_4\mathrm{Cl}\right]}{\left[{\mathrm{NH}}_4\mathrm{OH}\right]}=4.74+\log \frac{2}{2}\\ =4.74+\log 1=4.74\\ \therefore \mathrm{pH}=14-4.74=9.26\end{array}$$