${\mathrm{p}}^{\mathrm{H}}$ of mixture containing two moles of ${\mathrm{NH}}_4\mathrm{OH}$ and half mole of ${\mathrm{H}}_2{\mathrm{SO}}_4 \mathrm{is} \left({\mathrm{p}}^{{\mathrm{K}}_{\mathrm{b}}} \mathrm{of} {\mathrm{NH}}_4\mathrm{OH} = 4.8\right)$

The mixture acts like a basic buffer;

${\mathrm{NH}}_4\mathrm{OH}$ is a monoacidic base;

Number of gram equivalents = 2 x 1 = 2

${\mathrm{H}}_2{\mathrm{SO}}_4$ is a dibasic acid;

Number of gram equivalents = 0.5 x 2 = 1

$\underset{1. \mathrm{g}. \mathrm{eq}.}{\mathrm{SA} } + \underset{\left.\begin{array}{r}1. \mathrm{g}. \mathrm{eq}.\\ 2 \mathrm{g}. \mathrm{eq}.\end{array}\right\}\underset{\mathrm{excess}}{1 \mathrm{g}.\mathrm{eq}.} }{\mathrm{WB}} \stackrel{ }{\to } \underset{1 \mathrm{g}. \mathrm{eq}.}{\mathrm{Salt}}+{\mathrm{H}}_2\mathrm{O}$

$$\begin{gathered} {\mathrm{p}}^{\mathrm{OH}}={\mathrm{p}}^{\mathrm{kb}}+\log \frac{{\mathrm{V}}_{\mathrm{s}}{\mathrm{N}}_{\mathrm{s}}}{{\mathrm{V}}_{\mathrm{b}}{\mathrm{N}}_{\mathrm{b}}} \\ {\mathrm{p}}^{\mathrm{OH}}=4.8+\log \frac{1}{1} \\ {\mathrm{p}}^{\mathrm{OH}}=4.8 ; \\ {\mathrm{p}}^{\mathrm{H}}=\left(14-4.8\right)=9.2 ; \end{gathered}$$