A buffer solution contains 1 mole of ${\left({\mathrm{NH}}_4\right)}_2{\mathrm{SO}}_4$ and 1 mole of ${\mathrm{NH}}_4\mathrm{OH} \left({\mathrm{K}}_{\mathrm{b}} = {10}^{-5}\right).$  The pH of solution will be

$\mathrm{pOH} = {\mathrm{pK}}_{\mathrm{b}} + \log \frac{\left[\mathrm{Salt}\right]}{\left[\mathrm{Base}\right]} = {\mathrm{pK}}_{\mathrm{b}} + \log \frac{\left[\mathrm{Cation}\right]}{\left[\mathrm{Base}\right]}$

$$\begin{gathered} {({\text{NH}}_4\text{)}}_2{\text{SO}}_4\rightleftharpoons 2 {\mathrm{NH}}_4^{+}+{\mathrm{SO}}_4^{-2} \\ \left[{\mathrm{NH}}_4^{+}\right] = 2 \times 1=2 \\ {\mathrm{P}}^{{\mathrm{K}}_{\mathrm{b}}} =-\log {\mathrm{K}}_{\mathrm{b}} =-\log {10}^{-5} =5 \\ \therefore \mathrm{pOH} = 5 + \log 2 = 5.3 \\ \mathrm{or} \mathrm{pH} = 14-5.3 = 8.7 \end{gathered}$$