${\mathrm{K}}_{\mathrm{a}}$ for ${\mathrm{CH}}_3\mathrm{COOH}$ is $1.8 \mathrm{x} {10}^{-5}$ and ${\mathrm{K}}_{\mathrm{b}}$ for ${\mathrm{NH}}_4\mathrm{OH}$ is $1.8 \mathrm{x} {10}^{-5}$. The pH of ammonium acetate will be:

Since, ${\mathrm{K}}_{\mathrm{a}}={\mathrm{K}}_{\mathrm{b}};\mathrm{pH}=\frac{1}{2}{\mathrm{pk}}_{\mathrm{w}}$

$\begin{array}{l}\mathrm{pH}=7+\frac{1}{2}\left({\mathrm{pK}}_{\mathrm{a}}-{\mathrm{pK}}_{\mathrm{b}}\right)\\ \mathrm{Ka}=1.8\times {10}^{-5}\end{array}$

$\begin{array}{l}\mathrm{pKa}=-\log \left(1.8\times {10}^{-5}\right)=5-\log 1.8\\ =4.744\end{array}$

$\begin{array}{l}{\mathrm{pK}}_{\mathrm{b}}=-\log \left(1.8\times {10}^{-5}\right)=4.744\\ \mathrm{pH}=7+\frac{1}{2}\left({\mathrm{pK}}_{\mathrm{a}}-{\mathrm{pK}}_{\mathrm{b}}\right)=7\end{array}$