K a for CH 3 COOH is 1 . 8 × 10 - 5 and K b for NH 4 OH is 1 . 8 × 10 - 5 . The pH of ammonium acetate…

${\mathrm{K}}_{\mathrm{a}}$ of ${\mathrm{CH}}_3\mathrm{COOH}$$=1.8\times {10}^{-5}$

${\mathrm{K}}_{\mathrm{a}}$ of ${\mathrm{NH}}_4\mathrm{OH}$$=1.8\times {10}^{-5}$

We know that $\mathrm{pH}$ of solution of the salt made up of weak acid $\left({\mathrm{CH}}_3\mathrm{COOH}\right)$ and weak base $\left({\mathrm{NH}}_4\mathrm{OH}\right)$ can be found using formula

$\mathrm{pH}=7+\frac{1}{2}\left[{\mathrm{pK}}_{\mathrm{a}}-{\mathrm{pK}}_{\mathrm{b}}\right]$

${\mathrm{pK}}_{\mathrm{a}}=-\log \left({\mathrm{K}}_{\mathrm{a}}\right)$
$$\begin{gathered} =-\log \left(1.8\times {10}^{-5}\right) \\ =-\left[\log 1.8+\log {10}^{-5}\right] \\ =-\left[0.2553-5\right] \\ =+4.7447 \\ {\mathrm{pK}}_{\mathrm{b}}=-\log \left({\mathrm{K}}_{\mathrm{b}}\right) \\ =-\log \left(1.8\times {10}^{-5}\right) \\ =4.7447 \\ \mathrm{pH}=7+\frac{1}{2}\left[4.7447-4.7447\right] \\ =7+0 \\ =7 \end{gathered}$$

If ${\mathrm{K}}_{\mathrm{a}}={\mathrm{K}}_{\mathrm{b}}$, then $\mathrm{pH}$ of salt is always $7$.
Therefore, the $\mathrm{pH}$ of given solution is $7$.