K _ a for CH _ 3 COOH is 1 . 8 × 10 ^ - 5 and K _ b for NH _ 4 OH...

${K}_{a}$ of ${CH}_{3}COOH= 1. 8 \times {10}^{- 5}$

${K}_{a}$ of ${NH}_{4}OH= 1. 8 \times {10}^{- 5}$

We know that $pH$ of solution of the salt made up of weak acid \left({CH}_{3}COOH\right)$and weak base \left({NH}_{4}OH\right)$ can be found using formula

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

${pK}_{a} = - \log \left({K}_{a}\right)<$br/>$= - \log\left(1. 8 \times {10}^{- 5}\right) = - \left[\log 1. 8 + log{10}^{- 5}\right] = - \left[0. 2553 - 5\right] = + 4. 7447{pK}_{b} = - \log\left({K}_{b}\right) = --\log\left(1. 8 \times {10}^{- 5}\right) =4. 7447pH = 7 + \dfrac{1}{2}\left[4. 7447 - 4. 7447\right] = 7 + 0 = 7$

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