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Q.
$K_a$ for $CH_3COOH$ is $1.8 \times 10^{-5}$ and $K_b$ for $NH_4OH$ is $1.8 \times 10^{-5}$. The $pH$ of ammonium acetate will be
Equilibrium
Solution:
$CH_3COONH_4$ is a salt of weak acid $(CH_3COOH)$ and weak base $(NH_4OH)$.
$pH=7+\frac{1}{2}\left(pK_{a}-pK_{b}\right)$
$pK_{a}=-log\,K_{a}$
$=-log\left(1.8\times10^{-5}\right)=4.74$
$pK_b = - log \,K_b $
$= - log (1.8 \times 10^{-5}) = 4.74$
$pH=7+\frac{1}{2}\left(4.74-4.74\right)=7.0$