Question

$K_a$ 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

• A. $7.005$
• B. $4.75$
• C. $7.0$
• D. between $6$ and $7$

Answer 1

Answer: (C)

$CH_3COONH_4$ is a salt of weak acid $(CH_3COOH)$ and weak base $(NH_4OH)$.
$pH=7+\frac{1}{2}\left(pK_{a}-pK_{b}\right)$
$pK_{a}=-log\,K_{a}$
$=-log\left(1.8\times10^{-5}\right)=4.74$
$pK_b = - log \,K_b$
$= - log (1.8 \times 10^{-5}) = 4.74$
$pH=7+\frac{1}{2}\left(4.74-4.74\right)=7.0$