$K_a$ for $\mathrm{CH}_3 \mathrm{COOH}$ is $1.8 imes 10^{-5}$ and $K_b$ for $\mathrm{NH}_4 \mathrm

Question

$K_a$ for $\mathrm{CH}_3 \mathrm{COOH}$ is $1.8 	imes 10^{-5}$ and $K_b$ for $\mathrm{NH}_4 \mathrm{OH}$ is $1.8 	imes 10^{-5}$. The pH of ammonium acetate will be

Professor PC Thomas Classes · Answer

(c): $\mathrm{CH}_3 \mathrm{COONH}_4$ is a salt of weak acid $\left(\mathrm{CH}_3 \mathrm{COOH}ight)$ and weak base $\left(\mathrm{NH}_4 \mathrm{OH}ight)$. \ $\mathrm{pH}=7+\frac{1}{2}\left(\mathrm{p} K_a-\mathrm{p} K_bight)$\ $\mathrm{p} K_a=-\log K_a=-\log \left(1.8 	imes 10^{-5}ight)=4.74$\ $\mathrm{p} K_b=-\log K_b=-\log \left(1.8 	imes 10^{-5}ight)=4.74$\ $\mathrm{pH}=7+\frac{1}{2}(4.74-4.74)=7.0$