Question

What will be the $pH$ of $0.1MC{H}_{3}COON{H}_4$ ?
$K_a=1.8\times 10^{-5};K_b=1.8\times 10^{-5}$

Answer 1

Answer: 7

$pH=\frac{1}{2}\left[p{K}_w+p{K}_a-p{K}_b\right]$

$\because p{K}_a=p{K}_b$

$\therefore pH=\frac{1}{2}\times p{K}_w=\frac{1}{2}\times 14=7$