Question

Calculate the pH of a solution that is 0.20M with benzoic acid and .18with its salt, sodium benzoate

Answer 1

pH$=$ pk$_a$ $+log\frac{[A^-]}{[HA]}$

k$_a=6.5×10^{-5}$

pH$=-log(6.5×10^{-5})+log\frac{0.18}{0.20}$

$=4.1871-0.04576$

$=4.1413$