Question

What is the $pH$ of the solution when $0.20$ mole of $HCl$ is added to one litre of a solution containing
(i) $1M$ each of acetic acid and acetate ion,
(ii) $0.1M$ each of acetic acid and acetate ion?
Assume the total volume is one litre.
$K_a$ for acetic acid $=1.8\times 10^{-5}$

Answer 1

(i) $0.20$ mole $HCl$ will neutralise $0.20$ mole $C{H}_{3}COONa$, producing $0.20mol$ $C{H}_{3}COOH$.
Therefore, in the solution moles of $C{H}_{3}COOH=1.20$
Moles of $C{H}_{3}COONa=0.80$
$pH=p{K}_a+\log \frac{[\text{ Salt }]}{[\text{ Acid }]}$
$=-\log \left(1.8\times 10^{-5}\right)+\log \frac{(0.80)}{(1.20)}=4.56$

Now, the solution has $0.2$ mole acetic acid and $0.1$ mole $HCl$. Due to presence of $HCl$, ionisation of $C{H}_{3}COOH$ can be ignored (common ion effect) and $H$ in solution is mainly due to $HCl$.
$\left[H^{+}\right]=0.10$
$pH=-\log (0.10)=1.0$