Chemical and Ionic Equilibrium - Result Question 130
####72. What is the $pH$ of the solution when 0.20 mole of $HCl$ is added to one litre of a solution containing
(i) $1 M$ each of acetic acid and acetate ion,
(ii) $0.1 M$ each of acetic acid and acetate ion?
Assume the total volume is one litre.
$K _a$ for acetic acid $=1.8 \times 10^{-5}$.
$(1987,5 M)$
Show Answer
Solution:
- (i) 0.20 mole $HCl$ will neutralise 0.20 mole $CH _3 COONa$, producing $0.20 mol CH _3 COOH$. Therefore, in the solution moles of $CH _3 COOH=1.20$
Moles of $CH _3 COONa=0.80$
$$ \begin{aligned} pH & =p K _a+\log \frac{[\text { Salt }]}{[\text { Acid }]} \ & =-\log \left(1.8 \times 10^{-5}\right)+\log \frac{(0.80)}{(1.20)}=4.56 \end{aligned} $$
(ii)
$$ CH _3 COONa+HCl \longrightarrow CH _3 COOH+NaCl $$
$$ \begin{array}{lcccc} \text { Initial } & 0.10 & 0.20 & 0 & 0 \ \text { Final } & 0 & 0.10 & 0.10 & 0.10 \end{array} $$
Now, the solution has 0.2 mole acetic acid and 0.1 mole $HCl$. Due to presence of $HCl$, ionisation of $CH _3 COOH$ can be ignored (common ion effect) and $H^{+}$in solution is mainly due to $HCl$.
$$ \begin{aligned} {\left[H^{+}\right] } & =0.10 \ pH & =-\log (0.10)=1.0 \end{aligned} $$
