Chemical and Ionic Equilibrium 2 Question 74
72. What is the $\mathrm{pH}$ of the solution when 0.20 mole of $\mathrm{HCl}$ is added to one litre of a solution containing
(i) $1 \mathrm{M}$ each of acetic acid and acetate ion,
(ii) $0.1 \mathrm{M}$ each of acetic acid and acetate ion?
Assume the total volume is one litre.
$K_{a}$ for acetic acid $=1.8 \times 10^{-5}$.
$(1987,5 M)$
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Solution:
- (i) 0.20 mole $\mathrm{HCl}$ will neutralise 0.20 mole $\mathrm{CH}{3} \mathrm{COONa}$, producing $0.20 \mathrm{~mol} \mathrm{CH}{3} \mathrm{COOH}$. Therefore, in the solution moles of $\mathrm{CH}_{3} \mathrm{COOH}=1.20$
Moles of $\mathrm{CH}_{3} \mathrm{COONa}=0.80$
$$ \begin{aligned} \mathrm{pH} & =\mathrm{p} K_{a}+\log \frac{[\text { Salt }]}{[\text { Acid }]} \ & =-\log \left(1.8 \times 10^{-5}\right)+\log \frac{(0.80)}{(1.20)}=4.56 \end{aligned} $$
(ii)
$$ \mathrm{CH}{3} \mathrm{COONa}+\mathrm{HCl} \longrightarrow \mathrm{CH}{3} \mathrm{COOH}+\mathrm{NaCl} $$
$$ \begin{array}{lcccc} \text { Initial } & 0.10 & 0.20 & 0 & 0 \ \text { Final } & 0 & 0.10 & 0.10 & 0.10 \end{array} $$
Now, the solution has 0.2 mole acetic acid and 0.1 mole $\mathrm{HCl}$. Due to presence of $\mathrm{HCl}$, ionisation of $\mathrm{CH}_{3} \mathrm{COOH}$ can be ignored (common ion effect) and $\mathrm{H}^{+}$in solution is mainly due to $\mathrm{HCl}$.
$$ \begin{aligned} {\left[\mathrm{H}^{+}\right] } & =0.10 \ \mathrm{pH} & =-\log (0.10)=1.0 \end{aligned} $$
