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)
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Answer:
Correct Answer: 72. $\left(8.7 \times 10^{-4} \mathrm{gL}^{-1}\right)$
Solution:
- (i) $0.20$ mole $HCl$ will neutralise $0.20$ mole $CH_3COONa$, producing $0.20$ $mol$ $ CH_3COOH$. Therefore, in the solution moles of $CH_3COOH=0.20$
Moles of $CH_3COONa=0.80$
$\begin{aligned} pH & =p K_a+\log \frac{[\text { Salt }]}{[\text { Acid }]} \\ & =-\log \left(1.8 \times 10^{-5}\right)+\log \frac{(0.80)}{(1.20)}=4.56 \end{aligned}$
(ii)
$ \quad \quad 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{alignedat} {\left[H^{+}\right] } & =0.10 \\ pH & =-\log (0.10)=1.0 \end{aligned} $
BETA
