Chemical And Ionic Equilibrium Result Question 54
54. $500 \mathrm{~mL}$ of $0.2 $ $\mathrm{M}$ aqueous solution of acetic acid is mixed with $500 \mathrm{~mL}$ of $0.2 $ $\mathrm{M} $ $\mathrm{HCl}$ at $250^{\circ} \mathrm{C}$.
(i) Calculate the degree of dissociation of acetic acid in the resulting solution and $\mathrm{pH}$ of the solution.
(ii) If $6 \mathrm{~g}$ of $\mathrm{NaOH}$ is added to the above solution, determine the final $\mathrm{pH}$ (assuming there is no change in volume on mixing, $K _a$ of acetic acid is $1.75 \times 10^{-5 } \mathrm{~mol} / \mathrm{L}$.
(1984, 1M)
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Solution:
- (i) $\underset{C(1-\alpha)}{\mathrm{CH}_3 \mathrm{COOH}} \rightleftharpoons \underset{ C \alpha}{\mathrm{CH}_3 \mathrm{COO}^{-}}+\underset{C \alpha}{ \mathrm{H}^{+} }$
If no HCl is present,
$ \begin{aligned} {[\mathrm{HCl}] } & =\frac{0.2}{2}=0.10 \mathrm{M} \\ {\left[\mathrm{CH}_3 \mathrm{COOH}\right] } & =0.10 \mathrm{M} \end{aligned} $
The major contributor of $\mathrm{H}^{+}$in solution is HCl .
$ \begin{aligned} K_a & =\frac{C \alpha(0.1)}{C(1-\alpha)}=1.75 \times 10^{-5} \\ \alpha & =1.75 \times 10^{-4} \end{aligned} $
(ii)
$ \begin{aligned} \mathrm{mmol} \text { of } \mathrm{NaOH} \text { added } & =\frac{6}{40} \times 1000=150 \\ \mathrm{mmol} \text { of } \mathrm{HCl} & =500 \times 0.2=100 \\ \mathrm{mmol} \text { of } \mathrm{CH}_3 \mathrm{COOH} & =500 \times 0.2=100 \end{aligned} $
After neutralisation, mmol of $\mathrm{CH}_3 \mathrm{COOH}=50$
mmol of $\mathrm{CH}_3 \mathrm{COONa}=50$
$ \mathrm{pH}=\mathrm{p} K_a=4.75 $
BETA
