Chemical and Ionic Equilibrium - Result Question 110

Solution:

(i) $CH _3 COOH \rightleftharpoons CH _3 COO^{-}+H^{+}$

$ C(1-\alpha) \quad C \alpha \quad C \alpha $

If no $HCl$ is present,

$ \begin{alignedat} {[HCl] } & =\frac{0.2}{2}=0.10 M \\ {\left[CH _3 COOH\right] } & =0.10 M \end{aligned} $

The major contributor of $H^{+}$in solution is $HCl$.

$ \begin{aligned} K _a & =\frac{C \alpha}{C(1-\alpha)}=1.75 \times 10^{-5} \\ \alpha & =1.75 \times 10^{-4} \end{aligned} $

(ii) $mmol$ of $NaOH$ added $=\frac{6}{40} \times 1000=150$

$ mmol \text { of } HCl=500 \times 0.2=100 $

$mmol$ of $CH _3 COOH=500 \times 0.2=100$

After neutralisation, $mmol$ of $CH_3COOH=50$

$mmol$ of $CH_3COONa=50$

$ pH=p K_a=4.75 $