`500mL` of `0.2M` aqueous solution of acetic acid is mixed with `500mL` of `0.2HCI` at `25^(@)C`.
a. Calculate the degree of dissociation of acetic acid in the resulting solution and `pH` of the folution.
b. If `6g` of `NaOH` is added to the above solution determine the final `pH.[K_(a)` of `CH_(3)COOH =2 xx 10^(-5)`.

Explanation

(i)`CH_(3)COOHunderset("C(1-alpha)") rarr CH_(3)COO^(-)underset("Calpha") + H^(+)underset("Calpha")`
If no HCl is present, [HCl] = 0.2/2 = 0.10 M
`[CH_(3)COOH] = 0.10 M`
The major contributor of `H^(+)` in solution is HCl.
`K_(a) = Calpha(0.1)/C(1-alpha) = 1.75 xx 10^(-5)`
`alpha = 1.75 xx 10^(-4) (:' 1-alpha)`
(ii) mmol of NaOH added = 6/40 xx 1000 = 150
mmol of HCl = 500 xx 0.2 = 100
mmol of HCl = 500 xx 0.2 = 100
After neutralisation, mmol of `CH_(3)COOH = 50`
`mmol of CH_(3)COONa = 50`
pH = pKa = 4.75