`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

a. `[CH_(3)COOH]_("Just after mixing") = (0.2)/(2) = 0.1M`
`[HCI]_("Just after mixing") = (0.2)/(2) = 0.1M = [H^(o+)]_(From HCI)`
Note: Equal volumes added.
`{:(,CH_(3)COOH,hArr,CH_(3)COO^(Theta),+,H^(oplus)),(underset(("conc"))(t=0),C,,0,,0),(t=t_(eq),C-C alpha,,C alpha,,C alpha):}`
`K_(a) = ([CH_(3)COO^(Θ)][H^(o+)])/([CH_(3)COOH]) = (C alpha.(C alpha + 0.1))/(C-Calpha)`
[`:'[H^(o+)]_(Total) = [H^(o+)]_(FromCH_(3)COOH) + [H^(o+)]_(FromHCI)]`
`rArr C alpha + 0.1 ~~ 0.1, C - C alpha ~~C]`
[`:' alpha` will be small due to common ion effect]
`rArr K_(a) ~~ alpha xx 0.1`
`rArr alpha = (K_(a))/(0.1) = 2 xx 10^(-4) [Check: 1 - alpha = 1 - 2 xx 10^(-4) ~~1]`
and `[H^(o+)]_(Total) = Calpha + 0.1 = 0.1 xx2xx 10^(-4) + 0.1 ~~ 0.1M`
`rArr pH = 1`
b. `6gNaOH = (6)/(40//1) = 0.15mEq -= 150mmol NaOH`
`mmol HCI = 500 xx 0.2 = 100 mmol`
`NaOH` will first react with `HCI` and if there is a leftover then it will react with `CH_(3)COOH`.
`rArr` mmol NaOH left = 150 - 100 = 50
`{:(NaOH +,CH_(3)COOH,hArr,CH_(3)COONa,+,H_(2)O),(50,100,,-,,-),(-,50,,50,,):}`
`rArr` Formation of an acidic buffer
`rArr pH = pK_(a) + log.(["Salt"])/(["Acid"])`
`= 4.7 + log.(50//V_("Total"))/(50//V_("Total")) = 4.7`