`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)`.

Answer

A, B, D

Explanation

a. Upon mixing these solutions, the volume beocmes double and the concentrations are reduced to one half.
`[CH_(3)COOH] = 0.1`
`[H^(o+)]^(3)of HCI = 0.1`
`{:(CH_(3)COOH hArr,CH_(3)COO^(Theta)+,H^(o+)),(0.1,0,0.1),((0.1-a),a,(0.1+a)):}`
`K_(a) =([CH_(3)COO^(Theta)][H^(o+)])/([CH_(3)COOH])`
`K_(a) = ([CH_(3)COO^(Theta)]xx(0.1xxa))/((0.1-a))`
`1.75 xx 10^(-5) = [CH_(3)COO^(Theta)]`
Percentage of ionisation of acetic acid
`= (1.75 xx 10^(-5))/(0.1) xx 100 = 1.75 xx 10^(-2)`
`pH =- log [H^(o+)] =- log 0.1 = 1`
b. When `6g` of `NaOH` is added.
Concentration of `NaOH = (6)/(40) = 0.15`
`0.15M NaOH = 0.10 M HCI +0.05M CH_(3) COOH`
In reaction with `CH_(3)COOH`, acidic buffer is formed.
`[CH_(3)COOH] = 0.1- 0.05 = 0.05`
`[CH_(3)COONa] = 0.05`
`pH = pK_(a) +"log" (["Salt"])/(["Acid"])`
`pH = pK_(a) =- log 1.75 xx 10^(-5)`
`=5 - log 1.75 = 4.75`