Question

Determine the pH value of 0.001 M acetic acid solution if it is 2% ionised at this concentration. How can the degree of dissociation of this acetic acid solution be increased?

Answer 1

\[\ce{CH3COOH ⇌[water] CH3COO^- + H+}\]

Initial molar conc.	C	O	O
Eqborm molar conc.	C(1 - α)	Cα	Cα

α = Degree of ionisation

α = `2/100`

C = 0.001 M

(H⁺) = Cα

= `.001 xx 2/100`

= `2 xx 10^-5`M

pH = -log(H⁺)

= -log(`2 xx 10^-5`)

= -log 2 + 5 log 10

= - `0.3010 + 5 xx 1`

= 5 - 0.3010

pH = 4.699

The degree of dissociation of acetic acid can be increased by dilution.