pH of a Weak Acid

Hydrochloric Acid is a strong acid; in aqueous solution all of the HCl molecules present have become ionised.

HCl(aq) + H2O(l) ® H3O+(aq) + Cl-(aq)

Hydrochloric Acid is a monobasic acid, and in a 0.1 mol dm-3 solution the concentration of H3O+(aq) is also 0.1 mol dm-3.

The pH of a 0.1 mol dm-3 solution of HCl(aq) can easily be calculated:

pH = -log[H+]

\ pH = 1

Ethanoic Acid is a weak acid; in aqueous solution it is in equilibrium with its conjugate base. At any instant in time, only a small proportion of its molecules have become ionised.

CH3COOH(aq) + H2O(l) = H3O+(aq) + CH3COO-(aq)

Therefore, in a 0.1 mol dm-3 solution of CH3COOH(aq), the H3O+(aq) must be much less than 0.1 mol dm-3.

What is the pH of a 0.1 mol dm-3 solution of CH3COOH(aq) at 298 K?

For this calculation we need to know the acid dissociation constant for ethanoic acid at 298 K.

Ka = [CH3COO-(aq)][H3O+(aq)] / [CH3COOH(aq)] = 1.75 x 10-5 mol dm-3

The 'equilibrium table' below sets out the calculation.

 CH3COOH(aq) +H2O(l)   =  H3O+(aq) + CH3COO-(aq) 
Initial conc.
(before ionisation)
0.1   - -mol dm-3
Change-y   +y +y 
Conc at eqm. 0.1 - y    +y +y mol dm-3

In this calculation a 'simplification' can be made, since we have a weak acid with only a small amount of ionisation at equilibrium. This means that the value of y is so small compared with 0.1 that 0.1-y » 0.1.

Substituting the equilibrium concentration values into the equilibrium expression gives

y2/0.1 = 1.75 x 10-5 mol dm-3

\ [H3O+(aq)] = 1.3 x 10-3 mol dm-3

\ pH = 2.88