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.
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.
|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