For a solution prepared by dissolving a weak base in water to concentration c0(base), the conjugate acid and OH– are produced in reactant-favoured reaction (1).
(1) base + H2O
conjugate acid + OH–
At equilibrium,
for all bases: [base] = c0(base)0 - [OH–]
Assumption 1 for weak bases:
The extent of reaction 1 is low.
[base] ≈ c0(base)
Assumption 2 for weak bases:
Reaction 1 produces more OH– than reaction of H2O with itself
Substituting these into the Ka expression for the conjugate pair enables calculation of [H3O+] (and pH) if Ka is known.
(1) base + H2O
conjugate acid + OH–At equilibrium,
for all bases: [base] = c0(base)0 - [OH–]
Assumption 1 for weak bases:
The extent of reaction 1 is low.
[base] ≈ c0(base)Assumption 2 for weak bases:
Reaction 1 produces more OH– than reaction of H2O with itself
Then: [conjugate acid] = [OH–]
at 25°: Kw = [H3O+][OH–] = 10–14
[conjugate acid] =
at 25°: Kw = [H3O+][OH–] = 10–14
[conjugate acid] =10–14 [H3O+]
Substituting these into the Ka expression for the conjugate pair enables calculation of [H3O+] (and pH) if Ka is known.
| as above | unknown | |
| Ka = known | [conjugate base][H3O+] | |
| [acid] as above | ||