Worked example where both the acid and the conjugate base are major species

K_a =	[CH₃CH₂CO₂^–][H₃O⁺]
	[CH₃CH₂CO₂H]

A solution of 0.100

CH₃CH₂CO₂H and 0.100

CH₃CH₂CO₂Na
has a pH of 4.89. Use these data to calculate K_a(CH₃CH₂CO₂H).

[H₃O⁺] = 10^–4.89

[H₃O⁺] = 1.29 ×10^–5

Calculate [H₃O⁺]:

CH₃CH₂CO₂^–(aq) + H₂O

CH₃CH₂CO₂H(aq) + OH^–
CH₃CH₂CO₂H(aq) + H₂O

CH₃CH₂CO₂^–(aq) + H₃O⁺

Enter the second product of the reaction of each species to reach equilibrium.

	[CH₃CH₂CO₂^–]	[CH₃CH₂CO₂H]	[H₃O⁺]
initial (before reaction)	0.100	0.100	1.29 ×10^–5
change (due to reaction)	Both members of a conjugate pair are consumed in one reaction to reach equilibrium and produced in the other.Their change in concentration is smaller than if only one member of the conjugate pair is dissolved.
equilibrium	0.100	0.100	1.29 ×10^–5

Enter the initial concentrations and the calculated [H₃O⁺].
Deduce the other equilibrium concentrations.

K_a =	0.100 × 1.29 ×10^–5
	0.100

K_a = 1.3 × 10^–5

Substitute the equilibrium concentrations:

Note that K_a equals [H₃O⁺] for systems having the same concentration for both members of the conjugate pair.