Overview with all three cases

Two equilibrium conditions are satisfied in Solutions 1-3

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

Solution 1 is 0.100

CH₃CH₂CO₂H.
Solution 2 is 0.100

CH₃CH₂CO₂Na
Solution 3 is 0.100

in CH₃CH₂CO₂Na and CH₃CH₂CO₂H.

	initial concentrations			equilibrium concentrations deduced on previous pages
	c(CH₃CH₂CO₂^–)	c(CH₃CH₂CO₂H)	pH measured	[CH₃CH₂CO₂^–]	[CH₃CH₂CO₂H]	[H₃O⁺]	K_a
Solution 1	0	0.100	2.94	1.15×10^–3	0.099	1.15 ×10^–3	1.3 ×10^–5
Solution 2	0.100	0	8.94	0.100	8.77×10^–6	1.15 ×10^–9	1.3 ×10^–5
Solution 3	0.100	0.100	4.89	0.100	0.100	1.29 ×10^–5	1.3 ×10^–5

Irrespective of the initial concentrations of the acid and its conjugate base, [acid], [conjugate base] and [H₃O⁺] combine according to the K_a expression to give the same number. This number is a characteristic of the conjugate pair.
In each of the solutions [H₃O⁺][OH^–] = K_w= 10^–14
.

Dr Sheila Woodgate
© The University of Auckland 2002