Worked example where only the base is the major species

Ka = [CH3CH2CO2–][H3O+]
[CH3CH2CO2H]
The pH of 0.100 
CH3CH2CO2Na is 8.94.
Calculate Ka for the CH3CH2CO2H/CH3CH2CO2– conjugate pair.

[H3O+] = 10–8.94
 = 1.15 ×10 –9
Use the given data to calculate the equilibrium hydronium ion concentration.
pH = -log[H3O+]

CH3CH2CO2–(aq) + H2O → CH3CH2CO2H(aq) + OH–
Also need [CH3CH2CO2–] and [CH3CH2CO2H].
CH3CH2CO2H is produced to reach equilibrium.
  [CH3CH2CO2–] [CH3CH2CO2H] [OH–]
initial

(before reaction)		 0.100 0 0
change

(due to reaction)		 –[OH–] =
–8.70 ×10–6		 +[OH–] =
+8.70 ×10–6
+[OH–] = 10–14
1.15 ×10–9
equilibrium
0.100 8.70 ×10–6 8.70 ×10–6
Use
the reaction occurring to reach equilibrium,
the calculated [H3O+], and
the fact that in any aqueous solution
Kw = [H3O+][OH–] = 10–14,

to calculate the change in the initial concentrations of each member of the conjugate pair and hence their equilibrium concentrations.

The extent of reaction occurring to reach equilibrium is very small.
The initial concentrations of the reacting species (the base of the pair) are approximately equal to the equilibrium concentrations.
However, the increase (≈100 times) in both [OH–] and [conjugate acid] is significant.
Ka = 0.100
 × 1.15 ×10–9
= 1.3 × 10–5
8.70 ×10–6
Substituting the equilibrium concentrations:

Dr Sheila Woodgate
© The University of Auckland 2002