Problem summary

Ka = [CH3CH2CO2–][H3O+]
[CH3CH2CO2H]
The pH of 0.100 
CH3CH2CO2H is 2.94.
Calculate Ka for the CH3CH2CO2H / CH3CH2CO2– conjugate pair.

[H3O+] = 10–2.94
= 1.15 ×10–3 
Use the given pH to calculate the [H3O+].
pH = -log[H3O+]
CH3CH2CO2H(aq) + H2O → CH3CH2CO2–(aq) + H3O+
Deduce the reaction occurring to reach equilibrium from the pH:
The reaction occurring to reach equilibrium is the only source of CH3CH2CO2– and is the major source of H3O+.

given data (
)		 [CH3CH2CO2H] [H3O+] [CH3CH2CO2–]
initial
(before reaction)		 0.100 0 0
change
(due to reaction)		 -1.15 ×10–3 +1.15 ×10–3 +1.15 ×10–3
equilibrium 0.099 1.15 ×10–3 1.15 ×10–3
Use the reaction occurring to reach equilibrium,
the initial concentration of CH3CH2CH2CO2H
and the calculated [H3O+]
to deduce [CH3CH2CH2CO2H] and [CH3CH2CH2CO2–]
The calculated [H3O+] and the stoichiometry of the reaction occurring to reach equilibrium can be used to calculate the change in the concentrations of each member of the conjugate pair and hence their equilibrium concentrations.

 

Ka = (1.15 ×10–3)2
= 1.3 × 10–5
0.099 
Substituting the equilibrium concentrations:
 
Note that if the extent of reaction to reach equilibrium is small, the initial and equilibrium concentrations of species initially present in high concentration (major species) are nearly equivalent. The same value of Ka would have been calculated if [CH3CH2CO2H] had been taken as 0.100
.
Dr Sheila Woodgate
© The University of Auckland 2002