At equivalence, 25 mL 0.10
NaOH has been added to 25 mL 0.10
CH
3CO
2H.
The product–favoured reaction results in the major species shown below:
| For the reaction | CH3CO2H | + | OH– | → | CH3CO2– | + | H2O | amount added/mol:
| 0.0025 | | 0.0025 | | 0 | | 0 | amount reacting/mol:
| –0.0025 | | –0.0025 | | +0.0025 | | +0.0025 | | amount remaining/mol: | 0 | | 0 | | 0.0025 | | 0.0025 | | | At equivalence: c(CH3CO2Na) = | 0.0025 mol | = 0.05 | | 0.050 L | |
CH3CO2 – is the major species that is "pH active".
The pH of the resulting solution is calculated using:
The solution is alkaline due to:
CH3CO2 – + H2O  CH3CO2H + OH – | | Ka = | [CH3CO2 –][H3O+] | = 1.74 × 10 –5 | | [CH3CO2H] |
|
BUT CH3CO2Na is a weak base, thus [CH3CO2–] = 0.05
Through the reaction above,
[CH3CO2–] is related to [H3O+] AND
[CH3CO2H] = [OH –] = Kw/[H3O+] = 10 –14/[H3O+]
| | 1.74 × 10 –5 = | 0.05 [H3O+] | | [CH3CO2H] |
| 1.74 × 10 –5 = | 0.05 [H3O+] | | 10 –14/[H3O+] | [H3O+] = 1.9 × 10-9
pH = 8.73 |