Materials: - NaOH (0.1 mol dm-3), CH3COOH (0.1 mol dm-3), pH meter, stand, clamp, thermometer, beakers, pipette, magnetic spin, titration pipette.
Methods:-
- 25 ml of ethanoic acid was poured into a beaker and placed below the titration pipette.
- Then the pH meter was calibrated and placed in the beaker to measure the initial pH.
- 25 ml of NaOH was poured into the titration pipette.
- Then 1 ml sodium oxide was allowed to pour into the beaker. A magnetic spin was placed in the beaker to facilitate the reaction.
- The change in pH was recorded.
Results:-
Table 1. Change in the pH value of Ethanoic acid as 25 ml of NaOH is added to it.
It can be seen from Table 1 that the pH value of Ethanoic acid increases as more sodium hydroxide is added. The pH value also took an abrupt increase from 5.89±0.01 to 11.31±0.01 as the volume of NaOH increased from 15±0.05 to 16±0.05 ml.
Graph 1. Change in the pH value of Ethanoic acid as the volume of NaOH increases
As seen from Graph 1 the pH value gradually increases up to a point where it drastically show increase in the pH value. Then it begins to increase gradually again. All of this is due to the change of ethanoic acid to its conjugate base as more and more sodium hydroxide is added.
The pka value of ethanoic acid can be found by analyzing the above titration graph. From the graph it is possible to find the equivalence point. The equivalence point is a point where the acid-base solution is neutral. This means that all of the acid has been consumed and will have a pH value greater than 7 since ethanoic acid is a weak acid and NaOH is strong acid. It will also make it possible to find the volume of NaOH used to neutralize the acid. By dividing this volume by two it is possible to find the volume of NaOH at half the equivalence point. This is the point where half of the ethanoic acid has changed to the ethanoic ion conjugate base. This means that their concentrations are equal. Therefore ka will be equal to [H+] as shown below:-
[H+] = ka x [CH3COOH]
[CH3COO- ]
ka = [H+] , since [ CH3COOH]= [CH3COO- ]
ka = 10-pH
Then as pka = - log ka
pka = - log 10-pH
pka = pH
Since Pka is equal to pH, the pH value at the half titration point can correspond to the pka value of ethanoic acid. The equivalent point on Graph 1 is found to be 15.5±0.05 ml. Dividing this value by two will result in 7.75±0.32 ml. (The new uncertainty was found by calculating the percentage uncertainty of 15.5). The pH value at 7.75±0.32 ml is 4.60±0.01. The literature pka value of ethanoic acid is found to be 4.76. Therefore the experimental value calculated here is very close. The difference may be due to uncertainties and limitations of the experiment.
Conclusion: - according to this experiment the pka value of ethanoic acid was found to be 4.60. This is a very close result to the literature value which is 4.76. The difference may be a result of uncertainties and limitations.
Limitations: - the volume used for titration was 25 ml. instead using higher volumes of sodium hydroxide and ethanoic acid would allow the effect of one drop to be more accurate and it will also minimize the random/systematic errors on the final result.
Improvements: - using higher volumes such as 100ml of sodium hydroxide and ethanoic acid.
