Experimental procedure :
Preparation of the reactants ethanoic acid and propan-1-ol:
-
11.50 cm3 of glacial ethanoic acid and 15.00 cm3 of propan-1-ol were placed in a clean and dry 100 cm3 boiling flask from the two burettes, which were containing of propan-1-ol and glacial ethanoic acid respectively directly.
- The two reactants were then mixed thoroughly.
The first titration:
- The burette was rinsed, and then it was filled up with sodium hydroxide solution.
-
On the other hand, a 100 cm3 conical flask containing about 25 cm3 deionized water and two drops of phenolphthalein indicator solution was prepared.
-
1.00 cm3 of the mixture was then transferred to the 100 cm3 conical flask prepared.
- The initial burette reading was recorded. Then, the mixture in the conical flask was titrated with 0.4M NaOH until the end point is reached when the colour had changed from colourless to pink.
- The final burette reading was recorded.
Addition of sulphuric acid:
-
8 drops of conc. H2SO4 were added to the remainder of the acid-alcohol solutions in the boiling flask with continuously swirling.
The second titration:
-
Another 1.00 cm3 of the sample was pipetted out to a conical flask containing 25 cm3 deionized water and 2 drops of phenolphthalein indicator solution.
- The initial burette reading was recorded. The sample was titrated with NaOH immediately.
- The final burette reading was recorded.
The first reflux:
- The boiling flask was attached to a water-cooled reflux condenser. The joints were well-sealed by greasing. And a boiling chip was added to it.
- The apparatus were fixed to place vertically by using the clamps.
- Then, the boiling flask was put into a water bath and was heat gently.
- The system was left for one hour for reflux. During the reflux, the degree of heating was adjusted to make sure that the vapours did not rise above the lower half of the condenser.
The third titration:
- After 1 hour, the flask and its contents were cooled in an ice bath. As the mixture inside had turned milky, the flask and its contents were heated again until no more milky substance was found. Then, the flask and its contents were cooled again by running water instead of ice bath.
-
1.00 cm3 sample was removed from the flask for titration with the NaOH as before.
- The initial and final burette reading were recorded.
The second reflux:
- The refluxing was continued for an additional 1/2 hour, and then the flask and its contents were cooled as before.
The fourth titration:
-
Another 1.00 cm3 sample was removed from the flask for titration.
- The initial and final burette reading were recorded.
Data Collection and presentation :
Volume of glacial ethanoic acid (density = 1.05 g cm-3 )used : 11.40 cm3
Volume of propan-1-ol (density = 0.8 g cm-3 ) used : 15.00 cm3
Molarity of sodium hydroxide solution used : 0.4 M
Catalyst used : 8 drops of conc. H2SO4
Indicator used: about 2 drops of phenolphthalein indicator solution
Table 1(vol of NaOH solution used in each titration)
Calculations :
1. Equation for the esterification reaction between ethanoic acid and propan-1-ol:
CH3COOH (l) + CH3CH2CH2OH (l) CH3COOCH2CH2CH3 (l) + H2O (l)
2. Equation for the neutralization reaction between ethanoic acid and NaOH:
CH3COOH (l) + NaOH (aq) → CH3COONa (aq) + H2O (l)
3. Equation for the neutralization reaction between H2SO4 and NaOH:
H2SO4 (aq) + 2 NaOH (aq) → Na2SO4 (aq) + 2 H2O (l)
From the results obtained, and the three equations of the reactions, we have:
From the first titration,
we can find out the no. of moles of ethanoic acid present in 1 cm3 of mixture at the beginning.
No. of moles of NaOH added = 19.5 / 1000 × 0.4 = 0.0078 mole
From equation (2), no. of moles of ethanoic acid present in 1 cm3 of mixture at the beginning
= no. of moles of NaOH = 0.0078 mole
Relative molecular mass of ethanoic acid = 12×2 + 1×4 + 16×2 = 60
Volume of ethanoic acid in 1 cm3 mixture = 0.0078 X 60 /1.05 = 0.4457 cm3
Volume of propan-1-ol in 1 cm3 mixture = 1 - 0.4457 = 0.5543 cm3
No. of moles of propan-1-ol in 1 cm3 mixture = 0.5543 X 0.8/60 = 0.00739 mol
From the second titration,
No. of moles of NaOH added = 19.75 / 1000 × 0.4 = 0.0079 mole
No. of moles of NaOH reacted with H2SO4 = ( 0.0079 – 0.0078 ) mole = 0.0001 mole
∴ the amount of of H2SO4 present in 1 cm3 of mixture is 0.0001 mole
From the third titration,
No. of moles of NaOH added = 6.85 / 1000 × 0.4 = 0.00274 mole
No. of moles of NaOH reacted with CH3COOH = ( 0.00274 – 0.0001 ) mole = 0.00264 mole
From equation (2), no. of moles of ethanoic acid remained unreacted in 1 cm3 of mixture at the end of reflux = no. of moles of NaOH = 0.00264 mole
From equation (1), no. of moles of ester formed = no. of moles of water formed
= no. of moles of ethanoic acid reacted
= (0.0078 -0.00264 ) mole = 0.00516 mole
From equation (1), no. of moles of propan-1-ol reacted
= no. of moles of ethanoic acid reacted = 0.00516 mol
Equilibrium expression for the reaction:
∴ Kc = (5.16)(5.16)/ (2.23)(2.64) = 4.52
From the experimental result, the value of the equilibrium constant for the reaction is 4.52.
Note: In the calculation above, the assumptions that all the solutions are ideal and the volume
of conc. H2SO4 added can be neglected were made.
Answers to the Questions to be discussed:
-
What is the purpose of adding 25 cm3 of deionized water to the sample before each titration?
The purpose of this is to provide better observation for the colour change at the end-point during the titration since the volume of the sample to be titrated is so small, just 1 cm3. Moreover, the volume of deionized water added doesn’t affect the number of moles of reactants in the 1 cm3 sample. Thus, deionized water can be added to the sample before each titration.
-
Why should the sample be titrated immediately after the addition of conc. H2SO4 ?
After the addition of conc. H2SO4, which acts an acid catalyst, the esterification reaction is speeded up. The amount of ethanoic acid will gradually decrease. The titration should be carried out immediately so that the amount of the ethanoic acid would be the same as if the conc. H2SO4 is not added. Otherwise, we can not find out the volume of H2SO4 accurately by the results of the first and the second titration.
-
Why should the reflux process be repeated if the third and fourth titrations do not agree to within 0.2 cm3 ?
If the third and fourth titrations agree to within 0.2 cm3, it shows that the reaction has reached the equilibrium point. It is because as if the two titrations don’t agree to 0.2 cm3, it indicates that the concentration of ethanoic acid is still not a constant, the equilibrium position is still not reached. We can not take the reading of this titration to find the final number of moles of the reactants and products at equilibrium.
Overall Comment to the Experiment:
1. Uncertainty / Error Analysis :
A. Error in the loss of reactants and products
- Leakage of the system:
The joints in the system should be well-sealed by greasing to prevent any loss of volatile substance in the reaction mixture. However, during the experiments, the grease was found to have melted. This causes leakage of the system, which allows the volatile substance in the reaction mixture to escape away from the flask. The concentration of both the reactants and product will decrease. However, as the number of moles of them present in 1 cm3 of the mixture is different, the concentration decreases in different extent, and causes a deviation in the value of equilibrium constant.
- Transfer of the mixture:
There was a loss of reactants and products during the transfer of the 1 cm3 reaction mixture from the boiling flask to the conical flask. When the stopper of the boiling flask was removed, the volatile substance can escape away easily and caused errors to the results.
B. Error in volume of the reaction mixture for titration
-
The volume used for titration was so small, just 1 cm3.
In general, for the two samples from the same experiment, if the sample taken for titration is larger in volume, the percentage of deviation would be smaller than when the sample is smaller in volume.
- There was an error in measurement with the use of the pipette.
A pipette was used to obtain 1 cm3 sample of the reaction mixture. The pipette is accurate to 1cm3 sample of the reaction mixture giving an overall accuracy of 0.02cm3. Using the formula to calculate the percentage of uncertainty : 0.02/1 x 100 = 2%
C. After the reflux, the flask and its contents are cooled in an ice bath, however, solidification of ester occurred. This change caused the change of equilibrium position to shift towards the products. As the amount of liquid ester in the system decreased, the forward reaction rate increased to replenish the loss. As a result, the amount of ethanoic acid decreased to reach another equilibrium position. Although the equilibrium constant would not be affected. However, it takes some time to do so and if titration was taken at that time, when the new equilibrium position was not reached, we can not find out the number of moles of reactants and products at equilibrium position and this causes errors.
D. After the addition of sulphuric acid, the 1 cm3 sample should be titrated immediately or otherwise the results would not be accurate enough since the reaction is speeded up due to the presence of acid catalyst. The amount of ethanoic acid decreased gradually.
Therefore, the volume of H2SO4 obtained from the first and second titration in the experiment would be smaller than the actual volume added.
E. The boiling chip should not be added too much to the reaction mixture as it would absorb the substance inside the reaction mixture and causes error.
F. Since the equilibrium constant Kc depends on the temperature. Its value is fixed at a constant temperature only. As we obtained the initial number of moles of the reactants at room temperature, we should also get the final number of moles of the reactants and products at equilibrium position at room temperature to find out the equilibrium constant at this particular temperature (25℃). As heating of the reaction mixture is involved in the experience, we should cool down the reaction mixture to room temperature before titration after refluxing. However, as solidification of ester occurred when it is cooled by ice bath. It was difficult to maintain the temperature back to the room temperature.
This causes error to the result.
2. Discussion and Conclusion (ways to minimize errors) :
A. In order to reduce the loss of reactants and products, firstly, the joints of the system should all be sealed by vaseline instead of grease which can a much better performance in sealing the joints. This can prevent the volatile substance to escape away from the system through the leakage. Secondly, the flask should always be stoppered unless there is transferring of the reaction mixture. The transferring process should be done quickly and the flask should be stoppered at once after transferring.
B. As the percentage of deviation would be smaller when using larger collected volumes of sample, we should use a larger volume of sample each time for titration, say 5 cm3, so that the results are far more reliable. At the same time, the initial volume of the reactants used should also be increased.
C. As solidification may occur during cooling in an ice bath, the flask should be cooled by running water. However, sometimes solidification may still occur due to rapid cooling. Therefore, in order to cool down the flask as well as maintain the temperature of the reacture mixture back to the room temperature, the best method is to leave the flask behind until it is cooled down. Although it can ensure that it is at room temperature, it is time consuming.
D. Instead of carrying out two titration to find out the amount of H2SO4 added, the volume of H2SO4 can be measured before adding to the reaction mixture. This can prevent the errors that have mentioned before which existed in making two titrations.
In short conclusion, the experiment is not reliable and modification of the experiment is needed to obtain a more precise result.
Modification of the experiment:
- Use another method to determine the equilibrium constant for the reaction
- Gas Chromatography
We can obtain the ester product by the method of distillation and usually the product isolated from the distillation is not 100% pure ester. Contaminants can include unreacted alcohol, water, and solvent. A gas chromatograph of the product will tell the actual percent ester in the distillation mixture. The concentration of the different species can then be known, thus the equilibrium constant can be found.