To reduce the maximum % error of my results I have decided to make both the vinegars and distillates up to a 250ml solution. I will do this by placing my distillate (or 50cm3 of the vinegar) into a volumetric flask and filling it up to the mark with distilled water, making sure I use a fine-toothed pipette to improve accuracy. This will give me a diluted solution that I will use for the titration.
Firstly I will measure out 50cm3 of my vinegar or distillate using a pipette (I will use 2 x 25cm3 pipettes). I will then place it in a beaker and add a couple of drops of phenolphthalein indicator next I will place this onto a white tile, above which I will suspend my burette which I will fill with my NaOH solution. I will then take a reading of the level of the NaOH solution and then begin to slowly add my sodium hydroxide to my distillate or vinegar. When the colour of the solution changes to purple I will take another reading from my burette and from these values I will work out the amount of NaOH I have used in the titration. I will repeat this 3 times for each titrate and take an average. From this value I can then work out the number of moles of sodium hydroxide used in the titration and using the equation:
Carboxylic acid + NaOH = NaAcid + Water
I can work out the number of moles of acid I had in my titrate, and from this value I can work out the concentration of carboxylic acids in each of my samples.
PH Meter:
PH meter
Distilled water
Buffer solutions of known pHs
I can use the pH meter to give me values for the pHs of a number of my solutions. For my distillates, I can then work out the [H+] and then using ethanoic acids Ka Value I can calculate the [ethanoic acid] in my sample. However as the vinegars are a mixture of various acids I cannot obtain a Ka value for it and so will not be able to work out a value for the [Acid] from the vinegars from my pH readings. However I will still measure the pH of the vinegars, as it could prove useful to draw comparisons between the vinegars and the distillates.
To take pH readings I will use a pH meter. To make sure it is accurate I will have to calibrate it beforehand. I will do this by placing it in buffer solutions of known pHs e.g. 4 and 7 and adjusting the readings to give me the right readout on the screens. I will then place the pH meter in the original distillate (before its been watered down in the volumetric flask) and record the result from the screen. I will perform 2 repeats for each distillate and average these results. To maintain the accuracy of the pH meter I will wash it with distilled water after every test and this will ensure that my results are not affected by my previous readings.
Results:
Titration:
Malt Vinegar: Titration no. 1 2 3
Start position 23.4 31.5 40.3
End position 31.5 40.3 48.7
NaOH used 8.1 8.8 8.4
Average NaOH used = ( 8.1+8.8+8.4)/3 = 8.43 (3sf)
I can now convert use this to calculate the number of moles of NaOH used, using this formula:
Number of moles = (Concentration x Volume)/1000 = (1 x 8.433)/1000 = 0.00843
Since one mole of acid reacts with 1 mole of NaOH the number of moles of acid is also 0.00843 moles per 50cm3 sample. Since 5 samples of acid were made from my original 50cm3 sample of vinegar. The number of moles of acid in 50cm3 of vinegar is:
5 x 0.00843 = 0.0422 (3 sf) and from this I can calculate the [acid] in vinegar using the equation:
Acid Concentration = (1000 x number of moles)/volume = (1000 x 0.0422)/50 =
0.843 mol/dm3 (3sf)
I will now repeat this for each of my other vinegars following the same steps.
Cider Vinegar: Titration no. 1 2 3
Start position 18.9 28.2 37.5
Final position 28.2 37.5 46.8
NaOH used 9.3 9.3 9.3
Average NaOH used = (9.3 + 9.3 + 9.3)/3 = 9.3
Acid Concentration = 0.93 mol/dm3
White wine vinegar: Titration No. 1 2 3
Start position 15.8 26.6 37.5
Final position 26.6 37.5 48.2
NaOH used 10.8 10.9 10.7
Average NaOH used = (10.8 + 10.9 + 10.7)/3 = 10.8cm3
Acid Concentration = 1.08 mol/dm3 (3 sf)
Now I will titrate the distillates with the NaOH:
Malt Vinegar: Titration No. 1 2 3
Start position 6.4 12.8 19.4
Final position 12.8 19.4 26.4
NaOH used 6.4 6.6 7.0
Average NaOH used = (6.4 + 6.6 + 7.0)/3 = 6.63cm3 (3 sf)
No. Of moles in sample = (1 x 6.67)/1000 = 0.00663 (3 sf)
Moles in distillate = 5 x 0.00663 = 0.0332 (3 sf)
Ethanoic acid concentration in vinegar sample = ( 0.0332 x 1000)/50 . = 0.663mol/dm3 (3 sf)
I will now repeat this for the other distillates:
White Wine Vinegar: Titration No. 1 2 3
Start position 16.4 24.8 32.8
Final position 24.8 32.8 41.2
NaOH used 8.4 8.0 8.4
Average NaOH used = (8.4 + 8.0 + 8.4)/3 = 8.27cm3 (3 sf)
[Ethanoic acid] in vinegar sample = 0.827 mol/dm3 (3 sf)
Cider Vinegar: Titration No. 1 2 3
Start position 28.8 35.6 42.6
Final position 35.6 42.6 49.8
NaOH used 6.8 7.0 7.2
Average NaOH used = (6.8 + 7.0 + 7.2) = 7.0cm3
[Ethanoic acid] in vinegar sample = 0.7 mol/dm3
pH metering
I took 3 readings for the pHs of the distillates and took an average. From them I obtained these results:
Malt Vinegar = pH 2.5
Cider Vinegar = pH2.5
White Wine Vinegar = pH2.5
The equation of ethanoic acid disassociation is:
CH3CO2H = CH3CO2- + H+
Therefore the concentrations before disassociation are:
X = 0 + 0
After disassociation (at equilibrium) the concentration are:
X – Y = Y + Y
The [H+] is therefore equal to X and the [CH3CO2H] at equilibrium is equal to X - Y. Therefore to work out the value of Y I will need to add the concentrations of ethanoic acid and hydrogen ions at equilibrium together.
From my pH values I can now obtain values for Y following these steps:
PH = -log[H+] 2.5 = -log[H+] [H+] = 3.162 x 10-3
Using my value for [H+] I can now set up a Ka equation.
Ka of ethanoic acid = [CH3CO2-]eqm x [H+]eqm
[CH3CO2H]eqm
From my equation I know that [CH3CO2-]eqm = [H+]eqm, and also from my research I know the Ka value for ethanoic acid = 1.7 x 10-5.
Hence I can now write my equation in this form.
1.7 x 10-5 = [H+]eqm2
[CH3CO2H]eqm
1.7 x 10-5 = (3.162 x 10-3)2
[CH3CO2H]eqm
[CH3CO2H]eqm = (3.162 x 10-3)2
1.7 x 10-5
[CH3CO2H]eqm = 0.588 (3 sf)
Now I have obtained my values for X and X-Y however to get the concentration of ethanoic acid my distillate (Y) I will need to add the two values together.
[CH3CO2H] in vinegar (Y) = 0.588 + 3.162 x 10-3
= 0.591 (3sf)
Therefore the values obtained from my pH meter for all my distillates is that the concentration of ethanoic acid = 0.591 mol/dm3.
Now using the measurements I took of the volume produced by the distillation of the vinegar I can calculate the concentration of ethanoic acid in Vinegar.
Vol. Of Malt Vinegar distillate = 47.5 cm3
Vol. Of Cider vinegar distillate = 47.9cm3
Vol. Of White wine vinegar distillate = 48.1cm3
Concentration of ethanoic acid in vinegar =
Volume of distillate/ volume of vinegar x concentration of ethanoic acid in distillate.
Malt vinegar = 47.5/50 x 0.591 = 0.561 mol/dm3
Cider vinegar = 47.9/50 x 0.591 = 0.566 mol/dm3
White wine vinegar = 48.1/50 x 0.591 = 0.569 mol/dm3
Also So I will be able to compare the pHs of my distillates with that of my original acid I took 3 pH readings for all of my original acids and averaged them.
Malt vinegar = pH 3.0
Cider vinegar = pH 3.0
White Wine Vinegar = pH 2.8
During my experiment I used a number of techniques to try and ensure the results I obtained were as reliable and accurate as possible. Despite this it is very likely that qualitative and quantitative errors have had an effect on the accuracy of my results.
Quantitative Errors are due to the limitations of the equipment that I have used, and therefore if I know the accuracy of my equipment I can work out the maximum percentage effect quantitative error could have had on my results. To do this I will measure the maximum % error of each piece of equipment and add them together to form a total maximum % error. Since I performed two different techniques I will need a maximum % error for each one.
Maximum % error = Maximum possible error/ volume used x 100
Titration
Scales: Accuracy of equipment = 0.001g, Weight measured = 10g
Max possible error = 0.0015 Max % error = 0.0015/10 x 100 = 0.015
Volumetric flask: Since I used this twice for each trial I will multiply the result I obtain by two. Accuracy = 0.10cm3 Max possible error = 0.105 vol. measured = 250cm3
Maximum % error = 0.105/250 x 100 = 0.042%
Pipette: I used the pipette twice in my method so I will multiply the maximum % error by 2. accuracy = 0.04% Volume measured = 25cm3
Maximum possible error = 0.045
Maximum % error = 0.045/25 x 100 = 0.18% x
Measuring cylinder: accuracy = 1.0 Volume measured = 50cm3
Maximum possible error = 1.5
Maximum % error = 1.5/50 x 100 = 3%
Burette: accuracy = 0.1cm3, as the volume of the titres varied for each titration I will need to calculate a value for each titre.
MV = 0.15/8.43 x 100 = 1.78%
CV = 0.15/9.3 x 100 = 1.61%
WWV = 0.15/10.8 x 100 = 1.39%
MV distillate = 0.15/6.63 x 100 = 2.26%
CV distillate= 0.15/7 x 100 = 2.14%
WWV distillate= 0.15/8.27 x 100 = 1.81%
Total Maximum % error
MV = 1.78 + 0.015 + 0.084 +0.27 +3 = 5.15%
CV = 1.61 + 3.369 = 4.98%
WWV = 1.39 + 3.369 = 4.76%
MV distillate = 2.26 + 3.369 = 5.63%
CV distillate = 2.14 + 3.369 = 5.51%
WWV distillate = 1.81 + 3.369 = 5.18%
PH meter:
Measuring cylinder: Since I used this twice for each sample I will need to multiply the value I obtain for the maximum % error by two. Maximum % error = 3%
PH meter: As the pH readings were different for each vinegar and distillate I will need to perform separate calculations for each vinegar/ distillate.
Accuracy = 0.1 max possible error = 0.15
MV = 0.15 / 3.0 x 100 = 5%
CV = 0.15 / 3.0 x 100 = 5%
WWV = 0.15 / 2.8 x 100 = 5.36%
MV distillate = 0.15 / 2.5 x100 = 6%
CV distillate = 0.15 / 2.5 X100 = 6%
WWV distillate = 0.15 / 2.5 x 100 = 6%
Total maximum % error =
MV = 5 + 6 = 11%
CV = 5 + 6 = 11%
WWV =5.36 + 6 = 11.36%
MV distillate = 6 + 6 = 12%
CV distillate= 6 + 6 = 12%
WWV distillate=6 + 6 = 12%
Qualitative errors also may well of had an effect on the accuracy of my results, however they cannot be calculated so I cannot be sure how much of an effect they would of had. There are many possible qualitative errors that may have occurred. Firstly during my distillation of Ethanoic acid it is very likely that, as it was dangerous to evaporate to dryness some of the ethanoic acid was left in the round-bottomed flask. Also some may well have been left behind in the condenser or evaporated for example when a conical flask was left uncovered. All these would of lead to lower values for the concentrations of ethanoic acid being obtained.
During other procedures qualitative errors could have affected my results. For example whilst performing my titration it was difficult to get the end point of my titration exactly right especially as the vinegars were coloured, and so occasionally more NaOH may have been added then was necessary. Also as I read the levels of the NaOH from the meniscus lines on the burette with my eyes these readings could be affected by human error and therefore may well be inaccurate. As well as this in experiments acid is generally added from the burette, so there could have been acidic impurities left in it that could of neutralised some of the NaOH and decreased its concentration. This also could of occurred for any other glassware used in my experiment especially my round bottomed flask which I didn’t change between experiments, and so ethanoic acid could of remained in it from previous distillations.
When making my standard solution of NaOH as firstly the scales are very sensitive so pressure exerted on the surrounding surface could of affected my readings for the mass of NaOH. Also some of the NaOH may not of been transferred to the volumetric flask and so the concentration of NaOH could have been lower then we expected.
Whilst using my pH meter error could of occurred as firstly when I calibrated it the buffer solutions could have been of a slightly different pH then I expected which would of affected the results given by the pH meter. Also if the electrode of the pH meter was not properly rinsed between readings some of the H+ ions may have been transferred between solutions, which would of affected the pH value obtained for each solution.
To minimise qualitative and quantitative error I introduced a number of techniques to my method. For quantitative error I made sure I used equipment with the greatest possible accuracy. I also made sure that I tried to maximise the amounts I measured therefore although my equipment had limited accuracy the maximum percentage error was small.
To try and control qualitative error I used many procedures. During titration I diluted the vinegar to try and decolourise so to make spotting the end point easier as well as towards the end of the titration adding the NaOH drop by drop to minimise excess alkali being added. I also read the meniscus levels at eye level to try and ensure accuracy in my results. Finally I repeated the titration three times and took an average (remembering to ignore anomalous results) to try and increase the accuracy of my results.
Before I used any glassware I rinsed it with distilled water and before each experiment I washed my round bottomed flask and then rinsed it with distilled water. Thus I minimised the effect that acidic and alkaline impurities had on my results.
Other ways in which I tried to improve accuracy were to not lean on the bench whilst weighing out my NaOH and trying to evaporate the vinegar as close as possible to dryness in order to get as great a yield of ethanoic acid as possible. Also with my pH meter I made sure I rinsed the electrodes with distilled water well before taking any readings with it.
Results of my titration:
Vinegar [Acids] in vinegar [Acids] acid in distillate Difference
MV 0.843 0.663 0.18
CV 0.93 0.7 0.23
WWV 1.08 0.827 0.253
As you can see from my results the overall acid concentration and the concentration of ethanoic acid ([acids] in distillate) follow the same trend. This is that there is a higher concentration in White wine vinegar compared to Cider vinegar and the concentration in cider is higher than that in the malt. Also malt has the lowest difference between the values (amount of other carboxylic acids) and this is probably why it was colourless and why I observed little decomposition of impurities during its distillation. As you can see from my results, ethanoic acid is by far the dominant acid in vinegar but it also contains other acids as flavourings and colourings that lead to a difference between the titration results for the vinegars and the distillates.
Another trend show by all my acids is that the concentration of acid shown in my vinegars is higher then that shown in my distillates. This is because the vinegar contains the ethanoic acid plus all the other organic acids whereas the vinegar’s distillate contains only the ethanoic acid. Therefore the vinegar contains more acid then the distillate and so it requires more NaOH to neutralise the acid in the solution. The magnitude of the difference between the two values depends upon the number of other organic acids present in the vinegar and from my results it appears that the White Wine vinegar has the largest concentration of other acids out of all the vinegars.
pH Meter Results:
Vinegar pH of vinegar pH of distillate Difference
MV 3.0 2.5 0.5
CV 3.0 2.5 0.5
WWV 2.8 2.5 0.3
From My pH results you can see that the White wine vinegar has a lower pH and so a higher concentration of H+ ions then the other two vinegars. This supports my conclusion from my titration of the vinegars that the WWV has contains a higher concentration of both ethanoic and other carboxylic acids.
Another trend shown by my results is that the distillates have a lower pH then that of the original vinegars. I did not expect this to be the case as it is clear from my titration results that there is a higher concentration of acids in the vinegars compared to my distillates. So I presumed that this would lead to a decrease in the [H+] and therefore an increase in pH. However this is clearly not the case and therefore another explanation needs to be given. Whilst distilling the vinegar it lost roughly 4% of its volume, however it retained all its ethanoic acid. Therefore there is a larger concentration of ethanoic acid, which leads to an increase in the concentration of H+ ions in the solution, and therefore a decrease in the pH. This explanation is only possible however if the other acids left behind in the round-bottomed flask are very weak and so don’t disassociate readily. Therefore they do not produce many H+ ions and so have little effect on the pH of the vinegar but still effect the titration as it neutralises all the acids by shifting the position of equilibrium. Therefore in the vinegar the other organic acids increase the concentration of acid which is shown by the titration results but do not effect the concentration of H+ ions and so do not vary the pH. Therefore the distillate contains the vast majority of the H+ ions as in vinegar but not as much of the volume and so its pH is lower.
The conclusion that the pH is determined by the concentration of ethanoic acid and is not greatly effected by the concentration of the other organic acids is backed up by the values I obtained for the pHs of my vinegars. Two of my vinegars have identical pHs and these vinegars have similar concentrations of ethanoic acid (both 0.7 to 1 dp) whereas the values for the concentrations of other organic acids vary relitivly significantly.
The pHs of all my distillates were the same. This is surprising as my titration results showed that the concentrations of ethanoic acid in the distillates varied significantly. This could be due to the limited accuracy of the equipment as the pH meter can only measure to 0.1 and so may not be able to pick up subtle differences in the pHs of the distillates. Or it could be due to the fact that this method contains a lot of quantitative error that could of affected my results.
As you can see from the analysis I have been able to make trends in my results and most of them fit these trends. However as you can see by the differences between the two values for the concentration of ethanoic acid I obtained, there is clearly error in my results. However the results from both the methods seem to fit in with my scientific knowledge and as I haven’t obtained any anomalous results I believe that my results are fairly accurate. However there are improvements I could make to my method to improve accuracy. Instead of measuring out the vinegar in a measuring cylinder next time I would measure it out using a pipette as it is far more accurate. Also during titrations I would invert the burette to make sure that the NaOH does not collect towards the bottom. To ensure the accuracy of my titrations I would test the concentrations of the standard solution I have made by either titrating it with an acid of known concentration or using short-range indicator paper. This would ensure my titration results were accurate. This should improve the accuracy of my results if I were to perform the experiment again.
Josh Rowlands
March 2002
