No. mole of acid required: M . V / 1000 = 0.1 × 200 / 1000
= 2 × 10-2 mol
Therefore the volume 1.0 M acid required:
V = No. mole × 1000 / Molarity (M)
= (2 × 10-2) × 1000 / 1.0
= 20 cm3
To this 20 cm3 of acid 180 cm3 of de – ionised water must be added. In the real titration this will be done accurately with a volumetric flask, rinsing the flask out before hand with the water and using a pipette to add the vinegar. The mixture must then be mixed very thoroughly.
For the actual titration a burette will be set up (on the floor – so to avoid the risk of splashing sodium hydroxide in the eyes but safety specs will be worn as well) and sodium hydroxide poured into it after being rinsed by out by sodium hydroxide as well. Beneath it a conical flask will be set up with 20 cm3 of diluted vinegar measured accurately by the use of the volumetric flask (having been rinsed out with the diluted vinegar so to increase the accuracy of the experiment). There will be 20 cm3 of the diluted vinegar as this will be an easy value to deal with as it is a tenth of the amount of the original amount of diluted vinegar and it is also a convenient amount in relation to how much of sodium hydroxide will be added to it. The flask will be put on a white tile so the colour change of the indicator is more easily perceived. A few drops of indicator (phenolphthalein) must then be added to the vinegar.
Firstly a rough reading must be taken do to make sure that the accurate readings are not overshot and a value can be noted down for this.
The titration can now be started; the sodium hydroxide can be added quickly at first until it approaches an amount similar to that of what has been discovered from the rough reading and from then on it must be added drop by drop until the indicator changes colour so to obtain an accurate result, the amount of sodium hydroxide added will be noted. This will be repeated three times (or however many times required to get results which are close enough to each other) and an average (mean) of the results will be taken. The concentration of the ethanoic acid in vinegar can then be found out from this average this result by carrying out various calculations which will be demonstrated later.
Results:
The procedure as described in my plan was carried out and was done so for both vinegars. The values collected will be shown in a table and an average is shown in this table for the three values representing the number of centimetres cubed (NOT the rough) of sodium hydroxide added to neutralise the diluted vinegar for both vinegars to two decimal places.
The results were taken three times but had the three I had taken seemed completely random then I would have kept taking them until I found values which match up as being close to each other. The average values have been taken and used to find the concentration of both vinegars:
Analysis of Results:
Transparent Vinegar:
In this titration 18.77 cm3 of 0.1 M sodium hydroxide (NaOH) solution reacted with 20 cm3 of a solution of ethanoic acid of unknown concentration.
1 CH3COOH + 1 NaOH → 1 CH3COONa + 1 H2O
20 cm3 18.77 cm3 (sodium
? M 0.1 M ethanoate)
The first stage is to find the No. mole of the sodium hydroxide:
No. mole NaOH in 18.77 cm3 of 0.1 M solution =
Molarity × Volume / 1000
= 0.1 × 18.77 / 1000 = 1.877 × 10-3 mole
This now makes it possible to find out the number of moles of ethanoic acid because we know that one mole of the acid (CH3COOH) reacts with one mole of the alkali (NaOH). Therefore from this we now know that there are 1.877 × 10-3 moles in 20 cm3 of acid.
Knowing this, the concentration can now be found out by putting the information into this formula (the formula was derived from the “M V over a thousand” formula):
Molarity = No. mole × 1000 / Volume
= 1.877 × 10-3 × 1000 / 20
= 0.09385 mol/dm-3
This is the concentration of the diluted transparent vinegar.
To find out the molarity of the undiluted vinegar this answer has to be multiplied by ten because the vinegar was diluted by a factor of ten so this has to be compensated for:
0.09385 × 10 = 0. 9385mol/dm-3
Therefore giving the actual concentration of the vinegar. From this the percent by mass data can be calculated so it can compare more easily with the predicted value (5%). To do this the process demonstrated in the plan has to be reversed, so:
0.9385 ≡ to the no. mole in 1 dm3 (1000 cm3) and to be able to find out the percent mass data it is more convenient to know the no. mole in 100 cm3. And so to obtain this 0.9385 has to be divided by 10:
0.9385 / 10 = 0.09385
This value must now be put into the formula: Mass = No. mole × Mass of 1 mole to find the mass of ethanoic acid in 100g of solution (100 cm3 ≈ 100g). Therefore:
0.09385 × 60 (the mass of one mole of ethanoic acid) = 5.631g
5.631g is the mass of ethanoic acid in 100g of solution and therefore can be changed to a percentage, this being,
5.631%
Coloured (Brown) Vinegar:
In this titration 14.47 cm3 of 0.1 M sodium hydroxide (NaOH) solution reacted with 20 cm3 of a solution of ethanoic acid of unknown concentration. The same formula is used as shown before:
1 CH3COOH + 1 NaOH → 1 CH3COONa + 1 H2O
20 cm3 14.47 cm3 (sodium
? M 0.1 M ethanoate)
The No. of mole of sodium hydroxide in 14.47 cm3 of 0.1 M solution now has to be found, using the same formula as before:
Molarity × Volume / 1000 =
-
× 14.47 cm3 / 1000 = 1.447 × 10-3 mole
Using this it is now possible to work out the concentration of the ethanoic acid utilising the same method as before because the ratio of the two reactants in the formula is 1:1. So:
Molarity = No. mole × 1000 / Volume
= 1.447 × 10-3 × 1000 / 20
= 0.07235 mol/dm-3
This is the concentration for the diluted coloured (brown) vinegar.
To get the value of the molarity of the undiluted vinegar this has to be multiplied by ten (as before) because the original vinegar was diluted by a factor of this much:
0.07235 × 10 = 0.7235 mol/dm-3
The percent by mass can now be worked out (as shown previously).
0.7235 ≡ the no. mole in 1 dm3 (1000 cm3). It is more convenient to use 100 cm3 so 0.7235 will have to be divided by 10:
0.7235 / 10 = 0.07235
The percent by mass data can now be found out:
Mass = No. mole × Mass of 1 mole
= 0.07235 × 60 (the mass of one mole of ethanoic acid)
= 4.341g
This can therefore be changed to a percentage:
4.341%
Looking at these percentages by mass data for both vinegars it is evident that they reflect my prediction (5%), to help demonstrate this; an average can be taken of the two values:
(5.631 + 4.341) / 2 = 4.986 %
This obviously demonstrates the predicted percentage (taken from the internet) which would have likely to have been an average itself; taken from various commercial vinegars.
Evaluation:
The experiment which was carried was done so with as much care as possible but even so improvements could have been made
Looking at the main method: Only two different vinegars were chosen; if more had been chosen then it would increase the reliability of the results because it would be less likely that the results could be swung by a ‘freak’ vinegar and there are many more types of vinegar which could well have a notable difference.
The amount of sodium hydroxide added to the diluted vinegar was measured to one decimal place (of cm3) and this allows a certain margin for error. This margin can be illustrated thus: if say a reading was taken and it was 14.3 cm3 (to 1 d.p.) then the minimum this could possibly be would be 14.25 cm3 and the maximum would be 14.35 cm3 giving a possible error boundary of 0.1 cm3, which could well have quite an affect on the final results if the readings were up to this much out. This error boundary is likely to be added to by human error when stopping the sodium hydroxide being added once the acid has been neutralised (when the indicator changes colour).
The awareness of the colour change of the indicator is perhaps slightly impaired when the brown vinegar is being neutralised because it is more difficult to see past the brown colour when the actual indicator begins to change colour itself. This problem could have been overcome by filtering the brown vinegar through carbon to remove its brown colour and to make it transparent.
A similar type of error boundary as described previously can be applied to the measuring of the diluted vinegar in the volumetric flask and in the pipettes.
When the various artefacts (burette, conical flask and pipettes) were rinsed out ready for the titration, they had to be rinsed out with whatever they were going to contain before they could be used so to counteract whatever had been in there before hand. Even though they might have been rinsed out some of whatever had been there previously may still remain and adversely affect the results of the titration.
Finally, the results collected clearly reflect the prediction in the plan.
