Concentration of Vinegar
Introduction
In this investigation I am going to find out the concentration of ethanoic acid in vinegar. The concentration of acid is different in the product made by different manufactures. I will react the acid in the vinegar solution of sodium hydroxide of known concentration in a titration experiment.
Equipment
* Plastic beaker
* Glass beaker
* Pipette
* Funnel
* Phenolphthalein
* Vinegar
Method:
* Rinse the pipette with the diluted solution of vinegar.
* Rinse the conical flask with distilled water Pipette 25cm³ of diluted vinegar solution to the conical flask and add 4-5 drops of the indicator.
* Rinse the burette with the solution of Sodium Hydroxide and be sure to get rid of any air bubbles in the tip of the burette.
* Using the filter funnel, carefully fill the burette with the solution of Sodium Hydroxide.
* Record the initial burette reading in a results table.
* Set up the burette and the conical flask in the corresponding places for the titration.
* Make sure there are no bubbles in the burette before using it.
* Slowly and carefully run Sodium Hydroxide solution from the burette into the conical flask with swirling.
* Do this until the solution changes from colourless to light pink. This is the end-point.
* Record your final burette reading also to the nearest 0.05 cm³.
* Repeat this procedure until 2 concordant results are obtained making sure you rinses the conical flask between every titration.
Accuracy
All burette readings must be recorded to the nearest 0.05 cm³ (1 drop). Consecutive titrations should be concordant. This means that they are either identical or within 0.10 cm³ of each other. The titration must be repeated until this is achieved. Burette readings must be recorded to 2 decimal places. Results table for a titration: Burette Rough 1st Accurate 2nd Accurate Final (cm³) Initial (cm³) Titre (cm³) Titre: final burette reading - initial burette reading
We know that NaOH is in a one to one reaction with the Ethanoic acid:
NaOH + CH3COOH › CH3COONa + H2O
Safety
For safety, these rules should be followed: Wear safety goggles at all times. Make sure there is good ventilation as some solutions have strong smells. Sodium Hydroxide is corrosive and Hydrochloric Acid is irritants so wash your hands thoroughly if there is any contact with the chemicals.
* Thermometers contain mercury which is poisonous so be careful not to break it.
* Handle all glassware with care as they may be hot after a reaction takes place.
Risk Assessment
Sodium Hydroxide in its pure form is very corrosive and can cause severe burns. In its diluted form it is less harmful, but still an irritant. It can also cause severe permanent eye damage.
Pipette - always hold the pipette vertically close to the open end and not the tip.
If broken, do not touch, use parker and brush to pack into the broken glass container. If stabbed, seek medical advice.
Glass ware
If broken, do not touch, use parker and brush to pack into the broken glass container. If stabbed, rinse under running water and plaster. Seek medical advice.
Fair Test
To make sure that the test is fair I will measure the amounts of each liquid very carefully. Each time ...
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Pipette - always hold the pipette vertically close to the open end and not the tip.
If broken, do not touch, use parker and brush to pack into the broken glass container. If stabbed, seek medical advice.
Glass ware
If broken, do not touch, use parker and brush to pack into the broken glass container. If stabbed, rinse under running water and plaster. Seek medical advice.
Fair Test
To make sure that the test is fair I will measure the amounts of each liquid very carefully. Each time I start a new test, I will have washed and dried the beaker out as not to leave any of the liquids from the last experiment in. Instead of measuring the amount from a certain temperature, I will take the start temperature and then one at the end. The difference will then be taken. I will use a different measuring cylinder for each liquid, and the water will be taken from the tap. The Hydrochloric acid and the Sodium Hydroxide will be taken from a bottle. This should give time for the different liquids to mix together and hopefully neutralise. There are many things in the experiment that could be considered not fair but that is why we are doing the experiment, to see how the different amounts of each liquid react. In each of the experiments the total amount of liquid in the mixtures will be 100ml. One of the experiments does not contain water, but otherwise, the lowest amount in any one is 24.1ml and the highest 24.1ml. Having 50ml of liquid in each mixture will keep the experiment fair.
The test would be preformed again until it is accurate and so making this experiment a very fair test. I think that these are important factors because it can be detrimental to how accurate and reliable the results were on the whole and it could leave me with anomalous results that have no reason behind their appearance.
The amount of alkali should be used to neutralise vinegar that has not been diluted. However, we can see that for some of the other vinegar, the amount of alkali is less, which shows that it has been diluted. Having obtained the rough titre from this experiment we can be more accurate.
I will keep the alkali I use the same by always using Sodium Hydroxide, NaOH. I will keep the concentration of NaOH the same by ensuring I use a 1M solution each time. I will always keep the acid the same by always using Ethanoic acid, CH3COOH. Although the concentration will vary, the amount will always stay the same.
Results
Burette readings for the Vinegar solution
Rough titration
Run 2
Run 3
Run 4
Run 5
Run 6
Run 7
Burette readings at the end/cm³
24.00
24.20
24.00
24.10
24.00
25.00
25.10
Burette readings at the start/cm³
0.00 cm³
0.00cm³
0.00 cm³
0.00cm³
0.00cm³
.00cm³
.00cm³
Volume/cm³ of diluted vinegar added to 25.00cm³ of sodium hydroxide solution
24.00
24.20
24.00
24.10
24.00
25.00
25.10
Calculations
From this we can work out the percentage of acid in the vinegar. We have to first work out the relative formula mass of Ethanoic acid. To do this we just add up all the atomic weights in the formula CH3COOH.
Here are my calculations to find out the concentration and 1/concentration of each of the samples of vinegar:
Prediction
I predict that if any of the solutions have been diluted, the amount of alkali needed to neutralise them will be less. If they have not been diluted the amount of alkali used will be the same as for the first solution.
Conclusion / Evaluation
To ensure my experiment was a fair I had to cover a few things. I made sure the factors which needed to be constant were constant such as the volume of vinegar was measured accurately, the amount of indicator was added drop by drop, the concentration of sodium hydroxide was all ready supplied, I made sure any human errors such as not switching the tap off in time was switched of to the best of my ability and if the tap was not switched off and it overdosed the vinegar.
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 percentage error of each piece of equipment and add them together to form a total maximum percentage error. Since I performed two different techniques I will need a maximum percentage error for each one.
Qualitative errors also may well have 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 have 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 have neutralised some of the NaOH and decreased its concentration. This also could have 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 have remained in it from previous distillations.
This means the amount of NaOH is the same as the amount of acid.
MaVa / MbVb = 1
This means we can write this as MaVa =MbVb
Ma = 0.4
Va = 2.0
Mb = x
Vb = 10
0.4 X 2.0 = 10x
0.8 = 10x
0.08 = x
We then multiply this answer by 10 as the vinegars were diluted 10 times. This means that the amount of Ethanoic acid in the Dilute malt, according to our first test.
0(0.4 X 2.1) = 10x
0(0.84) = 10x
0.84 = x
Type of vinegar
Dilute malt vinegar
Dilute distilled malt
Ethanoic acid content (1)
0.8 mol/dm3
0.84 mol/dm3
Ethanoic acid content (2)
0.84 mol/dm3
0.88 mol/dm3
Practical
We decided to use the same set-up for the practical as we did for the preliminary. The only thing we changed was the strength of the NaOH. We changed this to 0.1M as this would give us a more accurate reading.
Results
Vinegar
st result
2nd result
Average Ethanoic acid content
Distilled vinegar
0.3
0.2
.025 mol/dm3
Cider vinegar
8.5
8.5
0.85 mol/dm3
White wine vinegar
8.6
8.6
0.86 mol/dm3
Red wine vinegar
0.3
0.3
.03 mol/dm3
Malt vinegar
9.1
9.0
0.95 mol/dm3
I took a rangefinder so that it was easier to get the exact volume when I did it for the three real attempts. This way it allowed me to drip the acid drip by drip until it was perfect. I took three reading rather than one so I would get sufficient evidence to support my conclusion. If I had only taken one reading then it may not have been quite right. I think that by presenting my results in a table it is visually easy to see what results I got for different things and the different attempts. Analysis
I then proceeded to find the molarity of each of the vinegars using the average values I had obtained from the experiment. To make these calculations I used the following equation
b x Ma x Va = a x Mb x Vb or bMaVa = aMbVb
NaOH +CH3COOH à CH3COONa + H2O
olution of ethanoic acid of unknown concentration. CH3COOH + 1 NaOH › 1 CH3COONa + 1 H2O 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 morality 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: CH3COOH + 1 NaOH › 1 CH3COONa + 1 H2O 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:
Molarities
Naz Mariwan
Science Coursework - Unit 1