Vitamin C Essay, Research Paper
Introduction
In this investigation, I will analyse by titration, the vitamin C (ascorbic acid) content of 7 different types of fruit juices purchased from a supermarket. I will test for the presence of vitamin C using an organic indicator solution called DCPIP (dichlorophenol indophenol), which will be reduced to a colourless form from dark blue, at the presence of vitamin C. Reduction is where a molecule losses an oxygen molecule, gains a hydrogen atom or gains electrons. Oxidation is the opposite, i.e. gains an oxygen molecule, loses a hydrogen atom or loses electrons. The amount of juices used to turn the DCPIP colourless will be recorded and the vitamin C content will be determined by measuring the amount of drops of a known vitamin C concentration solution, used to turn the DCPIP colourless.
Vitamin C
Vitamin C plays a vital part in the correct functioning of the living body (most animals synthesise their own vitamin C. we don t). There are 4 ways in which vitamin C considerably benefits the body:
1. Antioxidant. Oxidation of haemoglobin in the blood is necessary to survive, but too much oxidation can be damaging to tissues and organs. Vitamin C prevents too much oxidation from occurring.
2. Homeostasis. When the body is exposed to stress, the adrenal grand uses vitamin C to produce adrenaline in response to the stress by which blood will be moved to the muscles from the digestive system, etc. This is for flight, fright or fight.
3. Collagen. This is the principal component of tendons, ligaments, skin, bone, teeth, and cartilage, in addition to being the ground substance between cells. Inadequate vitamin C levels will lead to weakening of these areas leading to skin lesions, blood vessel fragility, loosening of gums and teeth, characteristics of scurvy.
4. Immune system. Vitamin C is involved in the production of white blood cells, which are the main defence against disease. Without vitamin C, the body will be very susceptible to disease.
Vitamin C has also been linked in the prevention of heart disease and cancer.
List of equipment
Volumetric flask to measure out the correct volume of DCPIP.
Beaker, 1 to containing the juice you wish to test and 1 for the DCPIP solution.
2ml syringe, to measure out the DCPIP.
1ml syringe, to measure out the quantity of juice added to the DCPIP.
10 test tubes, into which the DCPIP and the juice is added.
Test tube rack, to hold the test tubes.
Electronic balance to weight the correct amount of DCPIP.
Method
The experiment was performed at room temperature.
The juices used were:
Orange juice
Grapefruit juice
Lemon juice
Pineapple juice
Mixed fruit juice
Cranberry and raspberry juice
Apple juice
The first task was to measure out 1% DCPIP solution in the volumetric flask and to make sure it is completely mixed (distilled water is used so not to effect the accuracy of the experiment and a squeeze bottle can be used to top up the measurement for further accuracy). The level was read from the bottom of the meniscus. The DCPIP can then be put in to a beaker to ease the filling of the syringes (DCPIP must not be left overnight to avoid oxidation which may affect the results).
2ml of DCPIP are then put into each test tube using the 2ml syringe.
The 1ml syringe is then filled with the 0.05% vitamin C solution and is added drop by drop to each tube individually until the DCPIP has been reduced and the blue colour has disappeared. The amount of 0.05% vitamin C solution needed to reduce the DCPIP is recorded. This procedure was repeated 10 times to provide an adequate level of accuracy to the experiment.
The test was then performed on the fruit juices of unknown vitamin C content.
The relevance of testing the 0.05% vitamin C solution is to determine the vitamin C concentration in the fruit juices by dividing the amount of 0.05% vitamin C solution needed to reduce the DCPIP, by the amount of fruit juice needed to reduce the DCPIP. This is then multiplied by 0.05 (the known concentration of the first solution). By doing this, the vitamin C content of the fruit juices can be determined.
Any difficulties or anomalies were also recorded.
Calculation example:
Volume of 0.05% vitamin C solution needed to reduce DCPIP = 0.39ml
Volume of orange juice needed to reduce DCPIP = 1.55ml
Vitamin C content of orange juice = volume of 0.05% vit. C solution X 0.05%
Volume of orange juice
= 0.39 X 0.05
1.55
= 0.013%
Questionnaire
Put a tick in the box beside the juice that you think has the highest vitamin C content, and a cross in the box beside the juice with the lowest vitamin C content.
Orange juice Apple juice
Lemon juice Pineapple juice
Grapefruit juice mixed fruit juice
Cranberry and raspberry juice
Questionnaire
The questionnaire was given to a sample of 50 people to get an idea on general thoughts on fruit juices vitamin C content.
Juice with the most vitamin C:
52% orange juice
34% lemon juice
10% mixed fruit juice
2% cranberry and raspberry juice
2% grapefruit juice
(0% for the rest)
Juices with the least vitamin C:
32% apple juice
28% cranberry and raspberry
24% pineapple juice
10% mixed fruit juice
6% lemon juice
(0% for the rest)
Generally, people believe the citrus fruits contain the most vitamin C.
Graph showing questionnaire results
Juices with the most vitamin C content
Juice a = orange juice
Juice b = lemon juice
Juice c = mixed fruit juice
Juice d = cranberry and raspberry juice
Juice e = grapefruit juice
Juices with the least vitamin C content
Juice a = apple juice
Juice b = cranberry and raspberry juice
Juice c = pineapple juice
Juice d = mixed fruit juice
Juice e = lemon juice
Students T-test 1
Calculating the significant difference between the means of orange juice and pineapple juice. To see if there is more or less than 95% confidence that the difference between the means has not occurred by chance alone.
Orange juice mean = 1.55
Pineapple juice mean = 6.43
Difference between them = 4.88
Standard deviation of orange juice = Sx =0.0750
Sx2 = 0.005625
Sx2 = 0.0005625
N
Standard deviation of pineapple juice = Sy = 0.2744
Sy2 = 0.075295
Sy2 = 0.00753
N
Sx2 + Sy2 = 0.0080925
N N
Standard error = 0.08996
T = 54.247
There is 95% confidence that the difference has not occurred by chance alone.
In fact, there is 99.9% confidence that the difference has not occurred by chance.
Students T-test 2
Calculating the significant difference between the means of orange juice and grapefruit juice. To see if there is more or less than 95% confidence that the difference between the means has not occurred by chance alone.
Orange juice mean = 1.55
Grapefruit juice mean = 0.93
Difference between them = 0.62
Standard deviation of orange juice = Sx =0.0750
Sx2 = 0.005625
Sx2 = 0.0005625
N
Standard deviation of grapefruit juice = Sy = 0.0432
Sy2 = 0.0018662
Sy2 = 0.0001866
N
Sx2 + Sy2 = 0.00749
N N
Standard error = 0.2737
T = 22.653
There is 95% confidence that the difference has not occurred by chance alone.
In fact, there is 99.9% confidence that the difference has not occurred by chance.
Analysis
During the experiment, there was a problem because one of the volumetric flasks had been filled with an incorrect amount of DCPIP. This greatly affected the results. Fortunately, the problem was swiftly rectified and the experiments were tested again with the correct DCPIP volume. This problem could have been avoided with more care and attention put into the measuring of the DCPIP.
When the apple juice was tested, I found that after adding 15ml of juice to 2ml of 1% DCPIP, there was no change. 1ml was tested, but still there was no change. Eventually, 0.125ml of DCPIP was used in order for the complete reduction of the DCPIP. This alteration decreased the accuracy of the results because very small volumes of DCPIP had to be measured out.
The standard deviation was calculated for the results obtained. This measures the spread of the individual measurements about the mean. The larger the spread, the larger the deviation. The results found had relatively small standard deviations except for the apple juice. This had a significantly higher standard deviation because the apparatus was not accurate enough to measure such low levels of vitamin C. More repetitions could have been made with less amounts of DCPIP to reduce the spread and increase the accuracy. The levels of confidence found were 99.9% that difference between the two means were not by chance alone.
The mixed fruit juice and cranberry and raspberry juice (the two highest results), had added vitamin C. The carton did not state how much vitamin C was added so it was not possible to allow for the addition. This means that the results did not show how much vitamin C is in those two juices, but how much is in the cartons. Vitamin C is added to some juice as an antioxidant to prevent the deterioration of the substance. This means that the cranberry and raspberry juice and the mixed fruit juice did not have sufficient vitamin C in it originally, so more had to be added.
The lemon juice found to have low vitamin C content. However, it contained preservatives (e.g. E220 V E227) to prevent the growth of any bacteria and to be able to last a long time without decay. This preservation may have affected the results but damaging the vitamin C in the juice. Also, adding preservatives may prevent the decay of the juice, but it may not prevent the breakdown of the vitamin C. The modern diet may lack vitamin C because most produce is preserved.
Ways in which to improve:
Do more repetitions on different makes of juice in order to get a broader set of results.
Use freshly squeezed juices so not to contain preservatives.
Conclusion
This experiment has shown the vitamin C content of a small variety of juices, but different brands contain different amounts of vitamin C so it has not proved a great deal. A more extensive investigation has to be carried out in order to get a better idea of the general trends in vitamin C content in fruit juices.
I found from the data that I collected that my vitamin C reading were a great deal higher than the data shown in the table (Food Science, Nutrition and Health). This is probably from variations in procedure and the juices used. Their method in measuring the vitamin C content of juices could be very much different and so differences between the two experiments can occur.
Blackcurrant juice was going to be tested because it is a well-known, non-citrus fruit drink, with a high vitamin C content. The reason why this test was not performed on this juice was because of its colour. It would have been extremely difficult to accurately measure when all the DCPIP had been reduced to its colourless form because of blackcurrants deep colour.
The amount of vitamin C in different diets can vary tremendously. In MEDCs (more economically developed countries), the vitamin C content may slightly under the correct amount because of preservation techniques (explained earlier), but the ability to import fruits with high vitamin C content, and food supplements (vitamin tablets) make up for that lack. But, in LEDCs (less economically developed countries), where importation of fruit and food supplements are scarce, there is a high deficiency of vitamin C.
Generally, people do not understand the necessity of vitamin C (4 factors explained earlier) and the consequences of its deficiency (scurvy). The average person with a well balanced diet will be in the necessary requirement of vitamin C V between 75-100mg per day. However, a person with an unbalanced diet, may be under this level, e.g. 30mg of vitamin C per day. This will reduce the levels of vitamin C in the body eventually resulting in scurvy. Luckily, for a great deal of people, a lot of food contains vitamin C, but its deficiency could be reduced if people learned about its importance in the correct functioning of the human body.
Bibliography
Practical Biology for Advanced Level (Michael Roberts, Tim King and Michael Reiss) Nelson
cforyourself/overview/primer
Microsoft Word 97
Food Science, Nutrition and Health (Brian A Fox and Allan G Cameron)