Vitamin C Experiment, Essay Example

Objective

The aim of the experiment is to ascertain the saturation of Vitamin C in a variety of distinct fruit juices by means of t8itreation and to provide ranking for the vitamin C sources.

Introduction

Vitamins C (ascorbic acid) is a compound that dissolves in water and is one of the essential vitamins for life. Ascorbic acid is incorporated in a number of bodily functions, which are inclusive of the development of collagen in the joints, the production of dopamine, adrenaline and noradrenaline in the central nervous system. Vitamin C (ascorbic acid) is also responsible for the production of carnitine which is an essential component in the energy transference mechanism in the cellular mitochondria. A lack of ascorbic acid is the causal attribute of scurvy. It had been determined that protection from scurvy could be acquired by the consumption of citric fruits and sauerkraut. The research question that will be formulated is does the consumption of one daily fruit juice serving satisfy the recommended daily intake for Vitamin C in many of the industrialized nations including Australia, Canada, New Zealand, the United Kingdom and the United States. The minimum   recommended daily intake of vitamin C is 60 mg/ day. A Nobel prize award winning researcher, Linus Pauling recommended mega dosages of ascorbic acid on a daily basis in order to ward of diseases. Vitamins c is frequently applied as an antioxidant and antimicrobial in a number of food products. Ascorbic acid had been initially discovered in 1928 and in 1932 it had been determined as the biological agent which facilitated the deterrence of the dissemination of scurvy. The chemical structure had been documented in 1933 and had been verified by production.

Isomers which are enantiomers are reflections of one another. The L- enantiomer of Vitamin C is ascorbic acid. Ascorbic acid is a substance that has a stable solid quality which is inert with air and photoluminescence when found in an aqueous solution. The result of the oxidation process derives dehydroascorbic acid. When one of the species experiences oxidation, the other species experiences reduction. These are two processes which are complementary opposites. In the systems which have a biological quality, metabolic processes derive reactive oxidants. These oxidants have the capacity of reacting with molecules that are in their surroundings. consequently, causing damage to the cell. The human anatomy provides protection for its cells by applying a distinct collection of molecules which are designated antioxidants in order to detoxify and diminish the oxidants. The experiment applies this process in a redox/ oxidation titration procedure. In the process, Vitamin C decreases the orange hue of the iodine ion in order to produce the colorless solution. The reaction that is taking place in this experiment is the following:

H₄O₆ + I₂ ? H₂O₆ + 2I + 2H^*

The process of volumetric evaluation is an approach that applies the assessment of volumes in order to ascertain the quantity of a substrate in a solution. In any of the reactions which occur between one or more categories of ions, the reaction equation demonstrates the stoichiometric proportions of the reacting molecules. The chemical reaction that is being researched is the process that occurs between the solutions of iodine and ascorbic acid. The chemical reaction equation details that the proportion of molecules of iodine to the molecules of ascorbic acid is in the ratio of one to one.  This infers that one mole of ascorbic acid reacts with one mole of vitamin C.

Iodine as an Indicator

The presences of an iodine cause the creation of a blue complex when starch is detected. As a result, iodine can respond as its proprietary indicator. As there is an excess of ascorbic acid, the iodine molecule which has an orange hue that is being aggregated from the bn8urrettte is experiencing reduction and loss its orange hue. When oxidation has taken place for all of the ascorbic acid, the iodine molecules which have been aggregated will no longer experience reduction. As a result of the starch solution that had been added, the aqueous solution will acquire a blue hue.

Materials and Experimental Procedure

• Iodine Solution
• Ascorbic acid. Acetic acid (0.2 M).
• Starch solution

Iodine Standardization Process

An iodine solution can be standardized regarding titration in correlation to a determinate saturation of sodium thiosulfate in accordance with the following chemical equation:

I₂ + 2S₂O₃ ?2I + S₄O₆

The sodium thiosulfate solution had been prepared prior to the experiment. The precise concentration is inscribed ion the label. This is the solution that will be applied for ascertaining the precise amount of iodine solution that is to be applied for the approximation of the Vitamin C content in the fruit juice.

Step 1

Two hundred milliliters of iodine had been collected and placed into a flask which had a cap. The burette was prepared for the process of titration by rinsing and washing the burette with water and subsequently with a small amount of iodine solution. The cap was placed on the solution that was maintained in the flask.

Step 2

One hundred and twenty milliliters of the sodium thiosulfate had been collected and placed into another clean and dry, two hundred and fifty milliliter flask. The precise concentration of the sodium thiosulfate solution was recorded.

Step 3

A pipette was applied in order to transfer the sodium thiosulfate into a clan flask. Approximately half of a teaspoon of the starch in solid form had been applied in addition to 10 drops of acetic acid with a concentration of 0.2 M. The iodine solution was titrated to the point of realizing a color change for than interval of thirty seconds. The initial and final volumes were documented.

Step 4

Sodium thiosulfate in the amount of 25 ml was transferred into a flask. The starch solution and the acetic acid had been aggregated. The titration was performed with the iodine solution. This was realized incrementally until there had been a 2 ml amount titrated. The ital. and the final volumes had been documented in the logbook.

Step five

Step four was repeated until there had been three titration values which were similar. The similarities were maintained to a tenth of a milliliter.

Step Six

The calculation of the iodine solution was determined.

Outcome [Thiosulfate ions] = 2.040 x 10^{– 3} M (Derived from label)

Table 1: Titration Outcomes for the Standardization Process of Iodine (Mean Titer = 33.675 ml).

In the trails the clear iodine solution was transformed to a blue hue.

Chemical reaction:

I₂ + 2S₂O₃ ?2I + S₄O₆

Consequently, one mole of iodine responds to one mole of S₂O₃

25.00 ml of 2.040 x 10^{– 3}  M S₂O₃ possesses 2.040 x 10^{– 3}   x 0.025moles of S₂O₃. Consequently, 33.675 ml

of the mystery Iodine solution possesses (2.040 x 10^{– 3}  x 0.025)/ 2 moles of iodine. As a result, 100 ml of the mystery iodine solution possesses (2.040 x 10^{– 3}  x 0.025 x 1000)/ (2 x 33.675) miles of iodine. Consequently, the saturation of the mystery iodine solution is 7.57 x 10^-4 M.

Determination of Vitamin C

Step One

Tem milliliters of fruit juice was weighed on the loading balance.

Step Two

Five milliliters of fruit juice had been placed in the cylinder that had been applied for weighing.

Step Three

The fruit juice had been placed into a 250 ml flask and was diluted with 20 ml of denatured water. A spoonful of starch solution had been aggregated with approximately ten drops of acetic acid which had a concentration of 0.2 M.

Step Four

Steps one to three had been repeated for two different types of fruit juice.

Step Five

The weight of the vitamin C in the fruit juice sample had been calculated. The data was recorded.

Step Six

The recommended daily ingestion of Vitamin C in New Zealand and Australia is approximately 60 mg / daily. The mass of the fruit juice and the amount that would be needed to approximate 60 mg/ daily of vitamin C consumption.

Outcome: Fruit chosen- Orange

Weight of the fruit Juice One sample = 5.0 g

Weight of the fruit Juice sample two = 4.95 g

Table 2: Titration Outcomes for the Vitamin c determination of Orange Juice.

Molar weight of ascorbic acid is 176.12g/ mole

Chemical reaction: I₂ + Ascorbic acid ? @UI + Dehydroascorbic acid. Consequently, a mole of iodine responds to a mo9le of ascorbic acid

Fruit Juice One Sample

20.43 ml of 7.57 x 10^-4 M iodine solution possess 7.57 x 10^-4 M x 0.02043 mol of iodine.

As a result, 5.0 g of orange juice possesses 7.57 x 10^-4 M x 0.02043 mol of vitamin C.

As a result, 5.0 g of orange juice possesses 7.57 x 10^-4 M x 0.02043 x 176.12 g of Vitamin C

As a result, 5.0 g of orange juice possesses 2.7122 g of Vitamin C.

Consequently, the Vitamin C content in the fresh juice sample one is 2.7122mg/ 5.0 g = (54.2mg/g. + 54.0 mg /2) = 54.1 mg.

The hypothesis is disproven. The null hypothesis applies. The amount of vitamin c in one serving of juice is not sufficient to provide the recommended daily ingestion of vitamin C for New Zealand, Australia, Canada The UK and the United States.

Results

Which one of the single servings of fruit juices provides the recommended daily ingestion of vitamin C for New Zealand, Australia, Canada, the UK and the United States?

All of the assessments of the amount of vitamin C contained in the different fruit juice samples had been determined. There had been two distinct values for each of the categories of fruit juice. The average amount of ascorbic acid that was maintained in each gram of juice was documented. The fruit juice which had the most substantial vitamin C content was reviewed.

Discussion

In reality, it is discernable to know which of the fruit juices provides the greatest amount of vitamin C. The answerer derived may be different that the assessed values. The average value could have been skewed by two extremely elevated readings or one elevated reading for a sample. The portion of juice may have been outstanding. Usually, the measurements do not provide exact quantities, but instead they provide a distri9bution of numbers. When attempting to apply the information of which the distribution of numerical values in an actual real problem, the consideration of the baits of distribution which are the most relevant must be taken into account.

The two fruit juices which had the most elevated average vitamin C content were considered. In the discussion, they were delineated as Fruit Juice One and fruit Juice two. It would be easy to derive which of the juices was being examined from the data. In the event that fruit Juice one possessed the more elevated content of vitamin C, what is the statistical probability that if two new samples were to be analyzed, that fruit juice one would have a higher vitamin C content than fruit juice two. The method where the probability of the information derived from the collection of the fruit juices could be referenced. It is designated the term the sign test.

Step One

AS table had been drawn in the, logbook which detailed all of the collections’ assessments for fruit juice on the left side of the table. On the upper part of the table, the collections’ assessments for fruit juice two would be placed. The outcome is a grid with thirty-six boxes. An X had been drawn in each of the boxes where the reading of the fruit juice sample one had been greater than the value of fruit juice sample two. The statistical probability that the fruit juice sample one emerges as the victor in the comparison with the fruit juice sample two is equivalent to the proportion of the boxes in the table which have the cross. tabulate the fraction and document the fraction in the logbook.

Table 3: The statistical probability that fruit juice sample one would have a more solvated reading in the following sample is 24/ 67 = 0.6667