TITLE: THE AMOUNT OF VITAMI C IN THE FRIT JUICE IN DIFFERENT

CONDITION.

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ABSTRACT

Vitamin C also known as ascorbic acid, is a water-soluble vitamin needed by the body for

hundreds of vital metabolic reactions. Vitamin C is not produced in the human body, yet it is

essential for the formation of collagen, connective tissue, and immune system factors. This

experiment was conducted by putting each of the same amount of fresh pineapple juice in 5

different conditions for 10 minutes and the amount of vitamin C left was tested by using Iodine

Titration. The percentage of vitamin C loses is calculated and the vitamin C losses the most is

exposed to microwave while stored in refrigerator losses the least.

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TITLE………………………………………………………………………………….…..1

ABSTRACT………………………………………………………………………………2

CONTENT…………………………………………………………… ………………….3

1.0 INTRODUCTION

1.1 Problem Statement…………………...................................……………………...4

1.2 Objective………........…………………………………………………………….4

1.3 Literature Review

1.3.1 Previous Research………………......……………………...………………5

2.0 METHODOLOGY

2.1 Theory…………………………………………………………………….……...6

2.2 Procedure..........................…………………………………………………….…7

3.0 OBSERVATION, RESULT AND DISCUSSION

3.1 Data Presenation............…………………………………………………………8

3.2 Discussion.....................................……………………………………………….9

4.0 CONCLUSION………………………………………………………………............13

REFERENCES…………………………………………………………………………...14

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1.0 INTRODUCTION

Vitamin C or ascorbic acid is important to the body for prevention of scurvy. Scurvy is

caused by lack of vitamin C in the body which can affect muscles weakness, swollen and

bleeding of gums, loss of teeth and bleeding under the skin as well as tiredness and

depression. Good sources of vitamin C are fresh fruits and vegetables, especially citrus fruits.

Benefits of vitamin C (ascorbic acid) may include protection against immune system

deficiencies, cardiovascular disease, prenatal health problems, eye disease, and

even skin wrinkling. It also functions as an antioxidant and is necessary for the treatment and

prevention of scurvy. However, Vitamin c is a type of water soluble vitamin.Water-soluble

vitamins are less stable and can be destroyed by microwave, water, heat, air and light

1.1 Problem Statement

Which condition cause the lost of vitamin C content in fresh juices the most and the

least?

1.2 Objective

To determine the amount and percentage of vitamin C losses after exposed to heating,

sun light, refrigerator, normal room condition and microwave.

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1.3 Literature Review

1.3.1 Previous research

Vitamin C is a potent reducing agent, meaning that it readily donates electrons to

recipient molecules. Related to this oxidation-reduction (redox) potential, two

major functions of vitamin C are as an antioxidant and as an enzyme cofactor [1].

Vitamin C also participates in redox recycling of other important antioxidants; for

example, vitamin C is known to regenerate vitamin E from its oxidized form[2].

And due to vitamin C is so important for mankind, there are a lot of research has

been done by scientist. For example, diseases cause by vitamin C deficiency [3],

different fruits and vegetables vary in their vitamin C content[4], diseases cause

by vitamin C overdose [5] and preservation of vitamin C [6] Vitamin C is very

unstable in aqueous solutions. Tests showed ascorbic acid to be a weak,

monobasic acid and a strong reducing agent.Therefore, it oxidizes and the product

of such oxidation involving the loss of two hydrogens (C6H6O6), was named

dehydroascorbic acid.[7]Besides that, the factors that affecting the vitamin C is

also been discovered. Factor such as UV light that degrade the vitamin C via

wavelength[8], high temperature that decomposes the vitamin C due to increase of

enzymatic destruction[9] and microwave that hit the vitamin C with high energy

wavelength that directly causing the decomposition of the vitamin C.[10]

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2.0 METHODOLOGY

2.1 Theory

Vitamin C can be determined by acid-base reaction or oxidation-reduction reaction.

Vitamin C is a weak acid and a good reducing agent. Iodine is a weak oxidizing agent, so

that it will not oxidize substances other than the ascorbic acid in the sample of fruit juice.

As a strong reducing agent, vitamin C will reduce I2 to I- very easily. The excess of iodine

react the starch as indicator to perform the amount of vitamin C is finishing in redox

reaction. In this reaction, the ascorbic acid molecule gains oxygen (in the form of -OH

groups). Each iodine atom in the I2 molecule accepts an electron and become a negatively

charge to form iodide ion. Thus that, the ascorbic acid molecule was oxidized and the

iodine molecule was reduced.

C6H8O6 + I2 + 2H2O → C6H10O8 + 2I - + 2H+ …….equation 1

Excess iodine (I2) reacts with iodide ions (I-) to form triiodide ion (I3-) which forms a

very intense blue color when it reacts starch. This colour is due to incorporation of the

ions within the molecular structure of the starch.

I2 + I-→ I3- ………equation 2

To detect the end point, starch must be added at the beginning of the titration in the

conical flask. When all ascorbic acids have finished, the excess of iodine solution will

react the starch to form blue-black colour in the solution.

I3-+ starch → starch (I3

-) complex …….equation 3

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2.2 Procedure

350ml of fresh fruit juice were squized from the pineapple fruit. 50 ml of fresh fruit juice

A, B, C, D and E by using a measuring cylinder and poured saperately into 5 beakers.The

5 beakers were then placed under 5 different condition in the order of A-Hot water bath,

B-Under the sun, C-Refrigerator, D-Microwave and E-Room temperature for 10 minutes

measured by using stop watches. 1.27 g of iodine and 2.54 g potassium iodide (KI) were

dissolved in 200 ml of distilled water and added with 30 ml of 3 M sulfuric acid. The

solution was poured into a 1000 ml volumetric flask and diluted to a final volume of 1000

ml with distilled water. The solution was transferred to a 600 ml beaker labeled as iodine

solution. 0.50 g soluble starch was added to 50ml near-boiling distilled water and mixed

well and allows cooling before use. 15.00 ml of fresh squeezed pineapple juice standard

solution to a 125 ml conical flask.2 drops of 1% starch solution were added. Burette was

rinsed with a small volume of the iodine solution and then filled it. The initial volume was

recorded. The solution is titrated until the endpoint is reached which is the first sign of

blue color that persists after 20 seconds of swirling the solution is observed. The final

volume of iodine solution was recorded. The experiment is repeated by using fresh fruit

juice A, B, C, D and E .

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3.0 OBSERVATION, RESULTS AND DISUSSION

3.1 Data presentation

The chemical equation of the reaction:

I2 + C6H8O6 ---> C6H6O6 + 2 H+ + 2 I-

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Original Titration First Second Average

Final Reading/cm3 7.85 15.60

Initial Reading/ cm3 0.00 7.85

Volume of iodine solution used/ cm3 7.85 7.75 7.80

A Titration First Second Average

 Hot Final Reading/ cm3 6.50 12.90

 Water Initial Reading/ cm3 0.00 6.50

 Bath Volume of iodine solution used/ cm3 6.50 6.40 6.45

B Titration First Second Average

 Under Final Reading/ cm3 6.35 12.60

 The Initial Reading/ cm3 0.00 6.35

 Sun Volume of iodine solution used/ cm3 6.35 6.25 6.30

C Titration First Second Average

 Refri- Final Reading/ cm3 7.05 14.20

 gerator Initial Reading/ cm3 0.00 7.05

Volume of iodine solution used/ cm3 7.05 7.15 7.10

D Titration First Second Average

 Micro- Final Reading/ cm3 6.85 13.60

 wave Initial Reading/ cm3 0.00 6.85

Volume of iodine solution used/ cm3 6.85 6.75 6.80

E Titration First Second Average

 Room Final Reading/ cm3 5.50 10.90

 Tempe- Initial Reading/ cm3 0.00 5.50

 rature Volume of iodine solution used/ cm3 5.50 5.40 5.45

Iodine Ascorbic acid (vit C) dehydroascorbic acid hydrogen ioniodide ion

1 mole of I2 react with 1 mole of Ascorbic acid

Number of mole of iodine = 1.27/126.90

used in the iodine solution = 0.01mole

0.01 mole of I2 react with 0.01 mole of Ascorbic acid

Mass of ascorbic acid reacted= mole x molar mass of ascorbic acid

=0.01 x 176.12

= 1.7612g

Concentration of = 0.01mol/1000cm3

iodine solution = 0.01mol/dm-3

Original: 15 cm3 of pineapple juice required 7.80 cm3 of iodine solution to react completely

Mole of ascorbic acid in = MV/1000

15 cm3 of pineapple juice = 0.01x 7.80/1000

= 7.80 x 10-5mol

Concentration of ascorbic acid in = 7.80 x 10-5mol x 1000/15

pineapple juice = 5.20 x 10-3mol/dm-3

Condition A: 15 cm3 of pineapple juice required 6.45 cm3 of iodine solution to react completely

Mole of ascorbic acid in = MV/1000

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15 cm3 of pineapple juice = 0.01x 6.45/1000

= 6.45 x 10-5mol

Concentration of ascorbic acid in = 6.45 x 10-5mol x 1000/15

pineapple juice = 4.30 x 10-3mol/dm-3

Condition B: 15 cm3 of pineapple juice required 6.30 cm3 of iodine solution to react completely

Mole of ascorbic acid in = MV/1000

15 cm3 of pineapple juice = 0.01x 6.30/1000

= 6.30 x 10-5mol

Concentration of ascorbic acid in = 6.30 x 10-5mol x 1000/15

pineapple juice = 4.20 x 10-3mol/dm-3

Condition C: 15 cm3 of pineapple juice required 7.10 cm3 of iodine solution to react completely

Mole of ascorbic acid in = MV/1000

15 cm3 of pineapple juice = 0.01x 7.10/1000

= 7.10 x 10-5mol

Concentration of ascorbic acid in = 7.10 x 10-5mol x 1000/15

pineapple juice = 4.73 x 10-3mol/dm-3

Condition D: 15 cm3 of pineapple juice required 6.80cm3 of iodine solution to react completely

Mole of ascorbic acid in = MV/1000

15 cm3 of pineapple juice = 0.01x 6.80/1000

= 6.80 x 10-5mol

Concentration of ascorbic acid in = 6.80 x 10-5mol x 1000/15

pineapple juice = 4.53 x 10-3mol/dm-3

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Condition E: 15 cm3 of pineapple juice required 5.45 cm3 of iodine solution to react completely

Mole of ascorbic acid in = MV/1000

15 cm3 of pineapple juice = 0.01x 5.45/1000

= 5.45 x 10-5mol

Concentration of ascorbic acid in = 5.45 x 10-5mol x 1000/15

pineapple juice = 3.63 x 10-3mol/dm-3

Condition Concentration of ascorbic acid (moldm-3)

Original 5.20 x 10-3

Hot Water Bath 4.30 x 10-3

Under The Sun 4.20 x 10-3

Refrigerator 4.73 x 10-3

Microwave 4.53 x 10-3

Room Temperature 3.63 x 10-3

Table above shows the concentration of the ascorbic acid at different conditions.

The original pineapple juice (Initial) is the fresh juice of the same pineapple without go thought

any condition. Thus, the concentration of the ascorbic acid in the original pineapple juice is

assumed to be 100%.

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The formula:

[(Initial concentration – Final concentration)/Initial concentration] x 100%

is used to determine the percentage of vitamin C loss in the 5 condition after 10 minutes.

The pineapple juice in A is placed in hot water bath and the Vitamin C loss after 10 minutes is

[(5.20x10-3- 4.30x10-3)/ 5.20x10-3 ] x 100% = 17.31%.

The pineapple juice in B is placed under the sun and the Vitamin C loss after 10 minutes is

[(5.20x10-3- 4.20x10-3)/ 5.20x10-3 ] x 100% = 19.23%.

The pineapple juice in C is placed in refrigerator and the Vitamin C loss after 10 minutes is

[(5.20x10-3- 4.73x10-3)/ 5.20x10-3 ] x 100% = 9.04%.

The pineapple juice in D is placed in microwave and the Vitamin C loss after 10 minutes is

[(5.20x10-3- 4.53x10-3)/ 5.20x10-3 ] x 100% = 12.88%.

The pineapple juice in E is placed at room temperature and the Vitamin C loss after 10 minutes is

[(5.20x10-3- 3.63x10-3)/ 5.20x10-3 ] x 100% = 30.19%.

The table above shows the percentage of vitamin C loss in 5 different conditions.

4.0..CONCLUSION

The effect of cold, heat, oxygen, UV light and microwave on vitamin C is significant .The loss of

vitamin C is arranged in ascending order :

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Condition Percentage of Vitamin C loss (%)

Hot Water Bath 17.31

Under The Sun 19.23

Refrigerator 9.04

Microwave 12.88

Room Temperature 30.19

Refrigerator > Microwave > Hot Water Bath > Under The Sun > Room Temperature.

The vitamin C in pineapple juice which placed in refrigerator losses the least and the vitamin C

in pineapple juice which placed at room temperature losses the most. The vitamin C in pineapple

juice which treated with hot water bath can also shows the effect of vitamin C losses during

cooking process. The vitamin C in pineapple juice shows that placing the fruit juice under the hot

sun will lead to a great loss of vitamin C. The best way to preserve the vitamin C in the fruit

juice is keep it in an airtight container to prevent oxidation by the air in surrounding and store in

a cold condition to reduce the reactivity vitamin C. Lastly, my recommendation for future study

is to study the effect on enzyme present in the organism to the vitamin C in the organism.

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