Experiment No. 8
“Iodimetric Analysis for Vitamin C”
OBJECTIVES: 1.
To prepare and standardize an iodine solution;
2. To analyze an unknown sample containing vitamin C through titration method.
DISCUSSION: Ascorbic acid which is generally known as Vitamin C, is a water-soluble vitamin found in fruits and leafy vegetables or made synthetically and used as an antioxidant. Since humans are incapable of producing or synthesizing vitamin C, they need to have regular digestion to maintain healthy body and prevent scurvy. Scurvy is a disease caused by insufficient vitamin C, the symptoms of which include spongy gums, loosening of the teeth, and bleeding into the skin and mucous membranes. On the other hand, an onset of 60mg intake of vitamin C per day is the Recommended Dietary Allowance (RDA) for adults. Through oxidizing ascorbic acid, (C6H6O6), vitamin C content of fruits can easily be determined. C6H8O6
C6H6O6 + 2H+ + 2e-
Since this reaction can process immediately when in contact with moistness, an aqueous solution particularly I 2- a substance that is not very soluble in water and is good for oxidation, will be used. I 2 will be dissolved in potassium iodide (KI) solution in which I2 will exist as I3-, a complex ion. The reaction with ascorbic acid involves I2, which is reduced to I- ion. 2e- + I2
2I-
Overall, one mole of ascorbic acid (vitamin C) needs one mole of I 2 to totally oxidize. The I2 solution (red in color) was then added to the ascorbic acid solution which results to the disappearance of the iodine color due to the reaction that occurred. To attain better results, starch- a natural substance composed of chains of glucose units (particularly amylose and amylopectin) made by plants and providing a major energy source for animals, is added as an indicator to the solution. Starch responds with I2 specially with ascorbic acid. Its concentration remains very low until the ascorbic acid is all oxidized. At that point, the I 2 concentration begins to go up and the reaction with the indicator occurs.
I2 + starch yellow
Starch – I2 complex blue
Afterwards, I2 will be standardized with a primary standard substance which is the ascorbic acid. . During the standardization of I 2 solution, an approximately 0.10 g of three ascorbic acid samples were weighed and then dissolved in about 10 ml of water into a clean 250 mL Erlenmeyer flask. Then, 10 g of I 2 crystals were weighed in a 400 ml beaker which was added with approximately 150 mL distilled water. The solution was stirred thoroughly and then covered with a piece of aluminum foil. Afterwards, 1 ml of starch indicator was added to the ascorbic acid samples and the samples were titrated with the iodine solution. Through proper titration techniques, we were able to reach a distinct blue color of the solution. On the other hand, titration was also used to determine the vitamin C content of the fruit juice sample used. In this experiment we have used an orange fruit. We have extracted its juice and measure 10 mL for each trial (there were 3 trials). Then, for the titration, we have used the same process as in for standardization of I 2 solution.
RESULTS:
A. Standardization of Iodine Solution
Mass of ascorbic acid, g Moles of ascorbic acid
TRIAL 1
TRIAL 2
0.10 g
0.10 g
5.68 ×10−4
mol
5.68 ×10−4
Initial buret reading, ml
29.2 ml
34.4 ml
Final buret reading, ml
34.4 ml
39.8 ml
5.2 ml
5.4 ml
0.1092 M
0.1052 M
Volume of I2 delivered, ml Molarity of I2, M
B. Analysis of Orange Juice
mol
Volume of orange juice, ml Initial buret reading, ml Final buret reading, ml Volume I2 delivered, ml Conc. of vitamin C in liquid sample, mg/100 ml Average RDA, ml
TRIAL 1 10 ml
TRIAL 2 10 ml
TRIAL 3 10 ml
26 ml 26.5 ml 0.5 ml 9.3264 mg/100 ml
26.5 ml 26.9 ml 0.4 ml 7.4646 mg/100 ml
26.9 ml 27. 3 ml 0.4 ml 7.4646 mg/100 ml
74.21 ml
CALCULATIONS:
Average mL of I2 solution derived for standardization of I2 solution: 5.2 mL I2. + 5.4 mL I2 10.6 mL I2
10.6 mL I2 ÷ 2 = 5.3 mL I2
Trial 1:
5.3 mL I 2 0.10 g AA = 0.5 mL Vitamin C X
X ( 5.3 mL )=0.10 g ( 0.5 mL )
5.3 mL ( X ) 0.05 g( mL) = 5.3 mL 5.3 mL
X =9.43 ×10−3 g AA
9.43 ×10−3 g AA
1 mol =5.36 ×10 ( 176 g)
−5
mol AA∨mol Vitamin C
5.36 ×10−5 mol AA
g =9.3264 ×10 ( 174 1 mol )
9.3264 ×10−3 g AA
( 10001 gmg )=9.3264 mg AA∨Vitamin C
Trial 2:
5.3 mL I 2 0.10 g AA = 0.4 mL Vitamin C X
X ( 5.3 mL )=0.10 g ( 0.4 mL )
−3
g AA
5.3 mL ( X ) 0.04 g (mL) = 5.3 mL 5.3 mL
X =7.55× 10−3 g AA
7.55× 10−3 g AA
( 1176molg )=4.29 ×10
4.29 ×10−5 g AA
g =7.4646 ×10 ( 174 1 mol )
−5
7.4646 ×10−3 g AA
mol AA∨mol Vitamin C
−3
g AA
( 10001 gmg )=7.4646 mg AA∨Vitamin C
Trial 3:
5.3 mL I 2 0.10 g AA = 0.4 mL Vitamin C X
X ( 5.3 mL )=0.10 g ( 0.4 mL )
5.3 mL ( X ) 0.04 g (mL) = 5.3 mL 5.3 mL
X =7.55× 10−3 g AA
7.55× 10−3 g AA
( 1176molg )=4.29 ×10
4.29 ×10−5 g AA
g =7.4646 ×10 ( 174 1 mol )
−5
7.4646 ×10−3 g AA
mol AA∨mol Vitamin C
−3
g AA
( 10001 gmg )=7.4646 mg AA∨Vitamin C
Recommended Dietary Allowance (RDA):
Average mg Vitamin C :
9.3264 mg
+ 7.4646 mg
24. 2556 mg ÷ 3 =
8.0852 mg Vitamin C 7.4646 mg 24. 2556 mg
60 mg∶ X =8.0852 mg∶ 10 mL
X ( 8.0852 )=( 10 mL )( 60 mg )
X
8.0852mg 600 mL( mg) = 8.0852mg 8.0852mg
X =74.21 mLof Vitamin C content ∈orange juice is needed ¿ obtain 60 mg RDA . CONCLUSION: Through proper titration techniques together with the results gathered and calculations conducted, the vitamin C content of orange fruit is 74.21 mL which is needed to attain 60 mg Recommended Dietary Allowance (RDA).
ANSWERS TO QUESTIONS:
1. Write a balanced equation for the reaction between I2 and ascorbic acid. Identify the oxidizing agent and the reducing agent. C6H8O6 + I2 (ascorbic acid) C6H8O6
+
2I-
C6H6O6 + (dehydroascorbic acid) I2
C6H6O6
+
+ 2I-
2H+ +
2H+
The iodine oxidizes the ascorbic acid. The ascorbic acid reduces the iodine. OA: C6H6O6 RA: C6H8O6
2. A solution of I2 was standardized with ascorbic acid. Using a 0.100 g sample of pure ascorbic acid, 25.32 mL of I 2 were required to reach the starch end point.
a) What is the molarity of the iodine solution? 0.1 g C 6 H 8 O 6
1 mol C 6 H 8O 6 =5.65 x 10 ( 176.9 C 6 H 8O 6 )
−4
mol
5.65 x 10−4 mol =0.02 M 25.32ml 1000
b) What is the titer of the iodine solution?
1 mg AA 176.1 mg AA
= 0.005679 mmole AA
0.005679mmole AA mmol 0.02243 mL
= 0.25 mL
Titer =
1 mg ascorbic acid 0.25 ml I 2 sol ' n
3. A sample of fresh grapefruit juice was filtered and titrated with the above I 2 solution. A 100-mL sample of the juice took 9.85 mL of the iodine solution to reach the starch end point. a) What is the concentration of vitamin C in the juice in mg vitamin C/100 mL of juice?
25.32 ml 9.85 ml = 0.10 g x
0.0389 g
174 C 6 H 6 O6 1000 g =38.457 mg Vitamin C per 100 mL of juice ( 1176molg CC 66 HH 88 O6 O 6 )( 1 mol C 6 H 6 O6 )( 1 g )
b) What quantity of juice will provide the RDA amount of vitamin C?
60 mg =156.02mL of this fruit juice is needed 38.457
