The Assay of Ascorbic Acid

The Assay of Ascorbic Acid

Student Name: Med Chem-IV

Student ID :

Homework

THE ASSAY OF ASCORBIC ACID

OBJECTIVE:This method is used to determine the % purityof Vitamin C tablet in a solution by a redox titration using iodine.

Procedure: Titrate the solution of ascorbic acid with Iodine solution of known concentration using 0.5 % starch solution as indicator.

Solutions Needed:

1-Iodine solution: (0.01 M). Weigh 2 g of potassium iodide into a 100 mL beaker. Weigh 1.3 g of iodine and add it into the same beaker. Add a few mL of distilled water and swirl for a few minutes until iodine is dissolved. Transfer iodine solution to a 500 ml volumetric flask, making sure to rinse all traces of solution into the volumetric flask using distilled water. Make the solution up to the 500 mark with distilled water.

2-Starch indicator solution: (0.5%). Weigh 0.25 g of soluble starch and add it to 50 mL of near boiling water in a 100 mL conical flask. Stir to dissolve and cool before using.

3-Vitamin C solution: Weigh the tablet equivalent to 0.5 gm of Vitamin C and dissolve in 500 mL of distilled water (in a volumetric flask).

Titration

  1. Pipette a 10 mL aliquot of the sample solution into a 250 mL conical flask and add about1.0 mL of starch indicator solution.
  2. Titrate the sample with 0.01 Miodine solution. The endpoint of the titration is identified as the first permanent trace of a dark blue-black color due to the starch-iodine complex.
  3. Repeat step 1 and 2until you obtain concordant results (titers agreeing within 0.1 ml).

Calculation:

Theoretically

1 mole of iodine solution ≡ 1 mole of C6H8O6

1000 ml 0f 1M iodine solution≡ 176 g of vit c

1 ml 0f 1 M iodine solution≡ 0.176 g of vit c

1 ml 0f 0.01 M iodine solution≡ 0.00176 g of vit c

2 ml 0f 0.01M iodine solution≡ 2 x 0.00176 g of vit c

V(ml) of 0.01M iodine solution≡ V x 0.00176 g of vit c = X g of vit C

% Purity

For 7.93 ml of iodine solution (Practical reading from the lab.):

Amount of Vit.C (in 10 ml) = (7.93 ml) (0.00176 gm of Vit.C)…………………………Eq. 1

= 0.0139 gm in 10 ml of Vit.C solution

Calculation was based on 10 ml of Vit.C solution. same result should be obtained when doing the calculation in 500 ml or 1000 ml of Vit.C solution butremember to change the theoretical value to 500 mg or 1000 mg, respectively.

% Purity = (0.0139gm/ 10 mg) X 100 (Remember to unify the units)

% Purity = 139 %

Result: The ascorbic acid tablet was found to be 139 % pure.

For other groups ( 8.13 ml, 13.0 ml, 11.1 ml, and 12.9 ml ) follow Eq. 1 and do calculations.

Questions:

1)Potassium Iodide (KI) was added to make the Iodine solution, please explain Why?

Since I2 is not soluble in H2O, KI was added to form the tri-iodide salt and enhance the solubility of I2 to prepare the Iodine solution.

2)How many Iodine isotopes exist in nature, what is the most stable one and provide two medical uses of Iodine?

37 isotopes of Iodine exist in nature.

I127 is the most stable isotope.

Medical Uses:

  1. Hyperthyroidism.
  2. Disinfectant.
  3. Radiation emergency due to radioactive iodides.

etc…..

Other uses are also accepted.

3)Draw the structure of Ascorbic acid and name all the functional groups present in the structure?

4 Hydroxyl group (Alcohol)

1 Alkene

1 Ester