The isolation of vitamin C was carried out by monitoring the reducing properties of green pepper ext — Physical Chemistry — Kinetics Chemistry Question
Determination of ascorbic acid
The isolation of vitamin C was carried out by monitoring the reducing properties of green pepper extracts (1931, Szent-Györgyi). The determination of ascorbic acid can also be based on its reducing properties. This is often more convenient than titration as an acid, especially in the case of real life samples containing other acidic substances, e.g., citric acid.
A possible oxidizer is potassium bromate. Its use in direct titrations was suggested in 1872 by Győry. In strongly acidic solutions KBrO3 reacts with KBr to form bromine. This will oxidize ascorbic acid (C6H8O6) to dehydroascorbic acid (C6H6O6) in this titration. The end point of the reaction can be followed by a suitable redox indicator.
[VISUAL]
Chemicals and reagents
* Hydrochloric acid, 20 %
* Potassium bromate, 0.02 mol dm–3
* Potassium bromide, solid
* Para-ethoxychrysoidine, 0.2 % ethanolic solution
Procedure
Crush the vitamin C tablet with a few drops of water in a mortar. Wash the soluble parts of the mixture through a folded filter paper into a 200 cm3 Erlenmeyer flask. Do not use more than 60 cm3 of distilled water. Add 10 cm3 of 20 % (by mass) HCl solution and approx. 0.2 g of KBr to the sample. Titrate it immediately with the KBrO3 solution (c = 0.02 mol dm–3) in the presence of 2 drops of p-ethoxychrysoidine indicator (0.2 % in ethanol). The red solution will change to colorless (very light yellow) at the end point.
Write the equation for bromine formation from bromate and bromide ions.
Model Answer
BrO3– + 5 Br– + 6 H+ → 3 Br2 + 3 H2O
Give the vitamin C content of the tablet in milligrams.
Model Answer
If V cm³ potassium bromate solution is consumed then the tablet contains 10.57 × V mg ascorbic acid, as the following equation can be written:
m(C₆H₈O₆) = 3 × c(KBrO₃) × V(KBrO₃) × M(C₆H₈O₆)