Food Test for Vitamin C (DCPIP Test)

The DCPIP test detects vitamin C in food. A food extract is added drop by drop to blue DCPIP solution; decolourisation of the DCPIP shows vitamin C is present.

Aim

To test a food sample for the presence of vitamin C using DCPIP solution, and to compare the vitamin C content of different food samples.

What it tests for

The DCPIP test detects vitamin C (ascorbic acid) in food. DCPIP (2,6-dichlorophenolindophenol) is a blue dye that is reduced and loses its colour when it reacts with vitamin C, which is a reducing agent.

The number of drops of food extract needed to decolourise a fixed amount of DCPIP gives a rough, semi-quantitative comparison of vitamin C content between samples.

Variables

  • Manipulated variable: the type of food extract tested, for example orange juice, lemon juice, and guava juice.
  • Responding variable: the number of drops of food extract needed to decolourise a fixed volume of DCPIP solution.
  • Controlled variables: the volume and concentration of DCPIP solution, the size of each drop, and the temperature of the extract.

Materials and apparatus

  • Food extracts or fruit juices to be tested
  • 0.1% DCPIP solution (blue)
  • A standard vitamin C (ascorbic acid) solution for comparison
  • Test tubes and a test-tube rack
  • A graduated dropper or syringe, and a measuring cylinder
  • A white tile or white paper as a background to judge the colour
  • A filter funnel and filter paper for preparing extracts

Reagent and method

  1. Prepare a food extract, such as fruit juice, by squeezing or crushing the food and filtering if necessary.
  2. Place about 2 cm3 of blue DCPIP solution into a clean test tube.
  3. Using a dropper, add the food extract to the DCPIP solution one drop at a time.
  4. Shake or swirl the test tube gently after each drop.
  5. Continue adding drops and count them until the blue colour of the DCPIP just disappears.

Positive result

A positive result is shown when the blue DCPIP solution becomes colourless (decolourised) after a number of drops of food extract are added. Fewer drops needed to decolourise the DCPIP indicates a higher concentration of vitamin C in the food extract, while more drops needed indicates a lower concentration.

Expected results

The drop count for each extract can be tabulated and used to rank the samples by vitamin C content.

The fewer drops needed to decolourise the DCPIP, the higher the vitamin C content of the extract.
Food extractDrops needed to decolourise DCPIPInference
Guava juiceFew dropsHigh vitamin C content
Orange juiceMore dropsModerate vitamin C content
Boiled orange juiceMost drops, or no decolourisationLow vitamin C content (heat destroys vitamin C)

Inference and conclusion

DCPIP is decolourised because vitamin C is a reducing agent that reduces the blue dye to a colourless form. The fewer drops of extract needed to decolourise a fixed volume of DCPIP, the higher the concentration of vitamin C in that extract.

Comparing the drop counts therefore ranks the samples by vitamin C content, as long as the same volume and concentration of DCPIP is used each time. Because vitamin C is destroyed by heat, a boiled sample needs more drops than a fresh one, which shows why fresh fruit and gentle cooking preserve more vitamin C.

Paper 3-style questions

Question 1 (hypothesis). A student compares fresh orange juice with orange juice that has been boiled. State a suitable hypothesis.

Model answer. Fresh orange juice contains more vitamin C than boiled orange juice, so it needs fewer drops to decolourise the same volume of DCPIP.

Question 2 (variables). State the manipulated, responding and one controlled variable for this comparison. Model answer. Manipulated variable: whether the juice is fresh or boiled.

Responding variable: the number of drops needed to decolourise the DCPIP. Controlled variable: the volume of DCPIP solution (also its concentration, the drop size, and the temperature at testing).

Question 3 (tabulation and graph). The student tests four fruits. How should the results be recorded and displayed?

Model answer. Record the drops needed for each fruit in a table, then draw a bar chart of drops against fruit; a shorter bar means higher vitamin C content.

Question 4 (inference). Fresh juice needed 8 drops and boiled juice needed 20 drops to decolourise the DCPIP. What can you infer?

Model answer. The fresh juice contains more vitamin C than the boiled juice, because it needed fewer drops; boiling has destroyed some of the vitamin C. This supports the hypothesis.

Safety

  • Wear safety goggles, because DCPIP solution and acidic fruit juices can irritate the eyes.
  • Add the extract with a clean dropper and avoid touching your mouth with it.
  • Handle glassware carefully to avoid breakage, and clean up spills promptly.
  • Wash your hands after handling the reagents and dispose of used solutions as directed.

Common mistakes

Source:SRC-DSKP-EN

Frequently asked questions

What happens to DCPIP solution when vitamin C is present?
DCPIP solution is blue, but it loses its colour and becomes colourless when it reacts with vitamin C, which reduces the dye. This decolourisation is the positive result for the test.
How does the DCPIP test compare vitamin C content between two foods?
The number of drops of food extract needed to decolourise a fixed volume of DCPIP solution is counted for each food. Fewer drops needed means a higher concentration of vitamin C, so this allows a fair comparison as long as the same volume of DCPIP is used each time.
Why must the food extract be added drop by drop?
Adding the extract drop by drop, with shaking after each drop, allows the exact point of decolourisation to be identified and counted accurately. Adding it too quickly makes it impossible to determine how many drops were actually needed.
Why does boiled juice need more drops than fresh juice?
Vitamin C is destroyed by heat, so boiling lowers the vitamin C content of the juice. A sample with less vitamin C reduces less DCPIP, so more drops are needed to decolourise the same volume of dye.

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