Edexcel A-Level Biology A 9BN0
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Lifestyle, Health and Risk
Core practical 2: the vitamin C content of food and drink
- 3 things to know
- Papers 1, 2 and 3
Overview
How to find the vitamin C content of fruit juice by titrating it against DCPIP, which loses its blue colour as vitamin C reduces it: the calibration curve from known concentrations, the variables to control, and the sources of error in judging the end point.
What you need to know
Answer each one before you reveal it.
Why are repeats taken and anomalies discarded in the vitamin C practical?
What safety precautions are needed in the vitamin C practical?
Questions from this subtopic
Each one with the model answer underneath.
- 1
How can the vitamin C content of food and drink be investigated using DCPIP?
Model answer
By DCPIP titration, where vitamin C decolourises the blue dye.
- 2
In a DCPIP titration, what does a smaller volume of sample needed to decolourise the DCPIP indicate?
Model answer
A higher vitamin C concentration in the sample.
- 3
What is the principle behind using DCPIP to measure vitamin C?
Model answer
DCPIP is a blue dye that is decolourised by vitamin C, a reducing agent. The more vitamin C a sample contains, the less of it is needed to decolourise a fixed volume of DCPIP.
- 4
How is the DCPIP method calibrated using a standard vitamin C solution?
Model answer
Add a standard solution of known concentration, such as 1.0 mg cm⁻³, drop by drop from a burette or pipette to a fixed volume of DCPIP, shaking after each drop, until the blue colour just disappears. Record the volume and repeat to find a mean.
- 5
How is the vitamin C content of a fruit juice found once the method is calibrated?
Model answer
Repeat the procedure with the same volume of DCPIP, adding the juice instead of the standard, and record the mean volume needed to decolourise it. Compare this with the volume of standard solution.
- 6
1 cm³ of DCPIP is decolourised by 2.0 cm³ of a 1.0 mg cm⁻³ vitamin C solution and by 5.0 cm³ of orange juice. Calculate the vitamin C concentration of the juice.
Model answer
2.0 cm³ of standard contains 2.0 mg of vitamin C, so 5.0 cm³ of juice also contains 2.0 mg. The concentration of the juice is 2.0 ÷ 5.0 = 0.40 mg cm⁻³.
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