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I am doing an investigation in which I need to find out the vitamin C content in different fruits. I have got a method for doing a titration using iodine but need other experiments too. I have read many references to experiments using NBS and DCPIP but I need a method please?
Igloo writes
DCPIP method
First prepare the indicator solution
Dissolve about 0.4g of the dye 2,6-dichlorophenolindophenol (DCPIP) in 200 cm3 of warm water contained in a beaker. Stir well and then filter to remove insoluble material. Make up this solution to 1 dm3 in a volumetric flask.
Next you need to standardise this solution
1 Fill a clean burette with this dye solution.
2 Make up a standard solution of vitamin C containing exactly 200 mg dm-3 in a dilute solution of phosphoric acid in another volumetric flask.
3 Use a pipette – and safety filler – to transfer 25.0 cm3 of this solution to a conical flask and titrate rapidly with the dye solution. The end-point is taken when the solution remains faintly pink for about 10 seconds.
4 Repeat this titration to obtain consistent results.
5 It is conventional also to carry out a “blank” titration with the dilute phosphoric acid used to prepare the solution of vitamin C. (Refer to the React FAQs to find out the reason for carrying out this “blank” titration)
6 Carry out titrations with your vitamin C fruit extracts.
a Weigh out a known mass of your fruit sample (say 50g) and put this into a liquidiser, together with about 250 cm3 of the dilute phosphoric acid. Liquidise at high speed.
b Filter the mixture through fine muslin into a tall measuring cylinder using more of the phosphoric acid to wash out the liquidiser and the muslin filter.
c Finally make up the mixture of liquidised sample plus washings to a known volume (e.g. 500 cm3)
Carry out DCPIP titrations with 25.0 cm3 pipetted samples of this liquid extract.
Calculations
These are relatively straightforward and can be carried out via a method of arithmetic proportion.
For example, let us assume that 25.0 cm3 of the standard solution of vitamin C containing 200 mg dm-3 required 20.0 cm3 of DCPIP, but that 25.0 cm3 of the liquidised fruit sample required 30.0 cm3 of DCPIP.
Clearly the fruit sample is more concentrated in vitamin C than the 200 mg dm-3 standard solution (since it needed more DCPIP), and, more precisely, it must be 30.0/20.0 times more concentrated. In other words its concentration is (3/2 x 200) = 300 mg dm-3
Finally, if there are 300mg in every 1000 cm3, there must be 150mg in the 500 cm3 of total solution and this 150mg is all derived from the 50g of fruit.
I should mention that in each case the blank titre needs to be subtracted from the main one to arrive at a net figure. In practice it is usually found that the blank titrate is 0.0 cm3, or, very possibly, 0.1 cm3.
NBS method
If you haven’t already done so, use the search facility on the React website to find the FAQ I have already answered on the nature of the reaction taking place between NBS and vitamin C, using iodide ions as indicator.
Adopt the same sort of procedure as I have described above, substituting NBS for DCPIP but also using a solution of iodide ions (e.g. a 0.1 molar solution of KI) as an external indicator.
Starch could be used near the end-point since it will be easier to see the sharp appearance of a deep dark blue colour than a gradual change from colourless to yellow (which occurs in the absence of starch)
Always carry out a risk assessment and check with your teacher before starting any practical work.
Risk assessment
Before attempting any practical work based on the advice and suggestions on this website, you must do the following. Identify any hazards, assess the risks from these hazards, and then decide appropriate control measures to reduce the risks. You must have these approved by those in authority in your school or college laboratory. Do not rely on what is said on this website.
For further guidance see our tutorial on Risk Assessment.
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updated: 25 September 2007
