Food Science
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I am investigating the vitamin C content of fruits and fruit juices. I have so far used the method of titration with DCPIP. However, as a second method, I want to use colorimetry. I have read your FAQs with regards to using colorimetry with iodine, but have several questions about this.
Firstly, roughly what quantity of iodine should be dissolved in aqueous potassium iodide as a standard solution, and also, when making different concentrations, for the calibration curve, is it the amount of potassium iodide that is altered, or do you mix it with distilled water? Also (sorry, so many questions!) when adding the juice to the "stock solution of iodine," is this the original concentration of iodine in aqueous potassium iodide? Thank you very much.
Igloo writes
To answer your questions I’ll enlarge on the details given in the relevant FAQ on the React website which is at:
http://www.chemistry-react.org/go/default/Faq/Faq_22372.html
You need to use solutions of iodine which are similar to the concentrations of vitamin C in fruit. A typical fruit juice contains round about 200 mg dm-3, which equates to a concentration of about 0.001 mol dm-3. Since you will be diluting your iodine solution, you should perhaps start with a concentration of, for example, 0.002 mol dm-3. You will need to calculate what this represents in g dm-3, so that you can weigh out the relevant mass, transfer this to a volumetric flask, and use a solution of potassium iodide containing about 2 mol dm-3 to make up the iodine into solution. Unfortunately iodine is not a primary standard so you will have to standardise this solution against a standard solution of sodium thiosulphate in order to determine its accurate concentration.
Next comes the calibration stage. Make up, for example, 10 diluted solutions of the original iodine solution, using potassium iodide solution as the diluting agent, and use colorimetry (with an appropriate filter) to find the transmittance of each. For example:
| Volume of iodine solution / cm3 | Volume of potassium iodide solution / cm3 | |
| A | 0.0 | 0 |
| B | 9.0 | 1.0 |
| C | 8.0 | 2.0 |
| etc | etc | etc |
| etc | etc | etc |
| J | 1.0 | 9.0 |
| K | 0.5 | 9.5 |
Let’s assume that after standardisation the original iodine solution was found to have a concentration of 0.00192.
The concentration of solution A will obviously be 0.00192 mol dm-3.
The concentration of solution B will be 9.0/10.0 x 0.00192 = 0.001728 mol dm-3
The concentration of solution K will be 0.5/10.0 x 0.00192 = 0.000096 mol dm-3.
Next plot the calibration curve and you can use this to convert transmittance readings into concentrations of iodine for each of your experiments with the fruit juices.
I hope by now that you can make better use of our FAQ on our website, which explains the subsequent procedures. Use the stock solution of iodine to add to each of your fruit juice samples.
Please let me know if you have any outstanding queries.
Risk assessment
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updated: 07 January 2008
