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Determination of Vitamin C in juices

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In this experiment I am supposed to count the concetration of vitamin C in tested sample. Vitamin C (ascorbic acid) has the formula C6H8O6, and can be oxidized to dehydroascorbic acid C6H6O6, using iodine. Oxidizer comes from potassium iodate and potassium iodide, which react with each other in presence of hydrochloric acid.

KIO3 + 5KI +6HCl → 3I2 + 3H2O + 6KCl

Iodine reacts with ascorbic acid as shown below:

C6H8O6 + I2 → 2HI + C6H6O6

Due to this reaction, formed iodine is immediately reduced to iodide as long there is any ascorbic acid present. When ascorbic acid has ran out, the remaining iodine reacts with starch indicator. In order to make that reaction happen, linear triiodide complex is formed:

I2 + I- → I3-

The triiodide slips into starch, forming starch-iodine complex which has an intense black-blue colour:

2 I3- + 2C6H10O5 → 2 C6H8O5 +6HI


  1. Pipette 5 cm3 of the tested sample into a conical flask and add about 100 cm3 of distilled water, 5 cm3 of 0.6M potassium iodide (KI), 5 cm3 of 1M hydrochloric acid (HCl) and about 1 cm3 of 5% starch indicator solution.
  2. Titrate the sample with the 0.002M potassium iodate solution (KIO3). The endpoint of the titration is the first permanent trace of dark blue colour due to the starch-iodine complex.
  3. Repeat the titration three times.

Now my problem: I have counted, that the whole amount of iodine is 0.001503 moles. I assumed that density of starch is 1g/cm, so the amount of starch in 1 cm3 of 5% starch indicator is 0.05 g. Using proportions I have calculated the amount of triiodide, which was 0.000309 moles. As i know the reaction no. 2, I've also calculated the amount of iodide, that was made from the reaction no. 1 - also 0.000309 moles. Having this data obtained I was able to calculate the concentration of ascorbic acid. now it is 0.54% (not over 3%). It sould be 0.012%

Now the problem is: What am I doing wrong? Or the other question: I was using commercial juice, not ascorbic acid dissolved in water, so there were also some other substances present. Could they affect the readings?

Please, write if sth is unclear, I will explain it more deeply.

Edited by Marazm

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Yes, that's right, they're going to be affected. Try to disolve the vitamin C in water and check your results. If possible, repeat your readings with the orange juice (3 times) and with the dissolved vitamin C (3 times).

Check your results and try to see if they are similar. Perhaps you could use that data to prove that in fact, orange juice has some impurities that affect the readings. (Just what you said earlier).

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