I'm an IB chemistry SL student doing an IA on the vitamin C concentration of orange juice. We haven't learned about the titration process, so here is what I have done so far:

$\ce{5KI + KIO3 + 3H2SO4 -> 3I2 + 3K2SO4 + 3H2O}$

This is the equation for producing the iodine solution. Now using a titration process, I added the iodine solution to the orange juice. I measured the volume of Iodine solution used.

Here is the formula for the reaction between the vitamin C and the Iodine solution.

$\ce{C6H8O6 + I3- + H2O -> C6H6O6 + 3I- + 2H+}$

I'm not sure where the Triiodide comes from. Can someone explain this to me? I'm not a very good chemistry student.

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    $\begingroup$ "IB", "SL", "IA" mean nothing to me. Are they relevant to the question? $\endgroup$ – TAR86 Oct 19 '17 at 6:20
  • 1
    $\begingroup$ @TAR86 International Baccalaureate (IB) Standard Level (SL) chemistry student doing an Internal Assessment (IA), see owltutors.co.uk/chemistry-ib-ia-ideas-2017-2018 ... and not relevant to the question ... $\endgroup$ – Karsten Theis Dec 17 '19 at 15:14

Iodine and iodide will form triiodide in aqeous solution. This is a common way to increase the solubility of $\ce{I2}$ in water. For a titration, one will use excess $\ce{KI}$ and control the formation of iodide by adding $\ce{KIO3}$ using a burette.

In response to the comment:

$\ce{KI}$ contains iodide anions. I strongly suspect that an excess of $\ce{KI}$ was used (more than fivefold compared to $\ce{KIO3}$), leaving enough iodide anions to form the triiodide. Also note that effectively a $\ce{I-}$ can be subtracted from either side of the second equation of the question.

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  • $\begingroup$ Where does the iodide come from? What I'm seeing here is that the Iodine in the first equation is somehow changed into triiodide in the second equation. I included the course name, so people would understand the topics I have done so far, and explain things I would know about. Again, I'm not a particularly good chemistry student. $\endgroup$ – Omkar Vaidya Oct 19 '17 at 9:09
  • $\begingroup$ Thanks, that helped me. I did use more than five times the amount of KI. I'm just not sure how the I- can be subtracted from either side of the second equation. I am pretty behind in my course. $\endgroup$ – Omkar Vaidya Oct 19 '17 at 11:06
  • $\begingroup$ Also, how can you measure the mass of triiodide if you have an excess of KI? The reaction is reversible, so I'm not sure how much of the triiodide would be left. $\endgroup$ – Omkar Vaidya Oct 19 '17 at 12:11

One way to determine the amount of vitamin C in food is to use a redox titration. The redox reaction is better than an acid-base titration since there are additional acids in a juice, but few of them interfere with the oxidation of ascorbic acid by iodine.

Iodine is relatively insoluble, but this can be improved by complexing the iodine with iodide to form triiodide:

$$\ce{I2 + I- <=> I3-}$$

Triiodide oxidizes vitamin C to form dehydroascorbic acid: $$\ce{C6H8O6 + I3- + H2O → C6H6O6 + 3I- + 2H+}$$ As long as vitamin C is present in the solution, the triiodide is converted to the iodide ion very quickly. However, when all the vitamin C is oxidized, iodine and triiodide will be present, which reacts with starch to form a blue-black complex. The blue-black color is the endpoint of the titration.

This titration procedure is appropriate for testing the amount of vitamin C in vitamin C tablets, juices, and fresh, frozen, or packaged fruits and vegetables. The titration can be performed using just iodine solution and not iodate, but the iodate solution is more stable and gives a more accurate result.`

See: https://www.jacobtuition.co.uk/post/inspirational-ideas-for-your-chemistry-ib-ia

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  • $\begingroup$ (+1) We used to use a variant of this, using coulometric titration to produce iodine in situ, as described in the last paragraph of my answer here. The solution had excess KI. The endpoint was via the standard starch-iodine complex. $\endgroup$ – Ed V Jun 30 at 0:09
  • $\begingroup$ Balance the second equation $\endgroup$ – Adnan AL-Amleh Jun 30 at 6:33
  • $\begingroup$ Note that you copied an unbalanced equation from the OP. $\endgroup$ – Buck Thorn Aug 10 at 9:16

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