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.

  • 2
    $\begingroup$ "IB", "SL", "IA" mean nothing to me. Are they relevant to the question? $\endgroup$ – TAR86 Oct 19 '17 at 6:20

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.

  • $\begingroup$ 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

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