How to find the vitamin C content of orange juice? 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:

$\ce{5KI + KIO3 + 3H2SO4 -&gt; 3I2 + 3K2SO4 + 3H2O}$

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. 

$\ce{C6H8O6 + I3- + H2O -&gt; C6H6O6 + 3I- + 2H+}$

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