It's a well known and standard reaction : 3I<sub>2</sub> + 6NaOH --> 3H<sub>2</sub>O + 5NaI + NaIO<sub>3</sub>
@OP
When you initially did the reaction, you got a mixture(with perfect stoichiometry in fact for the oxidation) of sodium iodide and sodium iodate as
shown by the above equation as well as excess sodium hydroxide. Now, when you added phosphoric acid, under acidic conditions the iodate was able to
oxidize the iodide back to elemental iodine, a reaction which is only favored under acidic conditions. Think of it like an equilibrium in a way,
under acidic conditions, elemental iodine is favored. Under basic conditions, a mixture of iodide and iodate is favored.
Firstly, the phosphoric acid neutralized the excess sodium hydroxide : 3NaOH + H<sub>3</sub>PO<sub>4</sub> -->
Na<sub>3</sub>PO<sub>4</sub> + 3 H<sub>2</sub>O. Then, the free excess acid lowered the pH of the solution and
caused the following reaction to happen. 5NaI + NaIO<sub>3</sub> + 6H<sub>3</sub>PO<sub>4</sub> -->
3I<sub>2</sub> + 6NaH<sub>2</sub>PO<sub>4</sub> (Not sure if dihydrogen phosphate is a strong enough acid for the
reaction, it's possible that less phosphoric acid is necessary.) I'm slightly surprised that a chemistry teacher wasn't able to figure this out.
EDIT: This thread progressed quickly, there were two replies when I started
[Edited on 1-8-2016 by gdflp] |