wafflotron
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Sodium Iodide and Phosphoric Acid
So, first I should give a little backstory. Over the summer, I attempted to make Sodium Iodide with crystalline Iodine and Sodium Hydroxide. I had
done this before on a small scale with a bottle of Iodine tincture. So, at the time I had just finished first year chem (I'm taking AP right now in
High School) and wasn't really thinking about what I was doing. Rather than trusting the bottle which said there was 15g of Iodine, I went to weigh it
on a weighing tin. Well, my weighing tins are the disposable Aluminum ones, so you guys can probably guess what happened. After a brief panic and my
hurriedly dumping it out on suran-wrap (The only thing I had nearby) I added my Sodium Hydroxide, which I had calculated the stoichiometric amount
needed based off the 15g. What resulted was a majoratively Sodium Iodide sludge mixed with a bit of Sodium Hydroxide.
Now, fast forward seven months and I decide that with all the chemistry knowledge I've gained over the year I'll try and separate the Sodium Iodide
and Hydroxide. My first thought went to solubility, and I decided to make a saturated solution and then recrystalize, as according to Wikipedia Sodium
Iodide has three times the solubility of Sodium Hydroxide at 0C. The next day, I went to filter off my waste water, which should have contained
nothing but Sodium Hydroxide by my logic. In order to test that this was indeed true, I added a few drops of phosphoric acid to two drops of my
solution and had my infra-red thermometer at the ready in order to record any heat fluctuations, as acids and bases tend to do just that. What
happened next was not at all what I had expected. Very slowly, far from spontaneously, purple began to appear in the drops. Quickly following was what
was unmistakably Iodine vapor, and a few seconds later there was a small Iodine explosion, splattering Iodine on the side of my beaker. Since my
attempt at recovering my NaI was so obviously a failure, I figured, "Why not just recover the Iodine?" A little more Internet research however
revealed that this happens when you add H3PO4 to NaI. NaI + H3PO4 ----> NaH2PO4 + HI. (I checked other sources, and while they couldnt agree on
which Sodium-Phosphate was made, they all agreed on the HI) I asked my AP chem teacher and he too was baffled. Today I went and reacted the remainder
of my NaI/NaOH with Phosphoric acid, and several more unaccounted for events occurred. First, once the initial reaction died down, bubbles could be
seen floating to the top of my very purple/brown solution. Secondly, whenever I swirled the beaker and exposed the bottom before re-covering it with
solution the material on the bottom would float to the top, where it would then react with the solution again. It appeared to be Iodine, but at this
point I really can't be sure. When I went to top off my filtration (I was filtering out the solid iodine that formed) the solution from the beaker
reacted with the solution in my funnel, creating a plume of orange/brown and bubbles that at first had me thinking it was boiling. My chem teacher was
as confused as I was, and so I came here in the hopes that someone could explain what happened to me. I also looked up HI, but there is no mention of
its decomposition temperature or with or without a catalyst. Any insight would be greatly appreciated, and if there's any critical data I left out let
me know and I'll state my observations as soon as I can.
I know it can't be a reaction involving all three compounds, as I attempted to balance all three possibilities (for NaH2PO4, Na2HPO4, and Na3HPO4) and
none worked out.
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JJay
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Why would elemental iodine and sodium hydroxide react to make sodium iodide?
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Etaoin Shrdlu
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Why wouldn't they? Shouldn't this result in iodide and hypoiodate/iodate?
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JJay
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The sludge may have worked as a catalyst, but that doesn't mean it didn't contain any sodium iodate or other oxygen-sodium-iodine compounds. Those
would react with hydrogen iodide (created by the reaction of phosphoric acid and sodium iodide) to produce water and iodine.
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Etaoin Shrdlu
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Quote: Originally posted by wafflotron  | | the reaction actually forms Sodium Iodate, NaIO3, which then decomposes into NaI + O2 according to my 1st year chemistry teacher.
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Your first year chemistry teacher was a little lost here. Why would it decompose?
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wafflotron
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It actually doesn't. the reaction actually forms Sodium Iodate, NaIO3, which then decomposes into NaI + O2 according to my 1st year chemistry teacher.
My NaI + NaOH sludge worked as a catalyst for the decomposition of H2O2, and so far as I can research the Iodate ion does not do this. Hopefully this
is confirmation of my theory. Unfortuntely I cannot go ask my 1st year chemistry teacher (Who actually has his Masters in chemistry) as he got a new
job. My current teacher, by contrast, is an Evolutionary Biologist by trade, though hes been teaching basic chemistry for 12 years now and AP for 2.
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wafflotron
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He said that boiling the solution (Which I did to recover the solid salt) would be hot enough to decompose it. Unless the IO3 ion also acts as a
catalyst for H2O2 -----> H2 + O2 he was correct.
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gdflp
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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]
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Etaoin Shrdlu
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| Quote: Originally posted by wafflotron |
He said that boiling the solution (Which I did to recover the solid salt) would be hot enough to decompose it. Unless the IO3 ion also acts as a
catalyst for H2O2 -----> H2 + O2 he was correct. |
Oh, never mind then, he was amazingly lost. Give me a minute to pull together all the relevant reactions here.
EDIT: Never mind, gdflp got you. 
[Edited on 1-8-2016 by Etaoin Shrdlu]
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wafflotron
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| Quote: Originally posted by gdflp |
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] |
Thank you so much! This really clears up a lot. 
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JJay
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| Quote: Originally posted by gdflp |
It's a well known and standard reaction : 3I<sub>2</sub> + 6NaOH --> 3H<sub>2</sub>O + 5NaI + NaIO<sub>3</sub>
[Edited on 1-8-2016 by gdflp] |
That's interesting.... Iodine doesn't like to form hypoiodate ions.
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Etaoin Shrdlu
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| Quote: Originally posted by JJay | | Quote: Originally posted by gdflp |
It's a well known and standard reaction : 3I<sub>2</sub> + 6NaOH --> 3H<sub>2</sub>O + 5NaI + NaIO<sub>3</sub>
[Edited on 1-8-2016 by gdflp] |
That's interesting.... Iodine doesn't like to form hypoiodate ions. |
That's why it disproportionates to iodide and iodate.
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UC235
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It certainly does form hypoiodite. But like the other hypohalites, it is unstable on standing and when heated. Hypoiodite is the active species in the
iodoform test and is prepared immediately before use.
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