opfromthestart - 26-12-2018 at 09:39
I was trying to purify iodine from a 10% povidone iodine solution and I thought that bleach could oxidize the povidone without reducing the iodine.
When I did the reaction, the solution quickly became nearly clear. I spent some time trying to figure out the reaction and I had come up with:
2NaClO + I2 + H2O -> NaI + NaOI + 2HClO
I then tested this equation by placing 10mL of povidone iodine in a beaker and dripping NaClO into it, and stirring after every 2-3 drops. I expected
the solution to turn clear after .9 mL of bleach, but the color persisted until I added 2.75 mL of bleach, which corresponds to a mole ratio of .0004
mol I2 to .0022 mol NaClO, which is probably 5:1, 11:2, or 6:1 NaClO to I2.
My bleach was purchased one month ago and my iodine says it is 1% avaliable iodine. What would the corrsponding reaction be? Or did I do the reaction
incorrectly?
CouchHatter - 26-12-2018 at 09:48
Household bleach is not always close to the labeled concentration. Did you titrate it?
opfromthestart - 26-12-2018 at 13:03
Is there any way to titrate it using the chemicals in https://www.sciencemadness.org/whisper/viewthread.php?tid=11... along with 5% vinegar and sodium bisulfate?
TGSpecialist1 - 27-12-2018 at 10:48
There is a reaction: NaClO + H2O2 = O2 + NaCl + H2O
You could measure the volume of oxygen produced.
Sigmatropic - 28-12-2018 at 01:32
Since hypochlorite can oxidise iodate to periodate, I am not at all surprised you found more hypochlorite was needed.
The equation you provided is incorrect, it is actually the sum of a disproportionation and an oxidation.
2 NaOH + I2 --> NaI + NaOI + H2O (the disproportionation)
And
NaI + NaOCl --> NaOI + NaCl (oxidation)
Which combine to make
NaOCl+ 2 NaOH + I2 --> 2 NaOI + NaCl + H2O
This would suggest a 1:1 stoichiometry between hypochlorite and iodine, but since the oxidation of iodine doest stop there the following reactions
consume hypochlorite, which I believe explain why more bleach was needed.
NaOI + 2 NaOCl --> NaIO3 + 2 NaCl (oxidation to iodate)
Or
NaOI + 3 NaOCl --> NaIO4 + 3NaCl (oxidation to periodate)