Sciencemadness Discussion Board

Iodine did not precipitate from Potassium Iodide? (HCL+ H2O2)

LuckyWinner - 14-11-2020 at 05:02

I have a problem...
EDIT:

the first time I did this reaction i followd nurdrage...
his youtube vid is just eyeballing amounts...

I added an unknown weak strength
of HCL %, probably very weak to a potassium iodide water solution,
also diluted in lots of water. (way more then minimum)

then 3% H2O2 was added more then tripple the recommended amount.

this solution turned dark red but did not precipitate out any Iodine.
it is fizzeling like sparkling water and stops doing that after more then 45 min.


what did happen?


I repeated this process with a scale and exact measurements of reagents
and it worked perfectly.
source
https://sciencing.com/extract-iodine-potassium-iodide-564002...

but how can I rescue this first experiment?



[Edited on 14-11-2020 by LuckyWinner]

LuckyWinner - 14-11-2020 at 05:13

is this the answer

BUT the difference is that I did not use too much of a concentration but too much of my low
concentrated H2O2
Quote:



You used way too concentrated reagents. What happens is that excess H2O2 oxidizes HCl to Cl2 and water, the Cl2 in turn reacts with iodine, producing IO3(-) and ICl4(-) ions and you lose all of your iodine to highly oxidized species.

Try again with 3% H2O2 and 10% HCl. Dissolve your KI in 10% HCl.

The reaction you want is 2I(-) + 2H(+) + H2O2 --> I2 + 2H2O

In practice, assure that you have quite some excess of acid. E.g. 2 molecules of HCl for each KI is OK and non-critical. A little more or a little less HCl is no problem.
Also, use some excess H2O2. At these low concentrations, the H2O2 does not oxidize the HCl (or very slowly), while it immediately oxidizes the iodine. By using excess H2O2 you assure that all iodide is converted to iodine. Without any iodide ions present, the solubility of iodine is very low and nearly 100% of the iodine precipitates.



http://www.sciencemadness.org/talk/viewthread.php?tid=65425#...


how can these IO3(-) and ICl4(-) be converted to Iodine or sodium iodide?

IO3(-) and ICl4 will react with sodium thiosulfate to
NaIO3, Sodium iodate
NaICl4
?

[Edited on 14-11-2020 by LuckyWinner]

highpower48 - 14-11-2020 at 07:48

Two things...1st are you positive its Potassium iodine
2nd use concentrated hydrochloric acid

LuckyWinner - 14-11-2020 at 08:17

Quote: Originally posted by highpower48  
Two things...1st are you positive its Potassium iodine
2nd use concentrated hydrochloric acid


updated my first post.

the question now is how can I rescue my first attempt of this reaction.
I added waaaay to much H2O2.

how can I get NaI or Iodine out of this dark red H202 overoxidized mess?

ArbuzToWoda - 14-11-2020 at 08:21

I think an excess of some reducing agent (sodium thiosulfate or bisulfite) would suffice. Then quench the rest of your chosen salt, add a bit more water so that no precipitation occurs and try again. I don't think there are any good ways to isolate the iodide from the sulfate byproducts, but please correct me if I'm just being unimaginative.

LuckyWinner - 14-11-2020 at 09:22

Quote: Originally posted by ArbuzToWoda  
I think an excess of some reducing agent (sodium thiosulfate or bisulfite) would suffice. Then quench the rest of your chosen salt, add a bit more water so that no precipitation occurs and try again. I don't think there are any good ways to isolate the iodide from the sulfate byproducts, but please correct me if I'm just being unimaginative.



cause i overoxidezed i have now
IO3(-) and ICl4 floating around
this will react with sodium thiosulfate to
NaIO3, Sodium iodate
NaICl4

adding water will convert these 2 Na salts back into elemental iodine in H2O2 precence?

I just want to recover Iodine or KI
and do it the proper way.

clearly_not_atara - 14-11-2020 at 13:35

Use H2SO4/H3PO4, since these do not form complexes with I+ or I3+ ions. It may be possible to recover iodine as calcium iodate by adding CaCl2 (or Ca(OAc)2) and plenty of extra H2O2; Ca(IO3)2 has a low water solubility and will precipitate.

Ca(IO3)2 can be converted to the sodium or potassium iodate by metathesis with Na2CO3 or K2CO3 respectively. These may be used to comproportionate with the iodide salts and give iodine by acidification preferably not with HCl.

If necessary the alkali metal iodates decompose to the iodides on heating. In the case of sodium iodate a small amount of sodium oxide is usually produced as well.