Taslem - 23-1-2012 at 19:53
I was making iodine the HCl, alkali metal iodide way using .1mol solutions and 3% H2O2. It gave a very dilute iodine solution. When I heated it, it
turned clear and looked no different from when I first added the HCl and KI. Does anyone know what happened to the iodine?
neptunium - 23-1-2012 at 20:51
it got disolved....let it cool and add more peroxide..the reaction is exothermic so keep it slow and cool...by the end of it you should be left with a
clear solution with a precipitate of I2
weiming1998 - 23-1-2012 at 20:56
It also might have sublimed. Iodine is very easy to sublime, so be careful when heating. If you notice a faintly purple vapour, then it is subliming.
neptunium - 24-1-2012 at 15:21
if you heat it too much it could very much have sublimed very true!
AJKOER - 24-1-2012 at 16:39
OK, the reaction from start to end and the future:
HCl + KI --> KCl + HI
2 HI + H2O2 --> I2 + 2 H2O
Now, Iodine slowly reacts with water. However, with boiling the reaction proceeds:
I2 + H2O <---> HI + HIO
and very rapidly a disproportionation reaction:
3 HIO --> 2 HI + HIO3
So, the total reaction:
3 I2 + 3 H2O --> 5 HI + HIO3
Now, both HI and HIO3 are colorless in solution, hence the loss of color. However, with exposure to air (meaning O2), per Wikipedia:
"HI will undergo oxidation if left open to air according to the following pathway:[4]
4 HI + O2 → 2H2O + 2 I2
HI + I2 → HI3
HI3 is dark brown in color, which makes aged solutions of HI often appear dark brown."
[Edited on 25-1-2012 by AJKOER]