rstar - 7-10-2012 at 08:58
First of all don't ask me the amount of reactants i've added, i did not measure em out.
Here's wat i did:
1= boiled about half-cup water in the beaker
2= added a chunk of elemental Iodine to it
3= then, i added Sodium bicarbonate to it.
the solution turned red-yellow, i kept on adding NaHCO3 (to make it colourless 
) but it just didnt work.
finally i stopped adding nahco3 , but the final solution was still red. and then, i kept it in a covered testube,
and now (after about 10 hours) i can see some orange yellow stuff crystallizing out on the sides and bottom of testube.
now WTH is that ??
blogfast25 - 7-10-2012 at 09:33
Assuming you were trying to convert I2 to iodide, your base (NaHCO3) was probably not strong enough. You're left withsome I2, some NaI and some NaI3.
querjek - 7-10-2012 at 10:09
Did you use tap water or distilled?
rstar - 8-10-2012 at 00:56
boiled tap water
this is wat i expected:
1= NaHCO3 on heating will decompose to Na2CO3. this appears by the bubbling out of gases(CO2)
2= Na2CO3 +I2 --> NaI + NaIO3 + CO2 (more bubbling)
i hv done this experiment before some months, and was perhaps successful earlier. but this time it gave unexpected results 
[Edited on 8-10-2012 by rstar]
rstar - 8-10-2012 at 05:56
I has been more than 24 hrs now, and i think, a little bit of more stuff crystallized out.
no change in the colour of solution, still bloody red
i think to upload a picture of it after i borrow the better camera from my sister 
woelen - 8-10-2012 at 23:02
You made nothing really special.
NaHCO3 is a very weak base. Your iodine partially reacted.
Part of the NaHCO3 looses CO2 in the hot water:
2HCO3(-) --> CO2 + H2O + CO3(2-)
The CO3(2-) is a somewhat stronger base. It reacts with water:
CO3(2-) + H2O <---> OH(-) + HCO3(-)
Iodine reacts with OH(-):
I2 + 2OH(-) --> I(-) + IO(-) + H2O
More iodine reacts with I(-):
I2 + I(-) --> I3(-), this has a red/brown color in aqueous solution.
On cooling down, Na2CO3 crystallizes and inside these crystals, which mainly consist of Na2CO3.10H2O, small amounts of I3(-) are trapped, giving them
the orange/red color.
[Edited on 9-10-12 by woelen]
rstar - 9-10-2012 at 01:42
Wonder-full
So, that means i havent made enough NaIO3, nd the crystals r not naio3 ?
woelen - 9-10-2012 at 09:51
I assume that you used much more NaHCO3 than I2. So, indeed I expect you mainly made Na2CO3, with some I(-), I3(-) and also IO3(-) trapped in the
crystals.
If you tell more about the amounts, used in the experiments (e.g. specify how many grams or milligrams you used of each of the chemicals), then I can
tell more about your results, but given the observations, for now I assume that you mostly made Na2CO3.
elementcollector1 - 9-10-2012 at 10:32
On a side note, does this really lead to pure Na2CO3? One would think bicarbonate would still be present in significant amounts.