It's pretty simple stuff, but I'm stumped about assaying the amount of iodine I2 in a solution.

The first reaction was titrating sodium thiosulfate Na2S2O3 into a solution containing KIO3, KI and Hcl acid, so...

6S2O3(2-) +IO3(-) +6H(+) ---> 3H2O + I(-) +3S4O6(2-)

Anyway, the KIO3 was the standard, standardized the Na2S2O3 through titration.

Then we're given an unknown solution, that we just titrate with Na2S2O3 till end point. It was to know the mass of I2 in the unknown solution.

I basically need the number of moles between I2 and Na2S2O3... the only ratio I can think of (from the information I got) is where you add KI and Hcl acid to the KIO3 solution to liberate the iodine.

IO3(-) + 5I(-) + 6H(+) ---> 3I2 + 3H20

If the unknown solution was indeed the same as the solution I made... would the ratio of 6/3 be the figure I want to use in my calculations to find the mass of iodine.

Thanks in advance.

also, anyone with chemistry knowledge?

ive got all these cold and flu tablets i need borken down, coz ive got a really, really, really bad flu (honest)

ive got all these cold and flu tablets i need borken down, coz ive got a really, really, really bad flu (honest)

First, you need some road flares...

christ thats all coming back to me

i did this all about 1.5years ago, i woulda been able to do it for ya then.

i might have a think about it later, mentally drained atm

haha holy s*** nostalagia 2 the max..

cant remember at all tho. all i remember is titration was with those long f***en tube things and u had to do it just until the liquid color changed!

sup VHA3 in chemistry wasted to a life of binge drinking / drug abuse

mmmmm tit ration, sign me up for a daily dose please!

making speed is bad mmkay

I could be way wrong but lets guess:

Because the potassium iodate solution you make has no free iodine you add the potasium iodine and excess H+ to liberate iodine and for this you would use your initial equation.

The unknown has liberated iodine, which is what you are trying to find the concentration of. In that case the reaction between iodine and the sulfate ion would be I2 + 2S2O3(2-) -> 2I(-) + S4O6(2-) this is what gives you the stoichiometric coefficients for your unknown.

last edited by qmass at 23:41:00 09/May/07

http://metropower.co.uk/forum/forums/get-attachment.asp?attachmentid=15027

The unkown has liberated iodine, which is what you are trying to find the concentration of. In that case you are using the 6:1 ratio.

Though you said...

I2 + 2S2O3(2-) -> 2I(-) + S4O6(2-)

If the unknown solution was just liberated iodine... then you only need 2 moles of sodium thiosulfate to each mole of iodine, 2:1?

* Also we used Potassium Iodate as the standardizing agent.. that was 6:1


last edited by CHUB at 21:49:25 09/May/07

everyone who did chemistry is a fag imo

The unkown has liberated iodine, which is what you are trying to find the concentration of. In that case you are using the 6:1 ratio.

Though you said...

I2 + 2S2O3(2-) -> 2I(-) + S4O6(2-)

If the unknown solution was just liberated iodine... then you only need 2 moles of sodium thiosulfate to each mole of iodine, 2:1?

* Also we used Potassium Iodate as the standardizing agent.. that was 6:1

Yes, I had it around the wrong way cause my post reply box is too small and I had them mixed up.

The unkown has free iodine and reacts with the thiosulfate (hence I2 + 2S2O3)

You initially standardised your Sodium thiosulfate with free iodine that was liberated when you added the potasium iodine (KI) to the potasium iodate (KIO3) and H(+) (from HCL) prior to each titration. When the KI is added the solution turns brown becuase iodine has been liberated. The reaction is given by 6S2O3(2-) + IO3(-) + 6H(+) -> etc (net ionic equation, the spectator ions are left out)

So 6:1 for the standardisation of your thiosulfate and 2:1 for your determination of iodine in the unknown.

last edited by qmass at 23:43:35 09/May/07

just incase ppl browsing these forums didn't think we were nerds...

Faaaaaaark, all my chemestry knowledge from 5 years ago came back to haunt me.

right

Right on qmass! The reaction makes perfect sense now.

Cheers!