PHJchem: Iodine Adventures

(solution below!) I'm performing a few experiments with element 53; iodine. Physical appearance, solubility and complexes are the main subjects.

Derkozten, you've got the right answer to the question why adding bleach decolorizes the lower layer:
Since iodine will only dissolve in water when iodide ions is present, oxidizing these iodide ions with chlorine (2I(-) + Cl2 gives 2Cl(-) + I2) will result in a total migration of iodine to the organic layer.

Category:

Science & Technology

Tags:

• iodine
• experiment
• peter
• jacobse
• iodide
• triiodide
• starch
• complex
• gasoline
• halogen
• chemistry
• reaction

[kaviaari]

Iodine dissolves in pure water, not very much but it does. I'm making a new ChemCorner to show an interesting experiment with iodine.

[starshock01]

drop it in some alcohol

[phj1990]

Thanks friend,

I've updated the video description.

Peter

[derkozten]

Oh, I see... chlorine bleach consumes the iodide ion (I-).

Without iodide ions in the aqueous layer, all the elemental iodine migrates to the organic (gasoline) layer.

Hey, thanks for the hint XD

[phj1990]

Good thinking, derkozten,

but it's far more simple than that!

What chemical keeps iodine in the aqueous layer?

Peter Jacobse

[derkozten]

I think hypochlorite (ClO-) reacts with iodine as:

2ClO(-) + I2 --> Cl2 + 2IO(-)

Another plausible product is iodate (IO3-). As chlorine is generated, it also migrates towards the gasoline (organic) phase, leaving only the colorless iodine oxoanion in the aqueous phase.

Again, it's only my assumption. I could be wrong.

[IronMarshmallow]

Excellent. That'll work.

For a minute, I thought you meant Iodine salt as in the common table salt that has been mixed with small amounts of iodine in an attempt to supply people with iodine in their diets, but now I see you actually meant an iodine salt.

Hehe. Thanks.

[phj1990]

IronMarshmallow,

100 grams of potassium iodide costs about €12 in NL, so that's €12 for 70 g of I2, or $10 for 40 g.

Chemical regards,

Peter Jacobse

[phj1990]

The math is fairly easy:

Suppose you have 100 g of potassium iodide.

A(K) = 39.10 u; A(I) = 126.9 u

So M(KI) = 39.10 + 126.9 = 166 u = 166 g/mole.

n(KI) = n(I) = 100/166 = 0.602 moles of iodine.

This is 0.602 x 126.9 = 76.4 g of I2.

I guess you can achieve a yield of about 90%, which is .90 x 76.4 = ca. 70 grams of I2.

Peter Jacobse

[IronMarshmallow]

How much Iodine can be extracted from a normal sized canister of Iodine Salt?