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Chemistry Forums for Students => Inorganic Chemistry Forum => Topic started by: Fmeub on December 09, 2009, 02:32:25 PM

Title: What will the products of these reactions be?
Post by: Fmeub on December 09, 2009, 02:32:25 PM
Hi!

I'm not quite sure whether this should be posted in this forum or the nuclear one.
It seems that if there is a spill of radioactive iodine isotopes, then the contaminated area is sprayed with NaOH or Na3PO4.12H2O solutions. What would the products be?

1) NaOH + I2  :rarrow: ?
 
2) Na3PO4 +  I2  :rarrow: ?

Would it be 2NaI+2OH- for 1)

and PI3 or P2I4 + H2O + NaOH for 2) ? Though the balancing isn't complete.

Thanks!
Title: Re: What will the products of these reactions be?
Post by: Grundalizer on December 13, 2009, 08:43:58 PM
I'd make an educated guess of, NaOH + I2  :rarrow: IO3-

and actually found it on wikipedia 

3 I2 + 6 NaOH → NaIO3 + 5 NaI + 3 H2O

http://en.wikipedia.org/wiki/Sodium_iodate

and the second one produces, I'd guess, http://www.webelements.com/compounds/phosphorus/diphosphorus_tetraiodide.html

P2I4

OR, more likely, the metaperiodate ion , in the form of aqueous sodium metaperiodate

Na+ IO4-

http://www.piercenet.com/Products/Browse.cfm?fldID=02040122

Title: Re: What will the products of these reactions be?
Post by: BluRay on December 14, 2009, 06:55:04 AM

3I2 + 6NaOH → NaIO3 + 5NaI + 3H2O

Ok

Quote

and the second one produces, I'd guess, http://www.webelements.com/compounds/phosphorus/diphosphorus_tetraiodide.html

P2I4

OR, more likely, the metaperiodate ion , in the form of aqueous sodium metaperiodate

Na+ IO4-

Infact I doubt a lot that compounds as P2I4 could form in this case. Na3PO4 is very basic, so I'd say the reaction is analogous to the one with NaOH (iodine dismutates in basic solution at iodide and iodate).