[Fmeub]

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   ?
 
2) Na₃PO₄ +  I₂   ?

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

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

Thanks!

[Grundalizer]

I'd make an educated guess of, NaOH + I₂  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

[BluRay]

3I₂ + 6NaOH → NaIO₃ + 5NaI + 3H₂O

Ok

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).