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