LuckyWinner
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Registered: 27-8-2018
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How to Separate NaOH Impurities? (NaI, Sodium Iodide)
I made some NaI by reacting Iodine with NaOH.
molar equivalents were used with a bit excess of iodine.
the NaI+ NaIO3 is placed inside a furnace at 550C
to convert the NaIO3 to NaI.
this yields a solid white clean crystal mass.
it is grinded into fine power then dissolved in acetone
to filter out the NaOH. (should be insoluable in acetone according to
http://chemister.ru/Database/properties-en.php?dbid=1&id...
(NaOH is soluable in ethanol ethanol: 17.3g/100ml (28°C) [Ref.])
now whenever the NaI is dissolved in the acetone and
coffee filtered, evaporated Im left with some weird smelling
slightly discolored brownish mass.
did somehow NaOH get dissolved in the acetone that was
dried over calcium sulfate for several days?
I guess the acetone self aldol codensated and formed isophorone and
a gazillion other side reactions....
is there any way to get that NaOH out, it seems like I always
get this brown funk after extracting with my double distilled dry acetone...
even when using an excess of Iodine..
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Maurice VD 37
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Try to add ammonium chloride NH4Cl. If there is NaOH in your salt, it will produce NaCl and NH3 which is volatile. If there is a excess of NH4Cl, it
will soon vaporize by heating at much less than 500°C. And NaCl is not soluble in acetone. So you should get pure sodium iodide.
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LuckyWinner
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Quote: Originally posted by Maurice VD 37  | | Try to add ammonium chloride NH4Cl. If there is NaOH in your salt, it will produce NaCl and NH3 which is volatile. If there is a excess of NH4Cl, it
will soon vaporize by heating at much less than 500°C. And NaCl is not soluble in acetone. So you should get pure sodium iodide.
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thanks.
but there is no risk of
Nitrogen triiodide formation if there is some unreacted iodine in solution?
if some is produced it will be insoluble in water so best thing to do is
to filter the NaI water solution after adding the NH4Cl
or just heat the aqueous solution to boil it off.
Boiling point: sublimes at −20 °C
Solubility in water: Insoluble
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Maurice VD 37
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If you heat the mixture NaI + NaIO3 at 500°C, NaIO3 will be transformed into NaI, but any amount of remaining iodine I2 will be vaporized much before
this temperature. S the risk of reacting later on with ammonia is zero.
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