From: Erich Camerling (e.camerling@freeler.nl)
Date: 02/12/03-05:55:57 AM Z
The best way to reduce K2(PtCl6) (yellow) to K2(PtCl4) (red) is with K2C2O4
(potassium oxalate) according to the GMELIN.
You don't need SO2 (suffocating) and you have less risk to reduce K2(PtCl6)
too far.
K2(PtCl6) + K2C2O4 --> K2(PtCl4) +2 CO2 + 2 KCl
Start with 1000 ml H2O (dest).Solve 37 g K2C2O4.H2O in it and add 100 g
K2(PtCl6).Stir well.Heat till boiling.
The dissolved K2C2O4 reacts with the dissolved part of K2(PtCl6).Keep just
boiling for some hours till the solution is deep-red and no visible yellow
crystals remain.
Because of the reducing by evaporation it is possible you will already see
deep-red crystals.Let the liquid cool down and filter.The red crystals are
K2(PtCl4). Output ~ 80 %.
Prudent (slow) reducing by evaporation of the liquid can give you ~ 10 %
more K2(PtCl4) but there will be some KCl in it.Solve the crystals in the
smallest quantity of boiling water (dest) and cool down.Filter again and you
will have pure K2(PtCl4).Please let me know the results when used in
Platinotype.
Because of the strange sue culture in the USA :
I denie liability for any consequent damage,injury or loss resulting from
the use of the information contained in this e-mail.
Kind regards
Erich
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