In Bostick-Sullivan formulary pages, it is stated that Lithium Palladium
Chloride (Li2PdCl4) solution is made following the formula below:
Lithium Chloride 1.7gr (LiCl2, molecular weight = 77.8464)
+
Palladium chloride 2.3gr (PdCl2, molecular weight = 177.3254)
+
Water to make 25ml
What is the exact chemical reaction equation for making Li2PdCl4?
Is it:
2(LiCl2) + PdCl2 -> Li2PdCl4 + Cl2
or something else? Does water enter in the equation? If yes, how?
The mol value for 1.7gr LiCl2 is -> 1.7 / 77.8464 = 0.0218378... When
you mutiply this value with molecular weight of PdCl2 then you find that
you have to add around 3.9 gr of PdCl2 - Which obviously is not the
case. So, I assume my equation above is incorrect. Can you please
provide me the correct equation formula? I need this in order to mix
fresh Ammonium Iron(III) Oxalate sensitizer of right strength (I don't
want to waste Pd by preparing a weaker AFO sensitizer than what is
needed). Or maybe I shouldn't bother with all this and prepare a 60% AFO
solution (I guess this is almost a saturated solution)... What will be
the effects of preparing a sensitizer of extra strength?
Thanks in advance,
Loris.
Received on Fri Jun 24 06:26:21 2005
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