> I would think you
>loose some sensitivity with K2PtCl6 , am I right ? The ferrous oxalate you
>form while exposing as a given number of electrons to reduce the platinum
>ions. That number is proportional to the amount of light that you gave to
>the layer. The difference between K2PtCl4 and K2PtCl6 is that they contain
>respectively Pt2+ and Pt4+. To reduce Pt2+ to metallic black platinum, you
>need 2 electrons from the ferrous oxalate. To reduce Pt4+ you need 4
>electrons. So, for a given exposure, you will get twice the black reduced
>platinum with K2PtCl4 vs K2PtCl6.
>Does it match to your experiences ?
I guess I got a little wrapped up in the words here.
No you don't want the K2PtCl6 for pt printing as it is an only slightly
soluble yellow powder. petermarshall@cix.compulink.co.uk (Makes a nice ceramics glaze I'm told.)
It does match my experience. I've been able to get a very faint image with
K2PtCl6 by coating or rather smearing the yellow powder in water onto the
print and into the paper fibers. As it is very slightly soluble in water it
will dissolve out eventually if it is not converted to metal. I think the
question at bay here is wether or not some K2PtCl6 in the image will make
black spots. From your description it nof eeding 4 electrons instead of 2,
points out that it would go in the other direction, perhaps leaving tiny
white spots.
Dick
----------