RE: Lithium palladium vs. "normal" palladium
Witho, So you'd say you were getting a longer richer tonal scale using 60%? Eric Neilsen Photography 4101 Commerce Street Suite 9 Dallas, TX 75226 http://e.neilsen.home.att.net http://ericneilsenphotography.com Skype ejprinter > -----Original Message----- > From: Witho Worms [mailto:info@witho.nl] > Sent: Thursday, January 18, 2007 9:09 AM > To: alt-photo-process-l@usask.ca > Subject: RE: Lithium palladium vs. "normal" palladium > > Loris, > > I made a test using 40 and 60% Afo in the 2:1 ratio with LiPd. I had a > 0.20 > better Dmax using 60%. Dmax and tonal scale for LiPd and Amonium Pd were > the > same with 60% (LiPd used with excess of water). > > Witho > > -----Oorspronkelijk bericht----- > Van: Loris Medici [mailto:mail@loris.medici.name] > Verzonden: donderdag 18 januari 2007 15:11 > Aan: alt-photo-process-l@usask.ca > Onderwerp: RE: Lithium palladium vs. "normal" palladium > > Hi Camden, > > Let me remind you that the sample calculation I did for you was for > 60% AFO sensitizer. The norm is to use around 40-45%. Therefore the > amnt. of corresponding Li2[PdCl4] solution will increase if you use > 45% AFO instead of 60%. > > 45% AFO = 450g + water to make 1000ml > 450 / 428.067 = 1.05M > > Again, since molar ratio of AFO to Li2[PdCl4] should be 2:1 then you > need 1.05 / 2 = 0.52M Li2[PdCl4]. > > Remember that 2 Mole LiCl + 1 Mole PdCl2 gives 1 Mole Li2[PdCl4]. This > means: > > (a) 0.52 * 177.33 (1) = 92.2116 + water to make 1000ml = 9.22% > (b) 25g / 9.22% ~= 270ml > > Notes: > (1) Molar mass of PdCl2 > (a) Finding the solution % of 0.52M PdCl2 > (b) Interpolating the % found in (a) to calculate the final 0.52M > Li2[PdCl4] solution volume. > > In short, if you have 25g PdCl2 and > > a) you're going to use 60% AFO, then you can mix ~200ml of LiPd metal > solution > b) you're going to use 45% AFO, then you can mix ~271ml of LiPd metal > solution > > Since I use 40% AFO with good results (ask Sandy), there's no need to > go for 60%. > > Hope this helps (and is good news to you), > Regards, > Loris. > > P.S. Refer to my first message to find out how much LiCl and/or > (NH4)Cl you need. The amounts won't change since your PdCl2 amnt. > hasn't changed. You're just changing the final solution volume / the > water you add... > > > ---------- Quoting Camden Hardy ---------- > > Loris, > > Thanks for spelling this out for me. Most of this chemistry talk is a > little over my head (I should have paid more attention in those chemistry > classes...), but seeing some numbers in there really helped. :) > > I think my confusion on this matter is finally starting to clear up (we'll > see how long this "clarity" lasts, hehe). Thanks to those who attempted > to explain this stuff to me, on- and off-list.
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