RE: Lithium palladium vs. "normal" palladium
Loris, Your numbers are good and in many cases the small amount of volume change is not significant, especially in this endeavor, but it is just a correct statement to say "to make" rather than "in". You'll see that Mike uses the term, about 20ml, which is fine. Once again here we are talking about photochemistry and not life saving titrations of some drug. The small volume change that will happen at that small of a total solution volume is not significant. I try and make my batches no smaller that 40ml at a time. I have been using my Li, Na, NH4, palladium double salts at those concentrations for many years now. I have not ever mixed my solution of Lithium Palladium to the directions in Sullivan's text. With Lithium Chloride, the problem is that it is extremely deliquessant (sp) and it absorbs moisture from the air to the point of becoming a liquid. So to ensure that you have enough actual Li+ and Cl- ions, it would be better to error on the side of additional weight. Cheers Eric 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: Loris Medici [mailto:mail@loris.medici.name] > Sent: Wednesday, January 17, 2007 11:55 PM > To: alt-photo-process-l@usask.ca > Subject: RE: Lithium palladium vs. "normal" palladium > > Hi Eric, > > "molar solution" > Well, since I had my chemistry classes in Turkish, I may not use the > correct English terms. Please read my message again by reply "x in y" > expression with "x + water to make y" (because this is what I do when I > mix my solutions) and tell me if there's any mistake and/or am I right > in my confusion. Will look at the archives BTW... > > "knowledgeable person" > My calculations and the results of them matches the calculations/results > of a person you'll admit is very knowledgeable; see how M. Ware mixes > his metal solution in his article > http://www.mikeware.co.uk/mikeware/Platino-Palladiotype.html: > > "The palladium solution (Pd) > ... > > To make up: > > 1. Dissolve 1.8 g Ammonium Chloride in about 20 cc of hot (70-80 °C) > distilled water. > 2. Add 3 g Palladium(II) Chloride with stirring (which should be > well-powdered. HAZARD: wear a dust mask) > > ..." > > If you put 3g in place of 25g for PdCl2 in my calculations below, you'll > find that indeed you need 1.8g (NH4)Cl to make the (NH4)Pd double salt > solution by using 3g PdCl2. Probably there's something with Li which > necessitates using 50% in excess -> but this is surely confusing when > you look the process equations (because they don't tell us we'll need > excess salt!?). > > "misread" > I don't misread Jeffrey, I just compare the figures I found (my making > the calculations below) with the figures given by his "Metal Solution > Formula Calculator" using the same sensitizer strength, Pd double salt > and target volume... See for yourself. > > "rich print" > Well, I guess someone (including myself) has to mix the solutions > according to the calculations below and share their finding with the > list later. (I mean to report if their prints are as rich as they were > before...) > > Regards, > Loris. > > -----Original Message----- > From: EJN Photo [mailto:ejnphoto@sbcglobal.net] > Sent: 18 Ocak 2007 Perşembe 05:35 > To: alt-photo-process-l@usask.ca > Subject: RE: Lithium palladium vs. "normal" palladium > > > Loris, First a molar solution is not "in" but "to make". It would be a > normal solution not a molar solution. > > Your confusion about the information in TNPP book, was documented in the > archives. Why B&S/Sullivan stayed with that number I don't recall. And > any time someone starts writing down numbers and formula, it is always a > good idea to have another knowledgeable person read it, and read it > again because it is so easy to misplace a letter or period and make a > formula mistake. > > I also think that you are adding to the confusion by bringing > ingredients into the discussion in forms that are not being used. We > don't, at least I don't, typically mix our double salts from solutions > mixed to molar concentrations but by adding solids to water. i.e. X > grams of A chloride with Y grams of Palladium Chloride with water to > make Z ml of solution. It is at this point that we have our solution > hopefully made to our targeted molar concentration. > > AS to confusion #1, If Jeffrey wrote it down that way, it is wrong. I > don't recall his numbers being that far off in any correspondence that I > had with him several years back when he was working through those > numbers. It could also be that you misread Jeffrey? > > The idea behind his work was to get a solution that wasted as little > expensive metal salt as possible and to produce the richest print as > well. Where richest is not money, but tonally rich. > > Skype : ejprinter> -----Original Message----- > > From: Loris Medici [mailto:mail@loris.medici.name] > > Sent: Wednesday, January 17, 2007 2:22 PM > > To: alt-photo-process-l@usask.ca > > Subject: RE: Lithium palladium vs. "normal" palladium > > > > Hi Witho, > > > > Thanks for bringing this up. I did some calculations and I'm quite > > confused right now. See below: > > > > LiCl (Lithium Cloride) 1 Mole = 42.394g, > > 1.4M solution = 59.3516g in 1000ml (5.93%) > > > > (NH4)Cl (Ammonium Chloride) 1 Mole = 53.4912g, > > 1.4M solution = 74.88768g in 1000ml (7.48%) > > > > PdCl2 (Palladium(II) Chloride) 1 Mole = 177.33g, > > 0.7M solution = 124.131g in 1000ml (12.41%) > > > > Li2[PdCl4] (Lithium double salt of Pd) 1 Mole = 262.11g, > > 0.7M solution = 183.477g in 1000ml (18.34%) > > > > (NH4)[PdCl4] (Ammonium double salt of Pd) 1 Mole = 284.31g, > > 0.7M solution = 199.017g in 1000ml (19.9%) > > > > (NH4)3[Fe(C2O4)3].3H2O (AFO) 1 Mole = 428.067g, > > 1.4M solution = 599.2938g in 1000ml (59.92% - say 60%) > > > > 2 Mole salt + 1 Mole PdCl2 -> 1 Mole Pd double salt. Therefore > > > > 2 LiCl + 1 PdCl2 -> 1 Li2[PdCl4] > > 2 (NH4)Cl + 1 PdCl2 -> 1 (NH4)2[PdCl4] > > > > If you compare the atom counts on each side you'll see that the > > equations are correct. > > > > To Camden: > > > > Let's assume you have 25g PdCl2 (current special price for this amnt. > > for APUG.org members is $262.50 (+ 135g Ferric Oxalate for free > > - > > kudos B&S). .. > > > > You can make 200ml (201.45 actually) 0.7M Li2[PdCl4] solution with > > this amnt. of PdCl2 (25g / 12.41% - see info provided for > > PdCl2 above.) > > > > You'll need > > > > 25 (1) / 177.33 (2) * 2 (3) * 42.394 (4) = 11.95g LiCl. > > > > or > > > > 25 (1) / 177.33 (2) * 2 (3) * 53.4912 (5) = 15.08g (NH4)Cl > > > > Notes: > > (1) Amnt. of PdCl2 you have > > (2) Molar Mass of PdCl2 > > (3) 2:1 salt to PdCl2 ratio > > (4) Molar Mass of LiCl > > (5) Molar Mass of (NH4)Cl > > > > Calculate your costs according to the info I gave you above. >
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