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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