Re: Newbie in lith print
Thank, Loris and Alberto, Ryuji,
This makes more sense, because I got my decimal place wrong. It is 37.5g
paraformaldehyde in a LITER, so that is only 3.75% solution, not 37.5% as I
thought off the top of my head yesterday, so using close to 100ml is
logical. DOH. I should NEVER post to the list in a rush.
Ryuji, I revisited that post of yours and it looks like a good formula in
that the ingredients shouldn't be hard to find. I assume trisodium
phosphate is the same as that at a hardware store or do you think it needs
to be photo grade? The hot is problematic in our student darkroom, though,
as is the short life.
The Fotospeed, btw, was working in 5-15 minutes with my class so speed is
not a problem. But I am intrigued to see the image tones you might get out
of your lith developer as compared to Fotospeed. My guess is image tone is
paper dependent, and would you agree?
I did buy Tim Rudman's lith book and it is gorgeous but have only had time
to look at pictures. But, last week of classes and only one more critique to
Dean Kansky, what is the SUNDRE formula and what are Sundre
----- Original Message -----
From: "Alberto Novo" <email@example.com>
Sent: Friday, December 08, 2006 10:32 AM
Subject: Re: Newbie in lith print
I don't know where you found that page...
Paraformaldehyde is a polymer of formaldehyde: the former is (CH2O)n (CAS
n.30525-89-4), the latter is CH2O. Hence, you can use the weight of
paraformaldehyde like it were pure (100%) formaldehyde, but you will need
to take into account the weight/weight percentage of the solution and its
density in order to calculete the corresponding liquid amount.
As an example, if 37.5 g of paraformaldehyde are required and you have a
37% (w/w) formaldehyde solution @ 1.09 kg/dm3 (this should be clearly
written in the label), you will need 37.5x100/37x1/1.09= 99 cc of
I found a page saying "...A 10% formalin solution should be equivalent
to a 4% paraformaldehyde (by weight) solution...". Concentrated /
saturated formalin is 37% AFAIK, not 40%. Anyway, I'll do the math: Since
it says 100g 4% paraformaldehyde (4g paraformaldehyde) solution is
equivalent to 100g 10% formalin (10g formaldehyde) then you have to use
2.5 units of formaldehyde per 1 unit of paraformaldehyde (10 / 4 = 2.5).
Since Christina's formula calls for 37.5g paraformaldehyde, you have to
use 37.5 x 2.5 = 93.75g formaldehyde. In 37% solution terms this should
make 93.75 / 0.37 ~= 253ml 37% formalin, *if my logic is correct*... I
don't know if the information I got from that page is correct and most
importantly I'm not sure if my logic is correct. Therefore don't rely on
this info ;) Regards,