[alt-photo] Re: DESICCATE! DESICCATE! DANCE TO THE MUSIC!

[Loris Medici]

Hi David,

Let me give you an example to let you understand:

You have a formula which needs that you prepare 100ml of 40% citric acid
solution. (If not specified, that would mean 40g "anhydrous" - see below -
citric acid in total 100ml solution.)

You go to the chemistry store to find out they don't have anhydrous citric
acid but citric acid monohydrate only - which means the compound has an
additional one (= mono) molecule of water (= hydrate) in it. (Eventually,
anhydrous means that it doesn't have any water.) Anyway, you can't find any
other supplier that has anhydrous citric acid in stock. In our case, we can
actually substitute monohydrate in place of anhydrous (because we're already
going to dissolve it in water, therefore the extra water content will not
hurt our purposes), but how?

You seach for "citric acid" in google and find this:
http://en.wikipedia.org/wiki/Citric_acid
Somewhere in the page it says, Molar mass = 192.124g / mol (anhydrous),
210.14g / mol (monohydrate)

Or, if the page doesn't give information about the molar mass then you look
at the formula (C6H8O7), note it, and seach "molar mass calculation" to find
this: http://environmentalchemistry.com/yogi/reference/molar.html
You put the formula in the form, submit and find out that the anhydrous form
is 192.125g / mol. (The last digit is different probably due to a rounding
error - not much important...) For monohydrate you put C6H8O7 (citric acid)
+ H2O (water), which is written as following: C6H8O7.H2O into the form to
find out that the monohydrate form is 210.141g / mol. (Dihydrate = <compound
formula>.2H2O, trihydrate = .3H2O, tetrahydrate = .4H2O, pentahydrate =
.5H2O... And so on: hexa, hepta, octa, nona, deca...)

Now, let's go back to the original question: How much citric acid
monohydrate in substitution of 40g anhydrous citric acid? The logic is
simple: You just do the simple ratio calculation shown below:

40g (anhydrous) x ( 210.141g (monohydrate) / 192.125g (anhydrous) ) = 43.75g
~= 43.8g
(Because of the extra water molecule, the molar mass of monohydrate is
heavier than the molar mass of anhydrous. Therefore, the weight of the
monohydrate should be more than the weight of anyhdrous it is going to
replace...)

In other words, 40g anhydrous citric acid in total 100ml solution = 43.8g
citric acid monohydrate in total 100ml solution = 100ml %40 citric acid
solution.

If you like, I can further extend the explanation with the following
examples later:

1. Preparing 20% trisodium citrate (kallitype and pt/pd developer) from
citric acid and sodium carbonate. (Meaningful if: 1. You can't buy trisodium
citrate and/or more likely 2. You have excess citric acid and sodium
carbonate which you want to put into good use...)

2. Preparing ammonium tetrachloropalladate(II) (for Ware-Malde print out
palladium process) or sodium tetrachloropalladate(II) (traditional palladium
double salt - which can be used both with print out or develop out palladium
printing processes) or lithium tetrachloropalladate(II) (for Ziatype print
out palladium process) from palladium(II) chloride (PdCl2). (Having the
precursor salt - PdCl2 - on hands and use it in portions - according to the
actual need - is definitely more practical than buying the double salt
solutions directly, where you'll be stuck to the specific process variation
/ sensitizer strength the ready-made solution in question works best. With
the precursor salt, you'll have the added bonus of being able to play with
the sensitizer iron / metal strengths too...)


Hope this helps as an introduction.

Regards,
Loris.

P.S. I don't say you definitely need to know all this to make nice prints
and such, but extra knowledge doesn't hurt and proves very useful at
times... Plus, I'm not a chemist (actually I always *HATED* chemistry when I
was at school. I realized only after my start to alt. processes that how
useful, interesting and *fun* is chemistry - at least at this elementary
level...), nor a mathematician or scientist and all my formal chemistry
knowledge dates back to my highschool education. I'm just conveying my
subjective understanding and usage of this knowledge.



-----Original Message-----
From: alt-photo-process-list-bounces at lists.altphotolist.org On Behalf Of
David Ashcraft
Sent: Thursday, July 29, 2010 11:27 AM
To: The alternative photographic processes mailing list
Subject: [alt-photo] Re: DESICCATE! DESICCATE! DANCE TO THE MUSIC!

...

I do not have a background in chemistry and have no idea what this means.
Several posts have been made that state how easy it is to understand (dumb
me), I have no idea.  Surely someone could share the knowledge and explain
simply so we all could understand.
At the start of this I didn't care, but now feel like I need to know.
Thanks,
David