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

Thu Jul 29 16:24:44 GMT 2010

Loris, thanks for taking the time to explain this to me.  It's  
starting to sink in and make sense.  Long ago I used to mix my own  
paper developers and would get frustrated because I didn't have the  
"right" chemical.  Now I see how this knowledge would have kept me on  
track without running out to find the chemicals in the form listed in  
the formula (if I had it on hand).

Although, I admit that there was resistance on my part from going  
deeper into chemistry because I just wanted to try the formulas out  
and see how they affected the look of my prints.  As I eventually  
ended up here and find that life is better through chemistry ;`) and  
that knowing this information can make my life easier.

It's impossible to walk into the local camera store and find the  
materials for these processes.  I'm glad that the kits are available  
from the few suppliers that there are, so convenient to open the box,  
add water, and make a print.  But it wasn't until I tried the mix your  
own formulas did want to know more about chemistry.  It makes my  
darkroom time an adventure!  A dash of this and a pinch of that and  
magic starts happening.

There is the monetary investment I think that was also a deterrent to  
getting into mixing my own, all the safety equipment and lab  
equipment.  Although by mixing your own you save money I still factor  
in the time spent weighing and mixing instead of making a print.   
Don't forget safety!  I know people who should hire somebody to mix  
liquid hypo clear for them!

Thanks again,
David

On Jul 29, 2010, at 4:51 AM, Loris Medici wrote:

> 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
>
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