[alt-photo] Re: Stoichiometry for the nonscientist

[Loris Medici]

David, a little more clarification:

Potassium carbonate: K2CO3, 138.206g/mol (Deliquescent!)
Oxalic acid: C2O4H2, 90.025g/mol (Typically C2O4H2.2H2O - that's dihydrate,
126.066g/mol)
Potassium oxalate: K2C2O4, 166.216g
(All of the above can be extracted from the web...)

Reaction:
K2CO3 + C2O4H2*2H2O = K2C2O4 + 3 H2O + CO2
(See http://www.webqc.org/balance.php - it's a tool which facilitates /
speeds up balancing reactions. I'm sure there are many more similar pages /
tools in the web...)

>From the above reaction formula, you can clearly see that 1 mol of potassium
carbonate gets into reaction with 1 mol of oxalic acid, giving 1 mol
potassium oxalate + 3 mols of water + 1 mol of carbon dioxide.

Let's assume you need 30% potassium oxalate. (= Standard pt/pd developer
strength, right?) That's 300g in 1000ml, which is 300g (actual) / 166.216g
(molar mass) ~= a 1.8M (molar) solution.

Therefore, you'll need 1.8 mol potassium carbonate and 1.8 mol oxalic acid
(anhydrous or dihydrate, doesn't matter...) in order to make 1000ml 1.8M
potassium oxalate solution. (See the reaction formula above, proportions are
1 + 1 -> 1...)

1.8 mol potassium carbonate = 1.8mol x 138.206g/mol ~= 249g

1.8 mol oxalic acid dihydrate = 1.8mol x 126.066g/mol ~= 227g (1.8 mol
oxalic acid anhydrous = 1.8 x 90.025 ~= 162g)

Therefore, theoretically, when you combine 249g potassium carbonate with
227g oxalic acid dihydrate (or 162g oxalic acid anhydrous), giving a final /
total volume of 1000ml, you'll have a 30% (1.8M) potassium oxalate solution.

Unfortunately, you can't be precise with potassium carbonate, since it's a
deliquescent/hygroscopic substance. Also, not all salt solutions give
neutral pH. Therefore, after the reaction ends (and the solution homogenize
& stabilize at room temperature), you may need to add a little more
potassium carbonate (checking with pH control strips or a pH meter; if it's
on the alkali side, you add acid, if it's on the acid side, you add alkali -
in order to counteract / balance), or a little more oxalic acid dihydrate
(always adding in small amounts, stirring well and checking the pH after
each addition) to adjust the solution's pH to the target / desired level. (I
think most aim for pH 7 or maybe 6 - I'm not sure; I would use a neutral
developer myself. I'll let this detail to others that have much more
experience with pt/pd developers than I have. I have almost exclusively used
print-out processes myself...) You take good notes, keep them in a safe
place and then use the net / final amounts of reactants that you have worked
out in this experiment next time, w/o loosing any time (10 mins!?) doing all
of those calculations. OTOH, no big deal if you loose your notes; you'll be
able to do the math etc. from scratch, since from now on, you "know" it.
Also you're not bound / stuck to the specific variants of the reagents, you
can use any variant that is appropriate, adjusting the weight according to
the required molarity, using the molar mass of the specific reagent you have
on hands...

I think that this was indeed a pretty complete example to the usage of the
following concepts: molarity, molar mass and stoichiometry, in a real-life
situation / alt-process context.

Regards,
Loris.


-----Original Message-----
From: alt-photo-process-list-bounces at lists.altphotolist.org On Behalf Of
etienne garbaux
Sent: Monday, August 02, 2010 8:27 AM
To: The alternative photographic processes mailing list
Subject: [alt-photo] Re: Stoichiometry for the nonscientist

> ...
> By inspection, you can see that 1 mol of potassium carbonate reacts with
> 1 mol of oxalic acid to make 1 mol of potassium oxalate, with one mol of
> H2 and one mol of CO3 left over.  The H2 and one oxygen atom react to
> make one mol of water (H2O), leaving only carbon dioxide (CO2) in excess.
> So, we expect this reaction to evolve one mol of CO2 for every mol of
> potassium carbonate and oxalic acid we react.

> ...
> Note that if you started with potassium hydroxide (KOH) instead of
potassium
> carbonate, you would require two mols for each mol of oxalic acid and the
> only byproduct would be water.