[giuliano]

Hello,  I am not a chemist but a physicist, so I hope you will forgive me for the wrong terms..

I have started making some Madder Lake (used by painters and luthiers), by precipitating aluminium hydroxide from a K2CO3 + AlK 2(SO4) 12H20 solution. I ignore the presence of the madder roots, which contributes mainly alizarine, in the initial K2CO3 solution and of water impurities. I have also read that it is the calcium salts present in these impurities that allows AlOH3 to bind with alizarine, but cannot find a precise reference for that statement (cited in J Michelman, Violin Varnish, pag 85, freely available on-line).

I have done some stoichiometry to evaluate the correct proportion of the ?two initial salts, but my chemistry knowledge lacks... Could someone check my ?estimated reaction (in water)?

2 (AlK 2(SO4) 12H20) + 3 (K2 CO3) => 2 (Al (OH)3) + 4 (K2 SO4) + 18 H2O + ?3 (H2 CO3)
that gives a mass ratio of approx. 2.29 parts alum/ 1 part potassium ?carbonate.
1) The above formula looks numerically correct, but is it what is really happening?

2) As AlOH3 is insoluble (in absence of acid or strong alkali), and H2CO3 escapes, as it will mainly be in the form CO2 + H2O, I also suspect that if (1) is the correct reaction, then it will go wholly from left to right.
?3) as the solution is kept between 30 and 40C, I suppose that all the ?CO2 will leave the solution, is then the result a neutral ?solution?

Note that point (3), that is neutrality, would give me a better way of evaluating the amount of alum to add to the solution, as after filtration of the potassium ?carbonate/madder root solution, I do not really know how much carbonate ?has been discarded with the root pulp.

Finally, does anyone know how AlOH3 sequestrates dye molecules from a solution?
 
Thanks
Giuliano

[Borek]

I would put it this way:

2AlK(SO₄)₂^.12H₂O + 3K₂CO₃ -> 2Al(OH)₃ + 4K₂SO₄ + 21H₂O + 3CO₂

(note small mistake in your formula for alum)

Educated guesses follow

Your estimation of mass ratio 2.29:1 is stoichiometrically - perfect, although I would not treat it too seriously, rather as a guidance. It is very unlikely that your reagents are pure. The most likely problem is that they will not follow their formula when it comes to crystallization water, so the theoretical ratio of 2.29:1 will be wrong; IMHO you will need some more potassium carbonate.

Perhaps you could start mixing the reagents and once the system reaches the equilibrium (no idea how long it could take - can be minutes, can be hours; most likely system will be limited by the CO₂ escape ratio) check pH - if it is below 7, add some more carbonate. You don't have to hit 7.00000000, something between 6 and 8 will probably do, so narrow range pH stripes should be enough. Perhaps even universal ones.

As for sequestration - Al(OH)₃ have huuuuuge surface per mass unit, adsorption is the key word to what is happening.

[giuliano]

Thanks! Yes, there was some inconsistency in notation for the SO₄ multiplier... and I also agree that the CO₂ should have not been put as carbonic acid.
So, you confirm that I should aim for neutrality. Good. I have relatively narrow range (4 - 10 pH) papers and I have used those.
From your comment about cristallisation water I suppose you refer to the Alum, but my main source of error is, as I said, the separation of the root pulp from the potassium carbonate solution. The pH method comes thus very handy.

You say it is adsorption (I have just learnt the difference between this and absorption...), and I imagine that's because of the huge surface of AlOH3. But I estimated a concentration of up to 50% (IIRC) of dye over the total mass of precipitate. Might that point to absorption instead? Not that it matters... the resulting pigment must work and that's it!
Thanks again.

[Borek]

Cristallisation water is mostly present in alum, but it is also present to some extent in carbonate - there are several forms of hydrated potassium carbonate and in case of such salts their exact composition is often a function of temperature and air humidity.

As for ab/dsorption - IMHO it starts with adsorption, later it is very likely that adsorbed dye becomes occluded (not sure if that's proper English term) inside - at this moments you may refer to it as to absorbed. But I agree with you that as long as it works it doesn't matter - much

[giuliano]

Now you have one more snack...
For anyone interested in the procedure I followed in making Madder (Rubia/Rubea Tinctorum) Lake, I posted a few weeks ago to a forum the whole procedure in italian and english (some incidental posts are in italian only) with photos at: http://www.claudiorampini.com/php/modules.php?name=Forums&file=viewtopic&p=23251#23251.
I will soon update that procedure with the chemical considerations, even if that will probably be a bit out of place given that forum topic...