Is it possible to prepare Cryolite Na3AlF6 from sodium aluminate Na3AlOH6?

[Sand Grain]

I worked out a method for making Cryolite with easily available home chemicals (though I never tried it, and I don't recommend so), namely:

 

1) Sodium Hydroxide.

2) Any Aluminium Compound

3) Not really "easily" available: an adequate supply of hydrofluoric acid (HF)

 

The method should go on like this:

 

1) Aluminium Compound + NaOH => Al(OH)3 +.....

 

2) Al(OH)3 + 3NaOH => Na3Al(OH)6

 

3) Na3Al(OH)6 + 6HF => Na3AlF6 +6H2O.

 

 

Any viewpoints are accepted, it's just a question

Edited February 3, 2011 by Sand Grain

[hypervalent_iodine]

Okay, so there is a reason why (in Australia at least) there are specialised labs that handle HF. As a home operation, this is a very silly idea and I'm glad you're not planning to attempt it. In terms of the reaction, it depends on what sort of ligands are attached to the Al as to whether the hydroxides will be able to substitute them. AlCl₃ would work. In industry, the production of Al(OH)₃ is achieved using the Bayer process, which requires high temperatures and pressures. Another alternative be to use a sulphide, Al₂S₃, and react it with water. Problem is that you get SH₂ as one of the products, which is particularly nasty stuff.

Turning the Al(OH)₃ into the sodium salt also wouldn't work. You'd actually form the tetroxide, [Al(OH)₄]^- Na⁺.

 

To make cryolite, it would be much easier to combine an AlF3 with some sort of sodium salt solution. It can be done at reasonable temperatures and doesn't require HF. There is a problem in that trifluorides have a characteristically poor solubility in most solvents and there is also an issue of safety (very toxic when it comes into contact with your skin). The problem with your proposed reaction to cryolite (other than the HF thing) is that you may not end up with the hexafluoride. Ligand substitution is a stepwise process, by which I mean one ligand is replaced and then another (it doesn't happen all at once). With each successive substitution there are less reactive positions for the fluoride ion to attack, so statistically speaking the second, third, and fourth (etc.) substitution becomes progressively less likely to occur. This trend is supported by Gibbs free calculations as well (using Δ_rG°= -RT ln K_f). With each substitution comes a new formation constant, K_f. When we get to the second substitution, the value for K_f decreases, which correlates to a smaller Δ_rG° and thus indicates that it is energetically less likely to occur spontaneously. So I guess what I am trying to say is that you are better off not relying on 6 successive substitutions, because (at a guess) there is a chance you'd only end up with a tetrafluoride and two remaining hydroxyls in the trans position, or something to that affect. I could be wrong on that last bit though. In any case, HF is not something you want to use if you can avoid it.

Edited February 3, 2011 by hypervalent_iodine

[mississippichem]

So I guess what I am trying to say is that you are better off not relying on 6 successive substitutions, because (at a guess) there is a chance you'd only end up with a tetrafluoride and two remaining hydroxyls in the trans position, or something to that affect. I could be wrong on that last bit though. In any case, HF is not something you want to use if you can avoid it.[/font]

 

Everything else is right. But If two hydroxyls remained, they would most likely be cis to each other. This is the electronic "trans-effect" and would be vanishingly small for fluorine. So really you're looking at an almost statistically insignificant excess cis over trans.

Edited February 3, 2011 by mississippichem

[hypervalent_iodine]

Everything else is right. But If two hydroxyls remained, they would most likely be cis to each other. This is the electronic "trans-effect" and would be vanishingly small for fluorine. So really you're looking at an almost statistically insignificant excess cis over trans.

 

I was really just giving an example there. Thanks though. There is definitely a good reason why I never did inorganic after I finished undergrad .

[mississippichem]

I was really just giving an example there. Thanks though. There is definitely a good reason why I never did inorganic after I finished undergrad .

 

Sorry I wasn't trying to bust your chops. I saw a chance to talk about coordination and I jumped on it instinctively. Ignore me, I don't know any better .

[Sand Grain]

Well I don't even have the equipment to attempt the experiment, so dont mind warning me

 

However i do know HF's a lab-only chemical, and with lab-only i dont mean a homelab or a school lab (=

 

Just to clarify my "level of expertise", I'm a tenth grader doing an IGCSE in chemistry, so the Gibbs constant and other langrangian constants are far higher than my level.

 

I reached this method by pure theory and a curiousity to know whether this is the industrial method or not. It all came out when I compared the close structural formulae of Na3Al(OH)6 and Na3AlF6, and I browsed many websites in a quest to find if HF is used in preparing synthetical cryolite, and I found some info on wikipedia (not a really recommended source though) and some other chemistry sites which said HF is actually used:

 

http://en.wikipedia....drofluoric_acid look under the "production of fluorides" section.

 

Anyway that was one nice suggestion, and thanks for your replies (=

 

EDIT: I think your suggested sodium salt is sodium fluride:

 

AlF3 + 3NaF = Na3AlF6

Edited February 4, 2011 by Sand Grain

[Sand Grain]

bump (=