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Author: Subject: Aluminium Sulphate woes
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[*] posted on 30-5-2014 at 14:38


Quote:
.16H2 hydrate was the target, so there was 13mls too much water.

CHRIS25 just pointed out that the 13mls was what was required, and not what needed boiling off ...
Bugger.


[Edited on 30-5-2014 by aga]




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[*] posted on 31-5-2014 at 10:13


@Blogfast: """""You aren’t heating, right? Remember that higher temperature speeds up chemical reactions?""""
have I really projected such an ignoramous persona on this forum:D (I hear some say yes...) that's like asking why my potatoes aren't cooking - were they being boiled? (actually I forgot to switch the electric on....)

But thankyou for your explanation anyway. So acid hydrolysis occurs when the sulphate and Hydrogen ion bond is broken by water resulting in the taking of a proton by water resulting in the hydronium ion. The H ion reduces the Al ion which is oxidised by the loss of 3 electrons, each of these electrons is then snapped up by Hydrogen which then results in a neutral hydrogen atom escaping now as gas, and hydronium ions returning as water molecules, but naturally all this is self repeating and ongoing for as long as there is excess H ions from the acid.

Is this ok as far as trying to break it down? I am trying to visualize it as a picture, simplified I am sure.

[Edited on 31-5-2014 by CHRIS25]

[Edited on 31-5-2014 by CHRIS25]




‘Calcination… is such a Separation of Bodies by Fire, as makes ‘em easily reducible into Powder; and for that reason ‘tis call’d by some Chymical Pulverization.’ (John Friend, Chymical Lectures London, 1712)

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[*] posted on 31-5-2014 at 11:49


Chris:

A few points:

(I) The speed at which Al (as well as other metals) dissolves in acids (or alkali, in the case of amphoteric elements like Al) depends on several factors, mainly:

1) Specific surface area: grades with more m<sup>2</sup> surface area per kg will dissolve much faster than coarser materials
2) Alloying elements: since as we’re using scrap metal and not 99.9999 % aluminium, one type of scrap aluminium may dissolve more easily than another
3) Temperature
4) Acid/alkali concentration

So different experimenters will find different dissolution rates.

(II) For hydrolysis I prefer this explanation.

The hexaaqua aluminium cation (see above) tends to expel protons from its 6 ligand water molecules, these protons are then absorbed by solvent water molecules, to form oxonium ions. In watery solutions with high H<sub>3</sub>O<sup>+</sup> concentration this tendency is suppressed. But in alkaline conditions (small H<sub>3</sub>O<sup>+</sup> concentration) the tendency is favoured and hydrolysis can proceed all the way down to Al(OH)3.

(III) With regards to the oxidation of Al metal by the oxonium (aka hydronium) ion H<sub>3</sub>O<sup>+</sup>, when the latter ‘receives’ an electron (thus being reduced), a neutral hydrogen atom, here represented by H* (* indicates hydrogen's lone 1s<sup>1</sup> electron), is split off:

H<sub>3</sub>O<sup>+</sup>(aq) + e<sup>-</sup> === > H<sub>2</sub>O(aq) + H*(g)

Two H* then combine to form diatomic hydrogen gas:

2 H*(g) === > H<sub>2</sub>(g)

The notation that is most commonly used to represent the oxidation of a metal, e.g. here Al:

Al(s) === > Al<sup>3+</sup>(aq) + 3 e<sup>-</sup>

... which I've used much above is actually a serious simplification that doesn't correspond well to reality. Unfortunately that reality will have to wait a bit before I try and describe it here, because I don't want to create an 'information overload' on your system. Maybe later...

Does all that make any sense at all to you?


[Edited on 31-5-2014 by blogfast25]




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[*] posted on 31-5-2014 at 13:06


I would argue that Physical Reality doesn't fit well with Current Models at all.

However, current models do seem to work, mostly.

Certainly i cannot currently offer a Better model, so go with the flow.




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[*] posted on 1-6-2014 at 03:01
Success with a decent hydrate


It has the texture and hardness of a chocolate bar out the fridge when scraped, is actually whiter than the photo, and is 2.5M @ 0.125mol

Weight of this product is 76g
equivalent anhydrous would be 42.75
Logically therefore 33.25g is water

This is more like a .16H2O:
33.24g water = 2 mol approx
0.125/2 = 16


AlSulph.jpg - 85kB

[Edited on 1-6-2014 by CHRIS25]




‘Calcination… is such a Separation of Bodies by Fire, as makes ‘em easily reducible into Powder; and for that reason ‘tis call’d by some Chymical Pulverization.’ (John Friend, Chymical Lectures London, 1712)

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[*] posted on 1-6-2014 at 03:17


Congratulations !

Woe-lessness all round.




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[*] posted on 1-6-2014 at 04:01


@Aga:

The best models are the ones that work best, although they don't necessarily work very well. Further research and understanding then refines the theory, making it fit 'reality' better. I use quote marks because in the quantum world 'reality' is a far cry from what we normally understand by it.

@Chris:

I calculate yours more like 14.8 but that's due to your rounding off which is fine. Well done.

Bear in mind though that the number is kind of an average: not only does it probably contain multiple hydrates, it almost certainly contains small pockets of saturated solution (inbetween the small hydrate crystals). Assuming you refrigerated this material to obtain it, this small amount of solution will contain still about 25 w% of aluminium sulphate (expressed as anhydrate). Will try and elaborate a bit on this later on.

Those last bits of solution could probably be removed (if it was needed to do so) by vacuum drying, freeze drying or even prolonged drying in air.

Despite that, your product would definitely be usable. It would be interesting to determine Al content by, dare I say... titration? ;)


[Edited on 1-6-2014 by blogfast25]




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[*] posted on 1-6-2014 at 04:30


Interesting Blogfast, ""So small pockets of saturated solution and multiple hydrates."" I would not have considered this, but plenty to learn still I am interested, what I really need though is to see a visual of what you have described; I will titrate for Al content though. Obviously vacuum and freeze drying I can not do, unless I invest in a CO2 fire extinguisher, but air drying - I am sure that I read that this compound is hygroscopic.
I don't have EDTA and have no idea what else could be used.

[Edited on 1-6-2014 by CHRIS25]




‘Calcination… is such a Separation of Bodies by Fire, as makes ‘em easily reducible into Powder; and for that reason ‘tis call’d by some Chymical Pulverization.’ (John Friend, Chymical Lectures London, 1712)

Right is right, even if everyone is against it, and wrong is wrong, even if everyone is for it. (William Penn 1644-1718)

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[*] posted on 1-6-2014 at 05:08


My Al sulphate, after refrigeration overnight is now also a quite hard, white mass. Although I didn’t do a mass balance I suspect it too is in the n = 14 – 16 range. For the alum I only got about 60 % actual yield but that is largely because I used a weaker solution than I normally would.

Even though I cannot find a two phase system for Al sulphate + water, I can construct something that is kind of a ‘next best thing’. I used the Al sulphate solubility data from Wiki’s table and converted them to weight % aluminium sulphate and plotted them against temperature:



The data point line is called the ‘saturation line’. To the left all combinations of w% and temperature form completely liquid, non-saturated solutions. To the right, hydrates begin to form, in equilibrium with a saturated solution (noted as 'liquid' on the diagram).

Take the point A for instance (40 w% at 100 C), which is a completely liquid, non-saturated solution of Al sulphate in water. Now cool this down to about 70 C and it hits the saturation line, which means solid aluminium sulphate hydrate now starts forming, in equilibrium with saturated solution at 40 w% and 70 C.

Further cooling forces more aluminium sulphate hydrate out of the solution (unfortunately my diagram cannot tell which hydrates form) and the liquid phase composition now ‘slides down’ along the saturation line, until at 0 C it is about 25 w%. Because aluminium sulphate forms such high hydrates, the solid takes with it much of the solvent. That is why the solution solidifies but some solution remains nonetheless.


[Edited on 1-6-2014 by blogfast25]




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[*] posted on 1-6-2014 at 05:10


Quote: Originally posted by CHRIS25  
I am sure that I read that this compound is hygroscopic.
I don't have EDTA and have no idea what else could be used.

[Edited on 1-6-2014 by CHRIS25]


Actually, I don't think it is. Certainly my commercial grade isn't hygroscopic at all.

Yes, for Al, EDTA titration is great.

[Edited on 1-6-2014 by blogfast25]




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[*] posted on 1-6-2014 at 05:18


Quote: Originally posted by blogfast25  
I used the Al sulphate solubility data from Wiki’s table and converted them to weight % aluminium sulphate and plotted them against temperature:



The data point line is called the ‘saturation line’. To the left all combinations of w% and temperature form completely liquid, non-saturated solutions. To the right, hydrates begin to form, in equilibrium with a saturated solution.


Sorry, but normally I understand weight % of a chemical, but I am lost at what you mean here by converting the al sulphate from wiki into weight percent? and then on your graph 40%, what is this percent?




‘Calcination… is such a Separation of Bodies by Fire, as makes ‘em easily reducible into Powder; and for that reason ‘tis call’d by some Chymical Pulverization.’ (John Friend, Chymical Lectures London, 1712)

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[*] posted on 1-6-2014 at 05:31


Quote: Originally posted by CHRIS25  
[Sorry, but normally I understand weight % of a chemical, but I am lost at what you mean here by converting the al sulphate from wiki into weight percent? and then on your graph 40%, what is this percent?


For instance, Wiki says at 70 C, an aluminium sulphate solution in water can contain a MAXIMUM of 66.2 g Al2(SO4)3 per 100 g of water. Convert that to w% Al2(SO4)3:

w% Al2(SO4)3 = 66.2 g Al2(SO4)3 / (66.2 g Al2(SO4)3 + 100 g water) x 100 % = 39.8 w% Al2(SO4)3 (and 60.2 w% water).

Clear?




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[*] posted on 1-6-2014 at 06:08


Stupid me yes absolutely. Clear thankyou.

Aluminium chlorosulphate crystal - next side line project.




‘Calcination… is such a Separation of Bodies by Fire, as makes ‘em easily reducible into Powder; and for that reason ‘tis call’d by some Chymical Pulverization.’ (John Friend, Chymical Lectures London, 1712)

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[*] posted on 1-6-2014 at 06:12


Quote: Originally posted by AJKOER  
With respect to Al(OH)3(H2O)3, also called Aluminium triaquatrihydroxy complex, I repeat the interesting comments on this compound at http://www.chemthes.com/entity_datapage.php?id=4011 to quote:

"Real, long lived, electronically neutral reagent chemical.
Gram formula weight (molecular mass) = 132.04
Solubility in organics = insoluble
Solubility dependent upon pH. The trisaquatrishydroxyaluminium neutral complex is white, (even colourless), gelatinous material. Solubility is dependent upon pH."

The central, to be determined, question relates not to absolute solubility at any given pH, but relative solubility of this complex versus Mg(OH)2 at, say, pH 5.5 and the ability of the proposed reaction below to move to the right:

2 Al(OH)3(H2O)3 (aq) + 3 MgSO4 (aq) --?--) Al2(SO4)3 (aq) + 3 Mg(OH)2 (s) + 6 H2O


No reason to argue any more, I have good indications of a complete success!

First, I developed a fast route to Al(OH)3(H2O)3 by using my previous work on a galvanic cell of NaClO in the presence of Al and Cu with an electrolyte of NaCl (as one my references I have given in previous threads, please see http://www.exo.net/~pauld/saltwater/ ). After a few hours, just pour out some of the galvanic cell content leaving any undissolved Al, Cu or other solid residues behind.

What is actual created is a basic solution containing unreacted NaClO, NaCl and partially undissolved hydrated alumina, so the true magnitude of alumina in solution is not revealed until one adds a small amount of a weak acid (I used seltzer water which has a pH between 3 and 4). Immediately a thick suspension of hydrated alumina is revealed. Results of the creation of this compound can actually be obtained in minutes!

[Edit] Next, add dilute H2O2 to convert any unreacted NaClO to NaCl releasing O2 in the sample solution of hydrated alumina. On some runs, I did not perform this step, which can alter the final products.

Then, I recommend creating two working solutions of the hydrated alumina for comparison purposes. In the first solution, add MgSO4 dissolved in seltzer water. The second solution, add no MgSO4, just seltzer water. On standing in both test solutions, a similar precipitate is formed (I think the 1st solution precipitate is a little more fluffy).

Now to test the formation of Mg(OH)2, add vinegar as Alumina will not dissolve in weak vinegar, but Mg(OH)2 will. Results: the control solution remains cloudy, but the MgSO4 solution becomes completely clear as water. To test further, add some solid MgSO4 to the control solution. It eventualy becomes clear as well!

[Edit] First, I should mention that I used an excess of NaCl, which in the role of an electrolyte is essentially not significant, but for immediate side reactions and subsequent reactions involving H2CO3, amphoteric Al2O3 and MgSO4, this actually may have a significant impact on the 'activity coefficient' (which can significantly increase the functional acid strength of even weak acids as I have previously discussed on SM).

Also, caution, as while the presence of heavy metals in solution is much lower than direct routes employing especially impure (especially Iron) H2SO4, the longer the galvanic cell runs, potentially more toxic Copper is placed into solution. Allowing the final aqueous solution to sit in a piece of Aluminum (previously boiled and allowed to stand for a couple hours in vinegar) could mitigate this contaminant.

[Edited on 1-6-2014 by AJKOER]

[Edited on 1-6-2014 by AJKOER]
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[*] posted on 1-6-2014 at 06:31


JOKER:

You've proved nothing. You're just cackling on. Where's your proof of aluminium sulphate? There is none.

Chris:

I can even go further with this diagram, by making some simple (but probably inaccurate to some degree) assumptions. I’ll assume the main hydrate forming is n = 15 in my conditions. That has a molecular weight of 342 + 15 x 18 = 612 g/mol and a w% aluminium sulphate (AS) of 342/612 x 100 = 55.9 w%.

I’ll also assume I’m starting from a hot, 50 w% AS solution and allow it to cool to 0 C.

Say I start from 100 g hot solution and that after cooling it contains H g of AS hydrate and (100 – H) g of saturated solution. The diagram tells me that solution contains 26 w% of AS.

My original solution contained (100 x 50) / 100 = 50 g AS. Since as I didn’t add or remove any AS during the cooling I must find the same amount of AS in the cooled stuff:

Mass balance: 50 = (100 – H) x 26/100 + H x (55.9/100) or:
H = (50 – 26) / (0.559 – 0.26) = 80.3 g

So the 100 g of cooled solution would contain 80.3 g of solid hydrate and 19.7 g of saturated solution.




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[*] posted on 1-6-2014 at 06:54


Sorry Blogfast - it is hard for me to follow this because I just about 'cope' with working on stoichiometry and the maths involved in so many things, that to start trying to theorize with mathematical ideas stretches my capacity to nervous breakdown. I understand your chart, the principles and so on, and functional maths, but I simply can not think this way, (ex trucker and visual artist/photographer = wrong side of the brain).



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[*] posted on 1-6-2014 at 07:00


What the hell.

AJKOER : post a Full experiment, including volumes, masses, concentrations etc and i will give it a go and post the results.

Why not ?

Start a new thread for it though.

CHRIS25 no longer has any Alumnium Sulphate Woes.

[Edited on 1-6-2014 by aga]

[Edited on 1-6-2014 by aga]




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[*] posted on 1-6-2014 at 07:13


Quote: Originally posted by aga  
What the hell.

AJKOER : post a Full experiment, including volumes, masses, concentrations etc and i will give it a go and post the results.

Why not ?




Most of all, I'd like to see a tangible product, with proof that it is (or is mainly) aluminium sulphate. Any hydrate will do.

Joker's latest belief system is crutched up by one paragraph on Al(OH)3.3H2O:

"Real, long lived, electronically neutral reagent chemical.
Gram formula weight (molecular mass) = 132.04
Solubility in organics = insoluble
Solubility dependent upon pH. The trisaquatrishydroxyaluminium neutral complex is white, (even colourless), gelatinous material. Solubility is dependent upon pH."



He forgets there's nothing novel about that: Al(OH)3 dissolves in sufficiently acidic conditions to hexaaqua aluminium cations, in sufficiently alkaline conditions to aluminate, Al(OH)<sub>4</sub><sup>-</sup>, In neutral condition it is totally insoluble. Nothing new, nothing to see here, the wheels on the bus are turning.


[Edited on 1-6-2014 by blogfast25]

[Edited on 1-6-2014 by blogfast25]




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[*] posted on 1-6-2014 at 10:31
Various Hydroxides




AL_hydrolysis.png - 7kB




‘Calcination… is such a Separation of Bodies by Fire, as makes ‘em easily reducible into Powder; and for that reason ‘tis call’d by some Chymical Pulverization.’ (John Friend, Chymical Lectures London, 1712)

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[*] posted on 1-6-2014 at 11:41


Note that nearly all of these species are solvated, so for instance:

[Al(OH)<sub>2</sub>(H<sub>2</sub>O)<sub>4</sub>]<sup>+</sup>

[Al(OH)(H<sub>2</sub>O)<sub>5</sub>]<sup>2+</sup>

etc.

Note also extremely low concentration: 5 x 10<sup>-6</sup> mol/L


[Edited on 1-6-2014 by blogfast25]




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[*] posted on 1-6-2014 at 12:29


Quote: Originally posted by blogfast25  
Note that nearly all of these species are solvated, so for instance:

[Al(OH)<sub>2</sub>(H<sub>2</sub>O)<sub>4</sub>]<sup>+</sup>

[Al(OH)(H<sub>2</sub>O)<sub>5</sub>]<sup>2+</sup>

etc.

Note also extremely low concentration: 5 x 10<sup>-6</sup> mol/L


[Edited on 1-6-2014 by blogfast25]

Yes I know and that is the problem with Aluminium along with Iron, but aluminium is twice as bad due to its higher positive charge and the large size of its atom (though I confess I do not really understand the implications of the latter); anyway what is curious, is that I have read that Al + OH in soln becomes Al(OH)4- in alkali conditions and Al(H2O)6+3 in acidic. Question - Al(OH)3 is formed here in slightly acidic conditions plus the concentration is low, does this mean that the concentration of Al ions must be kept low by addition of plenty of free OH ions, since I am now dissolving more Al in sulphuric to make only the Hydroxide so that I don't have to use Al metal in other experiments + it is more efficient, I am curious.

[Edited on 1-6-2014 by CHRIS25]




‘Calcination… is such a Separation of Bodies by Fire, as makes ‘em easily reducible into Powder; and for that reason ‘tis call’d by some Chymical Pulverization.’ (John Friend, Chymical Lectures London, 1712)

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[*] posted on 2-6-2014 at 04:36


The tendency for hydrolysis of Al3+ and Fe3+ is actually fairly comparable. The Al3+ ion is smaller and has the same charge as Fe3+. For the former, being smaller the electrical field surrounding it is greater and it has a greater potential to expel protons from these 6 water ligands.

I'm not sure I understand the question. Can you rephrase?


Aluminate ions can form from hydroxide by so-called "ligand substitution":

Al(OH)<sub>3</sub>(H<sub>2</sub>O)<sub>3</sub> + OH<sup>-</sup> === > [Al(OH)<sub>4</sub>(H<sub>2</sub>O)<sub>2</sub>]<sup>-</sup> + H<sub>2</sub>O

One water molecule ligand substituted by a hydroxide ion ligand.




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[*] posted on 2-6-2014 at 04:55


Quote: Originally posted by blogfast25  


I'm not sure I understand the question. Can you rephrase?


I want to make Al hydroxide, I will precipitate it with 35% ammonia solution. Ph has to be carefully monitored and I have to arrive at 6 - 6.5. However you mentioned the percent of Al ions in solution that were given on this graph was extremely low. Is this really worth being too concerned about? (I have to confess in not being able to understand yet the maths nomenclature of what this concentration actually means).

[Edited on 2-6-2014 by CHRIS25]




‘Calcination… is such a Separation of Bodies by Fire, as makes ‘em easily reducible into Powder; and for that reason ‘tis call’d by some Chymical Pulverization.’ (John Friend, Chymical Lectures London, 1712)

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[*] posted on 2-6-2014 at 08:22


5 x 10<sup>-6</sup> = 5 times 10 to the power minus six = 0.000005. It's a convenient format to describe small numbers with.

With ammonia you don't need to worry about pH and aluminate formation: ammonia is far too weak an alkali to form aluminates with.

Just respect the following stoichiometry (simplified notation):

Al<sup>3+</sup> + 3 NH<sub>3</sub> + 3 H<sub>2</sub>O === > Al(OH)<sub>3</sub> + 3 NH<sub>4</sub><sup>+</sup>

If you start from Al sulphate, you'll need 6 mol of ammonia per mol of Al sulphate.

After the addition of the ammonia, check for pH, it should be pH > 7. If it isn't yet add more ammonia in small aliquots until pH > 7.

After filtering you can evaporate the filtrate to recover ammonium sulphate, if you want to.





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[*] posted on 2-6-2014 at 10:32


Quote: Originally posted by blogfast25  
Aluminate ions can form from hydroxide by so-called "ligand substitution":

Al(OH)<sub>3</sub>(H<sub>2</sub>O)<sub>3</sub> + OH<sup>-</sup> === > [Al(OH)<sub>4</sub>(H<sub>2</sub>O)<sub>2</sub>]<sup>-</sup> + H<sub>2</sub>O

One water molecule ligand substituted by a hydroxide ion ligand.


It's not actually a substitution; the hydroxide ion deprotonates the coordinated water molecule to form a hydroxy ligand.




Please remember: "Filtrate" is not a verb.
Write up your lab reports the way your instructor wants them, not the way your ex-instructor wants them.
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