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Author: Subject: Aluminium Sulphate woes
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[*] posted on 27-5-2014 at 23:46


http://www.researchgate.net/post/What_is_the_difference_betw...

Interesting discussion of what DIW actually is, and how it is produced.




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[*] posted on 28-5-2014 at 05:05


Quote: Originally posted by aga  
http://www.researchgate.net/post/What_is_the_difference_betw...

Interesting discussion of what DIW actually is, and how it is produced.



From that source:

"The H+ and OH- combine to form H2O, leaving only the residual H+ and OH- produced by autodissociation (autoprotolysis), H2O = H+ + OH-; the equilibrium constant of this reaction = 1 x 10^-14 at 25 °C"

Even though the H+/OH- is a misrepresentation, once you understand that statement, you'll understand why water is sooooooo NOT an ionic compound. Its dissociation is basically a zero followed by thirteen zeros. NOT ionic at all. The concentration of H3O+ (NOT H+, that's non-sensical) in neutral, distilled water is about 0.0000001 mol/L. Do you get it now?

Please get that filthy thought out of your heads AT ONCE. No kidding.

Will explain more later on.

Researchgate is a very low authority source, BTW. Careful with web pages: no place peddles more misinformation/disinformation than the Tinkerwebs.


[Edited on 28-5-2014 by blogfast25]




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[*] posted on 28-5-2014 at 07:36


At the risk of fueling a filthy thought:D I know that water is a polar covalent dipole molecule where the hydrogen is the weaker and has a slight positive charge and the oxygen has a higher pull over the hydrogen. But it has also the character of an ionic bond in that there is this metal + non-metal bond character. Surely, and withot wanting to appear pedantic, it can not be categorized into either box. I only actually started to investigate this because in all my reactions to date I have un-conciously treated water as if it did not exist, I mean it is water, and very few balanced reactions ever include water in their equations, yet water does play a huge role in affecting the direction of a reactions path. take the ferric chloride solution. Everything seems fine according to this:
Fe + 2HCl = FeCl2 + H; yet this is misleading due to the fact that you need to add water, and upon doing this you risk then hydrolysis of the ferrous ion to Fe(OH)3. So now we have to consider water as H+ and OH-

This is why I decided that water is not that neutral solution that one simply ignores in balancing equations, but that water is actually a chemical in its own right that needs to be considered. And then of course we have the same thing happening with aluminium sulphate, this too risks hydrolysis without careful attention to the amount of OH ions balanced against the amount of SO4 ions available. So my intention is not to argue a point, simply to understand a concept of chemistry that I have ignored because I perceived water, sub-conciously, as a non reactant in chemical equations.




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[*] posted on 28-5-2014 at 09:50


Quote: Originally posted by CHRIS25  
I only actually started to investigate this because in all my reactions to date I have un-conciously treated water as if it did not exist, I mean it is water, and very few balanced reactions ever include water in their equations, yet water does play a huge role in affecting the direction of a reactions path. take the ferric chloride solution. Everything seems fine according to this:
Fe + 2HCl = FeCl2 + H; yet this is misleading due to the fact that you need to add water, and upon doing this you risk then hydrolysis of the ferrous ion to Fe(OH)3. So now we have to consider water as H+ and OH-



Certainly water water cannot be ignored in aqueous reactions, yet very often it is nothing but a spectator solvent.

The dissolution of iron in hydrochloric acid to ferrous chloride (1), the oxidation of ferrous chloride to ferric chloride (2) and the precipitation of ferrous hydroxide (3) are three distinctly different phenomena.

(1) HCl is dissociated is water: HCl(aq) + H2O(l) === > H3O+(aq) + Cl-(aq)

Iron is oxidised by H3O+ but only to Fe(II):

Fe(s) + 2 H3O+(aq) === > Fe(2+)(aq) + H2(g) + 2 H2O(l)

(2) Fe(2+)(aq) === > Fe(3+) + e-

But that requires an oxidising agent and it isn't water. Oxygen, peroxide, etc etc can all act as oxidisers here.

(3) Hydrolysis of Fe3+, a multistep (equilibrium) reaction:

[Fe(H<sub>2</sub>O)<sub>6</sub>]<sup>3+</sup>(aq) + H<sub>2</sub>O(aq) < === > [Fe(OH)(H<sub>2</sub>O)<sub>5</sub>]<sup>2+</sup>(aq) + H<sub>3</sub>O<sup>+</sup>(aq)

A second and third step then leads to Fe(OH)<sub>3</sub>. Here water plays an active part.


Please (PLEASE?) do forget about H<sup>+</sup>/OH<sup>-</sup>, H+ cannot exist in watery solution. H+ is a proton, nothing more. In water it immediately bonds to a water molecule: H<sup>+</sup> + H<sub>2</sub>O === > H<sub>3</sub>O<sup>3+</sup>. There are NO free protons in watery solution.

[Edited on 28-5-2014 by blogfast25]

[Edited on 28-5-2014 by blogfast25]




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[*] posted on 28-5-2014 at 14:48


The most convincing argument would be to show how Aluminium Sulphate is made, and shown to be Aluminium Sulphate as the product.

Define the experiment, and i for one will do it and replicate it, and post the results.

I think CHRIS25 will too, so two verfications.

Once that works as defined, the ionic water which is not ionic argument may be safely culled for now.

[Edited on 28-5-2014 by aga]




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[*] posted on 28-5-2014 at 15:15


This is getting creepy now.

Given the sheer Wealth of knowledge and experience out there, and plugged into this board, it is incredible that nobody has jumped in and said 'Nah. This is how it's done. Here's where you went wrong.'

Prime derivations from the known facts are :-

1. Aluminium Suphate is too hard to do.
2. Ignore noobs doing pontless,unexplosive or psychoactive stuff.
3. Aluminium Sulphate is useless, so I as Guru, never tried to make it.
4. It is hard to make, and largely pointless so why bother.
5. I dunno. Never looked.

Thank you blogfast25 for your guidance, as you are our only guidance so far on this topic.

Silence speaks Volumes.

If Aluminium Sulphate synthesis is difficult, pointless and un-useful (regarding the reactions and ionic states) then somebody saying so, and Why, would be very much appreciated.

[Edited on 28-5-2014 by aga]




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[*] posted on 28-5-2014 at 23:44


Just a quick reply aga. My al solution is progressing with extreme interest. Placing clingfilm and poking 1 hole with a cocktail stick works wonders at preventing any visible loss of water, and so I have complete control now, the hole is simply for the hydrogen. Secondly after 25 hours (over three days) of cut aluminium metal pieces dissolving I have to switch it off every night. In the morning there is a precipitate but still with metal there of course. I have heated up and cooled several times and seen how the quality of the precipitate changes. this morning's precipitate was very different, it is most certainly al sulphate but it had a distinctive crystalline structure that I had not seen in other experiments, pressing lighly with a wooden stick you can feel it crumble, this is very different from the gooey mass of globulated porridge we have had before. There is still fluid in the flask and when you shake the flask vigorously it all but re-dissolves into a milky solution which is expected. So adding a touch more water you get the milkiness disappearing into a very clear solution, the solution now is no longer saturated. Solubility at 20c is about 36g in 100 mLs water so the trick here is to calculate amount of water left over after all is dissolved and do some maths. In view of the fact that this precipitate seems to be quite unique in that I read about the unpredicatability of the number of hydrated variations you can get that includes every single number from .2H2O to . 28H2O this explains part of the problem we are having (i am not yet aware of any hydrate compounds or crystals that can have such a variation of water molecules attached to their structure. Also I now realize that you really have to stoichiometrically calculate water content and make sure that the amount of water left in the flask when all is dissolved and you switch off the heat to cool, this water has to be very precise, something I have never needed to do in any other reaction to date. (In the past one just boils down and evaporates subjectively). So when my aluminium is fully dissolved I now feel confident about getting an exact hydrate (the .2H2O), after some maths. We shall see whether or not I have to humiliate myself by re-tracting the above later on.....

[Edited on 29-5-2014 by CHRIS25]

So I figured this out, it looks good, but only the ongoing experiment will confirm:
2Al + 3H2SO4 = Al2(SO4)3 +3H2

The net ionic equation: 2Al0 + 3H2+ + 3SO42- = Al23+ + (SO4)32- +3H20

But since we know that the sulphate comes in many many different hydrates (as we have experienced) we need to calculate water at the very end just before cooling and evaporate or add accordingly in thus manner:

Water = 18g/mol so if we want to insure aluminium sulphate as Al2(SO4)3.2H2O then calculating the theoretical yield of our product (will show you if you need help here) or just assuming the mols Al we already used (either way), we can do the following:

0.5 mol Al2(SO4)3 + 2 attached water molecules to each ionic molecule needs 0.5 x 2
thus: 0.5 x 2 = 1. And so this must be 18g/mol, in other words 18 mLs water must be left in the flask BEFORE any cooling off is allowed, but certainly no more than this. This is theoretical and I would leave a little less say 15 to allow for always a lesser yield than predicted. And this whole idea only came about since reading that this sulphate could precipitate out in dozens of hydrates ranging from 2 to 28. Allowing for the fact that an average room is 14c I would put it in the fridge, at 18 mLs at 0 degrees we have solubility of 31g per 100 mLs . So the fridge methinks would ensure absolute precipitation with just 18 mLs of water, Mmm...

Now at this juncture I am expecting a : "Chris all your reasoning is wrong, in fact it is absurd - with a slap on the wrist" But secretly I am hoping that there is sense to be found hidden somewhere in this brainstorming.

[Edited on 29-5-2014 by CHRIS25]

[Edited on 29-5-2014 by CHRIS25]

[Edited on 29-5-2014 by CHRIS25]




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[*] posted on 29-5-2014 at 04:52


Aga:

All in good time. I'm a bit stretched here but hope to do it tonight, if not Sunday.

You're reading far too much into too little.


Quote: Originally posted by CHRIS25  

So I figured this out, it looks good, but only the ongoing experiment will confirm:
2Al + 3H2SO4 = Al2(SO4)3 +3H2

The net ionic equation: 2Al0 + 3H2+ + 3SO42- = Al23+ + (SO4)32- +3H20



We would not write the equations the way you do. 3H2+ and (SO4)32- are definitely nonsense.

Will elaborate tonight. To write reaction equations (and not just stoichiometry equations) you need to break down what is happening and then write the corresponding equations, you're not doing that...

[Edited on 29-5-2014 by blogfast25]




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[*] posted on 29-5-2014 at 05:03


This is exactly how it is demonstrated on so many video tutorials, admittedly maybe I should have called it simply the ionic equation, or showing the charges, because I have not really done the "net" ionic I suppose, but the charges are correct and maybe on the right side I should have typed Al2(SO4)3 since this is the precipitate and all that actually changes is that the H reduces and the Al oxidizes. I really do not see what I am doing wrong, yet again?



‘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 29-5-2014 at 05:15


Many video 'tutorials' are made by people who couldn't tell beryllium from carbon. 'A little knowledge is a dangerous thing'.

No serious chemist would write these equations the way you do. They are plain wrong. 3H<sub>2</sub><sup>+</sup> or even H<sub>2</sub><sup>+</sup> e.g. doesn't exist although it would be a highly interesting quantum system if it did :D. Ditto your 'trisulphate' thingy.

Laters...

[Edited on 29-5-2014 by blogfast25]




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[*] posted on 29-5-2014 at 05:17


ah well then, I'll get back to my embroidery.

http://www.occc.edu/kmbailey/chem1115tutorials/Net_Ionic_Eqn...
just one of many tutorials, and most of the videos are from chem univerisities or schools, but they all follow this type of notation, The only thing I missed was to put aq, i, or s afterwards.

[Edited on 29-5-2014 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)

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

The very nature of Random, Chance development precludes the existence of Order - strange that our organic and inorganic world is so well defined by precision and law. (me)
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[*] posted on 29-5-2014 at 05:58


Here is an experiment that is on my to do list. Not sure if it will work, so if anyone else attempts it, please limit the size of the experiment.

1. Prepare a clear jelly like form of Al(OH)3 from say the action of dilute ammonia on Aluminum foil (slow, days).

2. Remove the solution over the precipiate, add distilled water and stir to dilute out the NH3 (I suspect this key step will have to be repeated several times).

3. Add the product, Al(OH)3 as a clear pure gel, to a concentrated MgSO4 solution.

If a white precipitate of Mg(OH)2 forms, a positive indication of the following reaction:

2 Al(OH)3 + 3 MgSO4 ----) Al2(SO4)3 + 3 Mg(OH)2 (s)

One could also more rapidly form Al(OH)3 from, say, boiling Al in boiling Na2CO3 to produce Al(OH)3 as a white precipitate. Filter and thoroughly wash as leftover Na2CO3 will react with the MgSO4. If it dissolves in aqueous MgSO4 with the formation of a visibly different (?) white precipitate, then possibly a more rapid path.

[Edited on 29-5-2014 by AJKOER]
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[*] posted on 29-5-2014 at 06:42


Quote: Originally posted by CHRIS25  
[...] but they all follow this type of notation, The only thing I missed was to put aq, i, or s afterwards.



It's not the type of notation that I object to (I use it too) but what you do with it: there is no such thing as (SO<sub>4</sub>;)<sub>3</sub><sup>2-</sup> and you have never seen this on a reputable site. N-E-V-E-R or in any half decent book. Please don't tell me otherwise.

Please ignore the village idiot AJKOER. Here he comes again peddling his Al + NH3 imbecility. He'll never learn.

[Edited on 29-5-2014 by blogfast25]




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[*] posted on 29-5-2014 at 08:01


The displacement reaction:

2 Al(OH)3 + 3 MgSO4 ----) Al2(SO4)3 + 3 Mg(OH)2 (s)

... cannot proceed: poorly soluble as Mg(OH)2 is, it is far, far more soluble than Al(OH)3 (in neutral conditions).

Replacing NH3 with sodium carbonate changes little: both are weak bases.




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[*] posted on 29-5-2014 at 09:18


Quote: Originally posted by blogfast25  
The displacement reaction:

2 Al(OH)3 + 3 MgSO4 ----) Al2(SO4)3 + 3 Mg(OH)2 (s)

... cannot proceed: poorly soluble as Mg(OH)2 is, it is far, far more soluble than Al(OH)3 (in neutral conditions).

Replacing NH3 with sodium carbonate changes little: both are weak bases.


Yes, I agree with your comment with the qualifier "neutral solution". However, Blogfast, please explain how a concentrated MgSO4 solution, the product of a strong mineral acid and a very weak base, Mg(OH)2, could produce a neutral solution?

My experience (in agreement with your unqualified statement) is that the white precipitated Al(OH)3 is still hard to dissolve in even non-neutral solutions, but the clear geltaneous form of Al(OH)3 (formula Al(OH)3(H2O)3 or commonly still expressed as Al(OH)3 ignoring water, see http://www.science.uwaterloo.ca/~cchieh/cact/applychem/alumi... ) in mildly acidic solutions as compared to solid Mg(OH)2 in the same medium?

I agree the odds are still not likely, but excuse me if I want to verify as it may provide a novel path to other Aluminum salts as well.

[Edited on 29-5-2014 by AJKOER]
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[*] posted on 29-5-2014 at 09:31


Magnesium hydroxide is a very weak base only because of its low solubility. The magnesium ion does not hydrolyze appreciably in aqueous solution. Merck Index says "Its aqueous soln is neutral. pH = 6-7"



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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[*] posted on 29-5-2014 at 09:35


It looks like I won’t be doing this today: it’s raining and part of it has to be done outside. So I’ll go back to theory a bit.

1) Sulphuric acid solution:

When H2SO4 is mixed with water, the acid deprotonates as follows:

H<sub>2</sub>SO<sub>4</sub>(aq) + 2 H<sub>2</sub>O(l) < === > 2 H<sub>3</sub>O<sup>+</sup> (aq) + SO<sub>4</sub><sup>2-</sup>(aq)

Although this is strictly an equilibrium, sulphuric acid is a very strong acid and the deprotonation is more or less complete (this is a slight simplification but OK for this discussion).

2) Oxidation and solvation of the aluminium:

The oxonium ions oxidise the Al to Al cations:

Al(s) + 3 H<sub>3</sub>O<sup>+</sup> (aq) === > Al<sup>3+</sup>(aq) + 3/2 H<sub>2</sub>(g) + 3 H<sub>2</sub>O(l)

Solvation of the Al<sup>3+</sup> ions:

Al<sup>3+</sup>(aq) + 6 H<sub>2</sub>O(l) === > [Al(H<sub>2</sub>O)<sub>6</sub>]<sup>3+</sup>(aq)

This hexaaqua aluminium cation is how Al exists in strongly acidic watery solutions.

3) Hydrolysis of [Al(H<sub>2</sub>O)<sub>6</sub>]<sup>3+</sup>(aq):

This cation is prone to hydrolysis acc.:

[Al(H<sub>2</sub>O)<sub>6</sub>]<sup>3+</sup>(aq) + H<sub>2</sub>O(l) < === > [AlOH(H<sub>2</sub>O)<sub>5</sub>]<sup>2+</sup>(aq) + H<sub>3</sub>O<sup>+</sup> (aq) (I)

A second and third step of deprotonation can give rise to solvated [Al(OH)<sub>2</sub>]<sup>+</sup> and even Al(OH)<sub>3</sub> (hydrated alumina, aka aluminium hydroxide).

But here’s an important note: when dissolving aluminium metal in acids, we use a strong excess of acid (I will be using 30 % in excess of the stoichiometric requirement) and the excess acid, which provides extra H<sub>3</sub>O<sup>+</sup>, pushes the equilibrium reaction (I) (as well as the two subsequent deprotonation reactions) strongly to the left and explains why no Al(OH)<sub>3</sub> forms. In short, in strongly acidic solutions there is no significant hydrolysis of the hexaaqua aluminium cation.

4) Crystallisation of aluminium sulphate hydrate:

As long as the solubility limit is not exceeded (for Al2(SO4)3 that is 36.4 g / 100 ml of water, at 20 C (Wiki) but it depends very strongly on temperature) the solution consist of [Al(H<sub>2</sub>O)<sub>6</sub>]<sup>3+</sup>, oxonium ions, sulphate ions, the excess of acid and of course, water.

When we exceed the solubility limit, solid aluminium sulphate hydrate must crystallise out. Assuming it is mostly octadeca (18) hydrate:

2 [Al(H<sub>2</sub>O)<sub>6</sub>]<sup>3+</sup>(aq) + 3 SO<sub>4</sub><sup>2-</sup>(aq) + 6 H<sub>2</sub>O(l) === > Al<sub>2</sub>(SO<sub>4</sub>;).18 H<sub>2</sub>O(s)

As you can see, only if crystallisation occurs, does the sulphate ion reappear in the equation, that’s why we call it a spectator ion. In essence it is there to ensure electrical neutrality of the solution/crystals.


[Edited on 29-5-2014 by blogfast25]




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[*] posted on 29-5-2014 at 09:38


Quote: Originally posted by AJKOER  
[...] if I want to verify as it may provide a novel path to other Aluminum salts as well.

[Edited on 29-5-2014 by AJKOER]


Twits like you should try SIMPLE things, not 'novel' paths that are likely not to work or prove highly impractical.

Thanks, DA. Faced with such stubborn stupidity help is appreciated.

[Edited on 29-5-2014 by blogfast25]




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[*] posted on 29-5-2014 at 10:10


Excuse my simplicity, but the compound we should be comparing for relative solubility to Mg(OH)2 at near neutral pH is (see my edited thread for clarity and reference link) is Al(OH)3(H2O)3 and not any published value for Al(OH)3. Note, higher aqua cations like the hexaaqua aluminium cation discussed above, are very soluble.

As such, I still argue there is some possibility of success.

If I do succeed, unless there is some modification of the dialogue on what l have clearly outlined as a speculative preparation, sorry, but I may not even consider sharing on this forum. But the debate has fueled my desire to perform this experiment for certain!

[Edited on 29-5-2014 by AJKOER]
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[*] posted on 29-5-2014 at 10:16


Quote: Originally posted by AJKOER  
If I do succeed, unless there is some modification of the dialogue on what l have clearly outlined as a speculative preparation, sorry, but I may not even consider sharing on this forum.


Could you make that: 'I won't share on this forum', please? With a firm signature?

Thank you.

[Edited on 29-5-2014 by blogfast25]




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[*] posted on 29-5-2014 at 10:30


Interestingly Blogfast, no chemistry based retort!

Please don't tell me I am starting to make sense, I am creating doubt, or worst you are starting to worry that my speculation is even possibly correct?!!


[Edited on 29-5-2014 by AJKOER]
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[*] posted on 29-5-2014 at 11:00


Quote: Originally posted by blogfast25  
It looks like I won’t be doing this today: it’s raining and part of it has to be done outside. So I’ll go back to theory a bit.

1) Sulphuric acid solution:

When H2SO4 is mixed with water, the acid deprotonates as follows:

H<sub>2</sub>SO<sub>4</sub>(aq) + 2 H<sub>2</sub>O(l) < === > 2 H<sub>3</sub>O<sup>+</sup> (aq) + SO<sub>4</sub><sup>2-</sup>(aq)

Although this is strictly an equilibrium, sulphuric acid is a very strong acid and the deprotonation is more or less complete (this is a slight simplification but OK for this discussion).

2) Oxidation and solvation of the aluminium:

The oxonium ions oxidise the Al to Al cations:

Al(s) + 3 H<sub>3</sub>O<sup>+</sup> (aq) === > Al<sup>3+</sup>(aq) + 3/2 H<sub>2</sub>(g) + 3 H<sub>2</sub>O(l)

Solvation of the Al<sup>3+</sup> ions:

Al<sup>3+</sup>(aq) + 6 H<sub>2</sub>O(l) === > [Al(H<sub>2</sub>O)<sub>6</sub>]<sup>3+</sup>(aq)

This hexaaqua aluminium cation is how Al exists in strongly acidic watery solutions.

3) Hydrolysis of [Al(H<sub>2</sub>O)<sub>6</sub>]<sup>3+</sup>(aq):

This cation is prone to hydrolysis acc.:

[Al(H<sub>2</sub>O)<sub>6</sub>]<sup>3+</sup>(aq) + H<sub>2</sub>O(l) < === > [AlOH(H<sub>2</sub>O)<sub>5</sub>]<sup>2+</sup>(aq) + H<sub>3</sub>O<sup>+</sup> (aq) (I)

A second and third step of deprotonation can give rise to solvated [Al(OH)<sub>2</sub>]<sup>+</sup> and even Al(OH)<sub>3</sub> (hydrated alumina, aka aluminium hydroxide).

But here’s an important note: when dissolving aluminium metal in acids, we use a strong excess of acid (I will be using 30 % in excess of the stoichiometric requirement) and the excess acid, which provides extra H<sub>3</sub>O<sup>+</sup>, pushes the equilibrium reaction (I) (as well as the two subsequent deprotonation reactions) strongly to the left and explains why no Al(OH)<sub>3</sub> forms. In short, in strongly acidic solutions there is no significant hydrolysis of the hexaaqua aluminium cation.

4) Crystallisation of aluminium sulphate hydrate:

As long as the solubility limit is not exceeded (for Al2(SO4)3 that is 36.4 g / 100 ml of water, at 20 C (Wiki) but it depends very strongly on temperature) the solution consist of [Al(H<sub>2</sub>O)<sub>6</sub>]<sup>3+</sup>, oxonium ions, sulphate ions, the excess of acid and of course, water.

When we exceed the solubility limit, solid aluminium sulphate hydrate must crystallise out. Assuming it is mostly octadeca (18) hydrate:

2 [Al(H<sub>2</sub>O)<sub>6</sub>]<sup>3+</sup>(aq) + 3 SO<sub>4</sub><sup>2-</sup>(aq) + 6 H<sub>2</sub>O(l) === > Al<sub>2</sub>(SO<sub>4</sub>;).18 H<sub>2</sub>O(s)

As you can see, only if crystallisation occurs, does the sulphate ion reappear in the equation, that’s why we call it a spectator ion. In essence it is there to ensure electrical neutrality of the solution/crystals.


[Edited on 29-5-2014 by blogfast25]

This raises so many more questions than it gives answers, i do understand mostly, and certainly get it. But I could spend a whole afternoon on asking questions. So I think the best thing to do is simply comment and say that I am astounded and yet suspected this as so, that WATER plays much more of part and poses more of a threat to this whole experiment. It would be absolutely impossible to work all this out (the water part) without experience. None of this could be predicted unless you are an experienced chemist I suppose. There is so much here that annoys me to be honest. Simply because I have watched literally dozens and dozens of ionic equations, reaction theories, goodness knows what else, and all from colleges and universities, not bumpkins on the side street, but yet not even one single paper or video has ever covered this topic of how water interferes with reactions not even mentioning reactions involving the H and O ions in water. This is why I am mad. I could never have worked this out. I simply slowly realised that water was a problem and I had to solve it in this reaction that I am still doing.

I guess to sum up this is an extremely complex reaction and too advanced for beginners really. Th reaction can be performed but the water involvement is a bit too difficult to understand and how you arrive at all those figures and formulas for the involvement of water. Yes I am still mad, mostly because nobody covers this topic in any thing I have read or watched.

[Edited on 29-5-2014 by CHRIS25]




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[*] posted on 29-5-2014 at 11:23


Quote: Originally posted by AJKOER  
Interestingly Blogfast, no chemistry based retort!

Please don't tell me I am starting to make sense, I am creating doubt, or worst you are starting to worry that my speculation is even possibly correct?!!


[Edited on 29-5-2014 by AJKOER]


I'll gladly put my hand in the fire that you are wrong.

The Ksp of Al(OH)3 = 3 x 10<sup>-34</sup>. That value may vary a bit from source to source and from product to product etc.

For Mg(OH)2 we have 5.6 x 10<sup>-12</sup>.

For your idea to work you'd have to find a form of Al(OH)3 that's roughly (AT LEAST) 10<sup>-12</sup>/10<sup>-34</sup> = 10<sup>22</sup> more soluble than what the tables tell you. A 1 followed by 21 zeros: some Al(OH)3 is that!

Hand in fire, thousands of dollars betted: betting against you succeeding is the safest bet I'd ever have placed.


http://www.ktf-split.hr/periodni/en/abc/kpt.html

[Edited on 29-5-2014 by blogfast25]




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[*] posted on 29-5-2014 at 11:29


Quote: Originally posted by CHRIS25  
So I think the best thing to do is simply comment and say that I am astounded and yet suspected this as so, that WATER plays much more of part and poses more of a threat to this whole experiment.


The water doesn't pose a 'threat'. You just need enough of it. Most of it gets recycled: turned into oxonium ions, which the in turn react away to hydrogen and water, for instance.

Hopefully the demo will shed some more light.




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[*] posted on 29-5-2014 at 11:49


Wow.

No wonder us noobs are confused.

In my ignorance, i made up 4 pots of some aluminium + sulphuric acid solutions today.
The quantities were carefully worked out, then adjusted - the acid volume increased by 30% and some more water added.
They are all reducing on the hotplate at the mo.

I'll post the results later, if there are any worth reporting.

AJKOER's suggestion almost had me throw some aluminium in ammonia, but it occurred to me that i have no idea what aluminium does in an organix mix, so decided against it.

As a great help, could someone say what the products look like ?
i.e.
is aluminium hydroxide Grey in solution (aka greysludge)
is aluminium sulphate .18H2O sort-of wet, a bit like toothpaste ?




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