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Offline Mr Andrew

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Alum crystals
« on: 28/10/2006 17:50:54 »
In my chemistry class, we synthesized alum crystals in a lab and then heated some of them to find the number of waters of hydration.  Anyway, once heated, we had anhydrous alum which isn't very useful so, wanting to rescue all the alum for a later lab (involving organic dyes), I heated an aqueous solution of anhydrous alum and cooled it overnight.  When I came back, I had a white residue in the bottom of the beaker (it looked a lot like the anhydrous alum I added the day before).  This residue was definitely not alum crystals.  What went wrong?  I can't think of anything and neither can any of my classmates or my teacher.  Please help me figure this out, I'd like to recover these crystals if at all possible.


 

Offline lightarrow

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Re: Alum crystals
« Reply #1 on: 28/10/2006 19:58:29 »
The white residue is Al2O3 (aluminum oxide). Actually, you didn't get pure anhydrous alum, but a mix of it and Al2O3.
All aluminum salts hydrolize in water:

Al3+ + 3H2O ↔ Al(OH)3 (aluminum hydroxide) + 3H+

Heating, aluminum hydroxide loses water and becomes oxide.

In order to make anhydrous alum (assumed it's possible at all), you should have acidified the initial solution with H2SO4.
This to invert the sense of the up reaction.
You need H2SO4 and not another acid (for example HCl) because alum is made of sulphate ions (SO42-).

« Last Edit: 28/10/2006 20:07:04 by lightarrow »
 

Offline Mr Andrew

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Re: Alum crystals
« Reply #2 on: 01/11/2006 03:31:45 »
lightarrow, thanks for responding.

So, if I understand you correctly, you are saying that I have aluminum hydroxide powder in my beaker instead of alum crystals because the sulfates left when I heated the crystals in the crucible.  If I wanted to make alum again could I add KOH to make [Al(OH)4]- ions and then add sulfuric acid to make alum?  I would then have to heat the solution up and cool it overnight again to make crystals appear, right?  I think I will try this this week.

Thanks.
 

Offline lightarrow

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Re: Alum crystals
« Reply #3 on: 03/11/2006 09:13:23 »
lightarrow, thanks for responding.

So, if I understand you correctly, you are saying that I have aluminum hydroxide powder in my beaker instead of alum crystals because the sulfates left when I heated the crystals in the crucible.  If I wanted to make alum again could I add KOH to make [Al(OH)4]- ions and then add sulfuric acid to make alum?  I would then have to heat the solution up and cool it overnight again to make crystals appear, right?  I think I will try this this week.

 If you have aluminum hydroxide, then you can just add sulfuric acid to make aluminum sulphate. If you have aluminum oxide, especially after having heated it, it's very difficult to dissolve it; hot KOH could do it, but which container would you use? I know only silver, for this purpose (everything else would dissolve as well!).
« Last Edit: 03/11/2006 09:15:02 by lightarrow »
 

Offline Mr Andrew

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Re: Alum crystals
« Reply #4 on: 04/11/2006 20:23:51 »
Alum is KAl(SO4)2.  You probably know that already but I want to emphasize the two sulfates.  I think you have to add an excess of OH- ions to make the Al(OH)4- anion which, when sulfuric acid is added, will form Al(SO4)2-.  If you used plain old aluminum hydroxide, you would get Al2(SO4)3, am I right?  This is not good because it is the Al(SO4)2- is an ion which would be attracted to the K+ ions in the KOH and form alum.  Aluminum hydroxide is not an ion so it would not react with the potassium.

As for if I have aluminum oxide, I think I'll just throw away the solution and live with the crystals I have left (which is a good amount so it wouldn't be a big deal if I couldn't recover the lost ones).  I have absolutely no idea how I would use hot KOH at all.  I have no silver to put it in.

I'm not much good at inorganic chemistry but wouldn't anything less reactive than potassium  be able to contain KOH?  Or because it's hot, does the activity series not apply anymore?  Hmm.... ???
 

Offline lightarrow

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Re: Alum crystals
« Reply #5 on: 05/11/2006 17:06:22 »
To form KAl(SO4)2 in water solution you only need K+, Al3+ and SO42- ions in the same solution, with the right pH (I think it should be ≈ less than 3). So you can dissolve Al(OH)3 with H2SO4 to have Al3+ and SO42-; of course you still need K+ ions to form the alum.

Al2(SO4)3 is water soluble, and dissociate in water, so you don't have to worry about it.

Quote
As for if I have aluminum oxide, I think I'll just throw away the solution and live with the crystals I have left (which is a good amount so it wouldn't be a big deal if I couldn't recover the lost ones).
Good strategy!

Quote
I'm not much good at inorganic chemistry but wouldn't anything less reactive than potassium  be able to contain KOH?  Or because it's hot, does the activity series not apply anymore?
A lot of things are less reactive than potassium, almost all, indeed.
The problem here is not KOH in itself, but strong basis. A hot, concentrated solution of a strong base as KOH, NaOH, ecc. or, even worse, melted KOH, NaOH, ecc. are very reactive and attack almost everything. Glass, for example, is one of their preferred "meal"!

Another, but more complicated, way to dissolve compact Al2O3 makes use of melted K2S2O7. You mix the powdered oxide (this method is also used for other very inert oxides as Fe2O3 TiO2 ecc.) with a greater amount of KHSO4 in a crucible and you heat it gently for some minutes. In this way, water is released from KHSO4 while it forms a foam, and K2S2O7 is left.

Then (you know there isn't water anymore because vapour doesn't form anylonger) you heat stronger. The K2S2O7 decomposes again, this time releasing very hot SO3 (extremely reactive!) which reacts with the oxide forming the sulphate. At the end, you add distilled water acidified with H2SO4 to dissolve the sulphate...et voilą, the extremely inert oxide is taken under solution.
 

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Re: Alum crystals
« Reply #5 on: 05/11/2006 17:06:22 »

 

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