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Author: Subject: Easy Sodium alum?
AJKOER
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[*] posted on 27-11-2013 at 06:42


Blogfast:

Stressing my use of an excess of Al is an important point, as is your depiction of the cold/slow part of the reaction.

I feel this scenario from the perspective of applying chemistry for the home chemist is also realistic. There is most likely more scrap Aluminum metal around than either NaOH or Baking Soda.

Now, as a point of disclosure, the chief product of my preparation with excess Aluminum is not so much Sodium aluminate (depending on when the process is halted), but instead a reactive insoluble Al2O3.3 H2O precursor to other Aluminum salts.

[Edited on 27-11-2013 by AJKOER]
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[*] posted on 27-11-2013 at 13:55


I've attached my simple model on alkaline Al(OH)<sub>3</sub> slurries, which basically shows that:

[Al(OH)<sub>4</sub><sup>-</sup> = K [OH<sup>-</sup>], with K a small value, fairly close to 1.

A reference cited towards the end of the *.docx confirms this model and gives a value of K = 0.059 (the reference is worth digging up, BTW).

For the carbonate/Al(OH)<sub>3</sub> slurry I created in my cold experiment, the value of [Al(OH)<sub>4</sub><sup>-</sup> ] would be in the order of 0.002 M. Negligible although slightly higher than what I expected. Especially since I started off with a carbonate solution of about 2.8 M!

This confirms what was already known: forget carbonates (or other weak bases) for preparing aluminate solutions, whatever the source of Al might be.

Attachment: On the Aluminate.docx (22kB)
This file has been downloaded 342 times

[Edited on 27-11-2013 by blogfast25]




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AJKOER
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[*] posted on 30-11-2013 at 16:13


OK, I repeated a hot run of dissolving Al in aqueous Sodium carbonate. What I observed upon repeating the restocking of the Al was that it was a little harder to dissolve all the Aluminum then initially. So, apparently, the reaction is not completely independent of the Na2CO3 concentration as would be impled by the following net reaction:

2 Al + 6 H2O --Heat, Na2CO3--> 2 Al(OH)3 + 3 H2 (g)

Next, adding a little vinegar did cause some bubbles and more Al(OH)3 precipitation. This implied that some Sodium aluminate was present in the solution containing an already evident Al(OH)3 precipitation.

An implied reaction chain consistent with these observations:

2 Na2CO3 + 2 H2O <---> 2 NaOH + 2 NaHCO3
2 NaOH + 2 Al + 6 H2O --> 2 NaAl(OH)4 + 3 H2 (g)
2 NaAl(OH)4 --Boiling--> 2 NaOH + 2 Al(OH)3 (s)
and with an excess of Aluminum:
2 NaOH + 2 Al + 6 H2O → 2 NaAl(OH)4 + 3 H2 (g)

Net:
2 Na2CO3 + 4 Al + 14 H2O --> 2 NaHCO3 + 2 NaAl(OH)4 + 2 Al(OH)3 (s) + 6 H2 (g)

or, upon rescaling:

Na2CO3 + 2 Al + 7 H2O --> NaHCO3 + NaAl(OH)4 + Al(OH)3 (s) + 3 H2 (g)


[Edited on 1-12-2013 by AJKOER]
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[*] posted on 1-12-2013 at 07:14


Quote: Originally posted by AJKOER  
OK, I repeated a hot run of dissolving Al in aqueous Sodium carbonate. What I observed upon repeating the restocking of the Al was that it was a little harder to dissolve all the Aluminum then initially. So, apparently, the reaction is not completely independent of the Na2CO3 concentration as would be impled by the following net reaction:

2 Al + 6 H2O --Heat, Na2CO3--> 2 Al(OH)3 + 3 H2 (g)


At least that is probably correct. The rate at which the Al dissolves depends on [OH<sup>-</sup>]. In the case of a weak base like a carbonate:

[OH<sup>-</sup>] = √(Kb) x √(Cb) (√ stands for square root) with Kb the equilibrium (Bronsted base) constant of the weak base and Cb the molarity of the weak base and for dilute solutions. As an aside this shows why weak bases are… erm, weak bases: for a strong bases,

[OH<sup>-</sup>] = α x Cb, with α near unity for dilute solutions.

But even for weak bases, [OH<sup>-</sup>] does depend on Cb. So higher concentration of carbonate should increase dissolution rate of Al.

Quote: Originally posted by AJKOER  
Next, adding a little vinegar did cause some bubbles and more Al(OH)3 precipitation. This implied that some Sodium aluminate was present in the solution containing an already evident Al(OH)3 precipitation.


Did you filter off the Al(OH)3 that was already present? If not judging how much extra precipitate formed would be difficult.

Quote: Originally posted by AJKOER  
Na2CO3 + 2 Al + 7 H2O --> NaHCO3 + NaAl(OH)4 + Al(OH)3 (s) + 3 H2 (g)


You make these firm claims about stoichiometry, yet your report doesn’t contain a single number of moles used or observed!

The ratio NaAl(OH)4 / Al(OH)3 = 1 that you claim here is simply impossible. Look at the www.wmsym.org/archives/2009/pdfs/9123.pdf‎ paper I linked to in my attachment. It found, as I did, that for low [OH<sup>-</sup>] (in the range of pH 12, typical of carbonate solutions):

[Al(OH)4(-)] = K x [OH<sup>-</sup>] with K ≈ 0.059, which holds<sup>*</sup> as long as there’s solid Al(OH)3 in the system.

That makes a ratio of NaAl(OH)4 / Al(OH)3 = 1 impossible, the amount of dissolved aluminate is much smaller.

Should you wish to prove me wrong you will need to determine the amounts of Al(OH)3 and aluminate accurately.

<sup>*</sup> The paper also shows that higher ionic strength and the presence of certain ions like nitrate, chloride and carbonate can increase the total amount of dissolved Al somewhat. But that suggests other complexed species, not aluminate.

[Edited on 1-12-2013 by blogfast25]

Edit:

I’ve also set up another experiment. 0.013 mole of Al(OH)3 was precipitated from ammonium alum solution with ammonia, then Buchnered and hot washed with boiling DIW plenty times. This gelatinous precipitate was then added to 50 ml of 2.8 M Na2CO3 solution. This contains about 0.14 mole of carbonate. Note the high ratio of carbonate to hydrated alumina.

I noticed no dissolution of the precipitate although it’s a little subjective of course. Certainly there was plenty precipitate left, after intense stirring. This is what it looks like:



I will allow this to equilibrate for about 24 h, stirring occasionally. Then the pH will be measured and the slurry filtered without any dilution. The filtrate will be acidified with approx. 20 ml of 100 % HOAc or until pH is close to 7. Any aluminate (or other alkali-soluble species) should precipitate to Al(OH)3.


[Edited on 1-12-2013 by blogfast25]




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[*] posted on 1-12-2013 at 10:52


I agree Blogfast with the comment "The ratio NaAl(OH)4 / Al(OH)3 = 1 that you claim here is simply impossible". It was not my intention to imply that the ratio was was most likely unity, but just that there was a possible mixture of the hydroxide and aluminate. Even the presence of CO2 (which may be a function of the volume of the reaction vessel) may alter the relativity causing Al(OH)3 precipitation. Interestingly, the form of Al(OH)3 from the action of the Acetic acid on the aluminate appeared different to me from that which was already present in the solution (a bit finer).

Good pictures, again. Mine are from my iPhone, and at times, are even upside down!
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[*] posted on 1-12-2013 at 11:16


Quote: Originally posted by AJKOER  
Even the presence of CO2 (which may be a function of the volume of the reaction vessel) may alter the relativity causing Al(OH)3 precipitation.

Interestingly, the form of Al(OH)3 from the action of the Acetic acid on the aluminate appeared different to me from that which was already present in the solution (a bit finer).



How can CO2 affect the 'relativity causing Al(OH)3 precipitation'? What do you mean by the latter anyway?

What these gelatinous precipitates ('hydroxides' in general) look like depends on many factors, notably concentration from which precipitation was forced and pH. These factors are often used to control filtrability of a precipitate.

[Edited on 1-12-2013 by blogfast25]




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[*] posted on 1-12-2013 at 13:13


To the question "How can CO2 affect the 'relativity causing Al(OH)3 precipitation'?", see http://www.usalco.com/wp-content/uploads/2013/05/Sodium_Alum...

To quote:
"1. The Introduction of Carbon Dioxide.
Carbon Dioxide in the air reacts with free caustic present in the Liquid Sodium Aluminate, lowers pH and disturbs the stability. All mixing of air and Sodium Aluminate solutions should be avoided.
2. Dilution with Water.
Dilution changes the product concentration and.."

So, the presence of CO2 can effect the 'relative' NaOH/NaAlO2 mix, lower pH,...

Interestingly, the same source also states:

"4. Excessive Agitation.
The mechanical effect of high shear while exposed to the atmosphere, may contribute to instability. Examples would include turbulent mixing with a propeller in an open tank or a leaking centrifugal pump seal / pump suction piping.
5. Prolonged Storage Time.
The recommended maximum storage time is three months for 38% Liquid Sodium Aluminate and one month for 45% Liquid Sodium Aluminate. Longer storage time is possible with optimum storage systems.
6. Excessive Heating.
Avoid hot spots exceeding 120°F.
7. Loss of Moisture.
Loss of moisture due to prolonged storage at elevated temperatures in open top tanks will lead to instability."

So boiling of the solution would appear to violate Points (4), (6) and (7) as well as possibly increasing CO2 contact.
----------------------------------------------------------

A more accurate (but less simple) reaction chain, which still ignores CO2 (which may not be just from the air, but from the action of undestroyed, by heating, Citric acid founded in common Baking Soda, used to prepare the Na2CO3, upon hydration, a path that occurs in baking a cake):

2 Na2CO3 + 2 H2O <---> 2 NaOH + 2 NaHCO3
2 NaOH + 2 Al + 6 H2O --> 2 NaAl(OH)4 + 3 H2 (g)
2x NaAl(OH)4 --Boiling--> 2x NaOH + 2x Al(OH)3 (s)
and with an excess of Aluminum:
2x NaOH + 2x Al + 6x H2O → 2x NaAl(OH)4 + 3x H2 (g)

Net:
2 Na2CO3 + 2(x+1) Al + (8+6x) H2O --> 2 NaHCO3 + 2(x+1) NaAl(OH)4 + 2x Al(OH)3 (s) + 3(x+1) H2 (g)

or, upon rescaling:

Na2CO3 + (x+1) Al + (4+3x) H2O --> NaHCO3 + (x+1) NaAl(OH)4 + x Al(OH)3 (s) + 1.5*(x+1) H2 (g)

where x, per Blogfast, is <1, and perhaps could be measured by recording the incremental increase in Hydrogen formed as a consequence of moving from a cold (or low heat) reaction state to rapid prolonged boiling (with possible water replenishment).


[Edited on 2-12-2013 by AJKOER]
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[*] posted on 3-12-2013 at 11:38


Yes, CO2 as an industrially used weak acid. But air CO2 is likely to have very little effect here.

I had to modify my experiment slightly because of a vacuum problem and because my GAA had frozen over again. So, gravity filtration overnight and neutralisation with about 20 ml HCl 37%, followed by accurate pH adjustment with drops of NH3 33 % to pH ≈ 7 – 8 was done today.

The resulting treated filtrate shows very slight turbidity, but only very slight. This photo is less focussed than the one above but I can assure you that you can see through this liquor easily. Just below the meniscus you can see the embossing pattern of the roll of kitchen towels placed behind the beaker, through the liquor. There is almost no Al(OH)3 present, as expected:



This shows simply that carbonate solutions can solvate small amounts of Al(OH)3 to aluminate. Those same carbonate solutions dissolve aluminium fairly slowly, but mostly with formation of Al(OH)3, not aluminate.


[Edited on 3-12-2013 by blogfast25]




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[*] posted on 4-12-2013 at 21:04


Quote: Originally posted by AJKOER  
.....
A more accurate (but less simple) reaction chain, which still ignores CO2 (which may not be just from the air, but from the action of undestroyed, by heating, Citric acid founded in common Baking Soda, used to prepare the Na2CO3, upon hydration, a path that occurs in baking a cake):

2 Na2CO3 + 2 H2O <---> 2 NaOH + 2 NaHCO3
2 NaOH + 2 Al + 6 H2O --> 2 NaAl(OH)4 + 3 H2 (g)
2x NaAl(OH)4 --Boiling--> 2x NaOH + 2x Al(OH)3 (s)
and with an excess of Aluminum:
2x NaOH + 2x Al + 6x H2O → 2x NaAl(OH)4 + 3x H2 (g)

Net:
2 Na2CO3 + 2(x+1) Al + (8+6x) H2O --> 2 NaHCO3 + 2(x+1) NaAl(OH)4 + 2x Al(OH)3 (s) + 3(x+1) H2 (g)

or, upon rescaling:

Na2CO3 + (x+1) Al + (4+3x) H2O --> NaHCO3 + (x+1) NaAl(OH)4 + x Al(OH)3 (s) + 1.5*(x+1) H2 (g)


I just noticed an error in summing to the net reaction, which would be significant in any quantitative determination for a value of x. The corrected net equation:

Net:
2 Na2CO3 + 2(x+1) Al + (8+6x) H2O --> 2 NaHCO3 + 2 NaAl(OH)4 + 2x Al(OH)3 (s) + 3(x+1) H2 (g)

or, upon rescaling:

Na2CO3 + (x+1) Al + (4+3x) H2O --> NaHCO3 + NaAl(OH)4 + x Al(OH)3 (s) + 1.5*(x+1) H2 (g)

My second experiment suggests, however, that there is, in fact, a limit to the Aluminum excess in the amount of 'x' that can be timely converted to Al(OH)3 via boiling.


[Edited on 5-12-2013 by AJKOER]
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