Sciencemadness Discussion Board - Reacting NaOH(aq) with Al(s)

Reacting NaOH(aq) with Al(s)

NiK - 23-11-2013 at 11:43

So in a successful attempt to capture hydrogen gas in a balloon and blow it up, I reacted NaOH with Al foil in a 250mL beaker (forgetting that strong bases can dissolve glass, but I seem to have done alright).

My left over solution is black and slightly thicker in viscosity than water. I'm under the impression that my solution left over is Na₃AlO₃, but it's black...is it supposed to be?? Google images of "Sodium Aluminate" don't show any black liquid.

When I was making my NaOH solution I had a little bit of contamination from some CuSO₄ particles, but extremely small amounts..however I did notice they had turned black, could this have to do with the black color of the solution?

I reacted 30g NaOH (in 200ml of d-water) and 10g Al foil; I added the foil to the solution in a flask...the balanced formula had a 3:1 ratio of NaOH to Al and for some reason I completely spaced the whole mole concept, maybe I started too soon after just waking up! Looks like I just reacted too much Al foil..but it all reacted sooo idk what's up.

Anyways, let me know what you guys think!

blogfast25 - 23-11-2013 at 11:51

The colour is almost certainly due to some contamination or other. Pure sodium aluminate solutions are clear and colourless.

CuSO4 and alkali give Cu(OH)2 which on heating quickly dehydrates to black CuO, so it's a likely cause of the dicolouration.

Strong bases don't dissolve glass. Very strong, hot alkaline solution superficially attack glass but I wouldn't call that 'dissolve'. Conditions dependent, of course.

papaya - 23-11-2013 at 12:08

Impurities in Al, and/or aluminate hydrolyses to Al2O3xnH2O, ..
Better ask how much base is needed to get m grams of Al dissolved (and hydrogen liberated), since real answer is not obvious (to me also). Because if aluminate can hydrolyze(and Al2O3 is quite inert) then some base is recovered and more Al can react.. If correct, then with relatively small amount of base one can get unexpected amounts of hydrogen if waits very long.. correct?

NiK - 23-11-2013 at 12:15

Quote: Originally posted by blogfast25

The colour is almost certainly due to some contamination or other. Pure sodium aluminate solutions are clear and colourless.

CuSO4 and alkali give Cu(OH)2 which on heating quickly dehydrates to black CuO, so it's a likely cause of the dicolouration.

Strong bases don't dissolve glass. Very strong, hot alkaline solution superficially attack glass but I wouldn't call that 'dissolve'. Conditions dependent, of course.

Thanks for the feedback. Yea it was definitely a very exothermic reaction so that could have heated the CuSO4 and made it turn black...I'm just reluctant to believe that such a small amount would turn the whole solution black, given there's only about 50mL of it but there was probably about .005g or less, probably less because I doubt it would have even come close shown up on my scale that reads in .01. They were so small I call them particles and I could almost not tell if they were black or if it was just lack of light; I of course answered that question by putting a tiny amount of CuSO₄(s) with a drop or two of my NaOH solution and the whole crystal turned black and got soft.

So in the future it is safe to use my glassware to work with conc. NaOH solutions?? What do you mean by superficially attack glass?

Lastly, is my black solution worth keeping for any kind of fun/demonstrative reactions? It's got Al in it so maybe some kind of replacement reaction? (Though Al is pretty high on the reactivity series isn't it?)

Thanks again for the feedback!

NiK - 23-11-2013 at 12:22

Quote: Originally posted by papaya

Impurities in Al, and/or aluminate hydrolyses to Al2O3xnH2O, ..
Better ask how much base is needed to get m grams of Al dissolved (and hydrogen liberated), since real answer is not obvious (to me also). Because if aluminate can hydrolyze(and Al2O3 is quite inert) then some base is recovered and more Al can react.. If correct, then with relatively small amount of base one can get unexpected amounts of hydrogen if waits very long.. correct?

I'm not sure I quite understand all that you said; I'm pretty amateur when it comes to chemistry. But I was using the balanced equation:

2Al (s) + 6NaOH (aq) → 3H2 (g) + 2Na3AlO3 (aq)

So with 30g of NaOH I should have used 6.xx grams of Al; I used 10g. Which would give me about 3g of hydrogen gas. But I'm not sure why the excess Al still reacted; it sounds like you offered an explanation for that, but like I said I'm pretty amateur, I only have 1 year of high school chem and 1 semester of college chem 101 plus what I have learned on my own time.

papaya - 23-11-2013 at 12:33

I meant
2Na3AlO3 + 3H2O -> 6NaOH + Al2O3
which may or may not be reversible process. I never saw Al to dissolve completely in NaOH and give no precipitate, which means some hydrolysis I mentioned takes place. Also a thing like Na3AlO3 most probably won't exist in water solution, it's just a simplification, more complex (even polymeric) anions containing Al3+ will be involved, sorry I can't explain better (nor I have expertise on this).

blogfast25 - 23-11-2013 at 12:50

Quote: Originally posted by papaya

If correct, then with relatively small amount of base one can get unexpected amounts of hydrogen if waits very long.. correct?

Complete baloney.

The reaction is:

NaOH (aq) + Al(s) + 3 H2O (l) === > NaAl(OH)4(aq) + 3/2 H2(g)

… and nothing else. Aluminates hydrolyse to Al(OH)3 at lower pH but that has nothing, nada, zilch to do with anything here.
So 40 g of NaOH and 27 g of Al would (ideally!) form 3 g of hydrogen, about 1.5 x 24 L of gas at STP.

Quote: Originally posted by NiK

2Al (s) + 6NaOH (aq) → 3H2 (g) + 2Na3AlO3 (aq)

So with 30g of NaOH I should have used 6.xx grams of Al; I used 10g. Which would give me about 3g of hydrogen gas. But I'm not sure why the excess Al still reacted; it sounds like you offered an explanation for that, but like I said I'm pretty amateur, I only have 1 year of high school chem and 1 semester of college chem 101 plus what I have learned on my own time.

No, he hasn’t.

Superficial attack means just that: a thin layer of glass, imperceptible to the eye even, may be dissolved off. In my case, even with hot 50 w% NaOH the attack on a Pyrex sauce jug gave no reason to throw away the jug.

[Edited on 23-11-2013 by blogfast25]

papaya - 23-11-2013 at 12:56

Proofs, blogfast?

blogfast25 - 23-11-2013 at 15:02

Idiots like you demand proof the sky is blue.

Consult any decent text book on Al amphoterism.

Now please f*ck *ff or stop trolling. You're spreading misinformation here. STOP IT. :mad:

Quote: Originally posted by papaya

(nor I have expertise on this).

Is about the only thing that's true from you on this thread.

If you do want to use an anhydrous notation of Na aluminate it is NaAlO2.

[Edited on 23-11-2013 by blogfast25]

papaya - 23-11-2013 at 15:34

Not going to continue, your behavior is unacceptable, detritus?

weiming1998 - 23-11-2013 at 16:06

I don't think the precipitate has anything to do with Al2O3, which is white instead of black. Instead, the answer is probably the aluminium foil, which has a small about (1-0.5%?) of iron/manganese/other metals added in. Iron and some of the others will not dissolve in a basic solution, so it is left behind as a black precipitate. The solution has a higher viscosity because a lot of solute is dissolved within.

Filtering the black solution results in a clear solution that remains clear unless you add acid to it (I've done this before).

Crowfjord - 23-11-2013 at 17:05

The black stuff is probably silicon (along with other metals). I get it a lot at work when I digest cast aluminum (in acid, mind you) for ICP-AES. I don't know if Al foil is cast, though.

[Edited on 24-11-2013 by Crowfjord]

blogfast25 - 24-11-2013 at 06:20

Quote: Originally posted by papaya

Not going to continue, your behavior is unacceptable, detritus?

Learn to write. This has been moved to beginnings because this chemistry is well understood and uncontroversial. Unlike your brainfarts.

Quote: Originally posted by Crowfjord

The black stuff is probably silicon (along with other metals). I get it a lot at work when I digest cast aluminum (in acid, mind you) for ICP-AES. I don't know if Al foil is cast, though.

Silicon is also amphoteric and thus also dissolves in strong sodium hydroxide solutions, forming silicates. It is not soluble in or reactive towards most acids (although very strong HNO3 may oxidise it to silica). I've dissolved Al scrap in both acids and alkalis on practically industrial scale and always get clear solutions, although the odd black speck may occur.

The colour is most likely due to the contamination he's already confessed to: dirty glassware with CuSO4 on it.

[Edited on 24-11-2013 by blogfast25]

TheChemiKid - 24-11-2013 at 06:39

Well, I must say the copper sulfate seems reasonable, but I think it could also be because of impure aluminum. When I have used aluminum foil for this reaction, it turns black. When I have used pure aluminum powder, it stays colorless.

Moderator Warning

bfesser - 24-11-2013 at 11:06

blogfast25, I've had enough of your arrogance and rudeness in this thread. You are not to reply here again, unless you wish to apologize to papaya for your insults and to everyone else for your general rudeness. The attitude you've displayed here is appalling, and your behavior is unacceptable. If you do choose to apologize, I expect to see some <del>goddamn</del> references—the real reason why I moved this to Beginnings. I strongly suggest that you review The ScienceMadness Guidelines before making any more posts on this site. Finally, in the future, don't take it upon yourself to try explaining actions or views of moderators. If I felt that it was necessary, I would have posted the reason myself.

[edit] I apologize for my use of what some consider to be profanity. I was clearly angry at the time, but realize now that it was hypocritical and inappropriate behavior for a moderator. Sorry to anyone who may have been offended.

<hr width="80%" />
[edit] Back to the Chemistry...

NiK, I will post some equations and references for your consideration.

First, <a href="http://en.wikipedia.org/wiki/Aluminium_oxide#Amphoteric_nature" target="_blank">disruption of</a> <img src="../scipics/_wiki.png" /> the <a href="http://en.wikipedia.org/wiki/Passivation_(chemistry)#Aluminium" target="_blank">Al2O3 passivation layer</a> <img src="../scipics/_wiki.png" />:

   Al2O3 + 6 NaOH + 3 H2O → 2 Na3Al(OH)6

Then, <a href="http://en.wikipedia.org/wiki/Sodium_hydroxide#Reaction_with_amphoteric_metals_and_oxides" target="_blank">dissolution of Al</a> <img src="../scipics/_wiki.png" />, forming <a href="http://en.wikipedia.org/wiki/Sodium_aluminate#Reaction_of_aluminium_metal_and_alkali" target="_blank">hydrated aluminate</a> <img src="../scipics/_wiki.png" />:

   2 Al + 2 NaOH + 6 H2O → 2 Na[Al(OH)4] + 3 H2

<a href="http://books.google.com/books?id=EvTI-ouH3SsC&pg=PA224#v=onepage&q&f=false" target="_blank">7.2.4. Chemical reactivity and trends</a> <img src="../scipics/_ext.png" /> (Chemistry of the Elements by A. Earnshaw & N. Greenwood @ Google Books)

[Edited on 25.11.13 by bfesser]

NiK - 24-11-2013 at 12:20

bfesser, thank you! I appreciate those links; I will review them to further my understanding of what went on in my reaction!

bfesser - 24-11-2013 at 17:17

As Crowfjord suggested, the black residue is probably Si. <del>If I recall correctly (from How It's Made

, it's added to the Al to prevent it from sticking during the cold rolling process which is used to transform it from an ingot to a foil.</del> [I didn't recall correctly.]

Attachment: al_foil_msds.pdf (85kB)
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<a href="http://www.esabna.com/us/en/education/knowledge/qa/How-and-why-alloying-elements-are-added-to-aluminum.cfm" target="_blank">How and why alloying elements are added to aluminum</a> <img src="../scipics/_ext.png" />

[edit] I guess they never said why, after all: <a href="http://www.youtube.com/watch?v=6OVIhXn3I3E">How It's Made: Aluminum Foil</a> <img src="../scipics/_yt.png" />

[Edited on 25.11.13 by bfesser]

NiK - 25-11-2013 at 07:19

Quote: Originally posted by bfesser

<a href="http://www.esabna.com/us/en/education/knowledge/qa/How-and-why-alloying-elements-are-added-to-aluminum.cfm" target="_blank">How and why alloying elements are added to aluminum</a> <img src="../scipics/_ext.png" />

Thanks, neat link! Says that one of the most common uses for the 1xxx series aluminum is foil and that's the series that's 99.00% Al or greater. I guess that 1% can still make a difference though.