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Why can't you remove hydroxide from Al(OH)3 with alkali metals?

  1. Nov 10, 2013 #1
    So I've tried to research a bit on refining aluminum without fluorine, and I came up with the idea of reacting the hydroxide off aluminum hydroxide with sodium or potassium. Only problem is, the reaction doesn't work, according to what I've found. I'm curious why, as the electronegativity table seems to indicate that Na or P should preferentially react with the hydroxide ions.

    I appear to have a gap in my reasoning somewhere, so what am I missing?
     
  2. jcsd
  3. Nov 10, 2013 #2

    Borek

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    Staff: Mentor

    No idea what kind of reaction you are talking about, please write the reaction equation.
     
  4. Nov 10, 2013 #3
    If it worked, it would be 3Na + Al(OH)3 → Al + 3NaOH.
     
  5. Nov 11, 2013 #4

    Borek

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    Staff: Mentor

    OK

    Electronegativities are not the only thing that matters. What matters is the thermodynamical stability of products. In this case mixture of Al and NaOH is not stable - Al nicely dissolves in the alkali producing sodium aluminate. In the presence of water reaction is (more or less)

    2NaOH + 2Al + 2H2O → 2NaAlO2 + 3H2

    When there is no water around situation gets more complicated, but you can still expect hydrogen to evolve and some kind of aluminate or mixed oxide to be produced. That's because Al is amphoteric, and its hydroxide can react both as an acid and as a base.
     
  6. Nov 11, 2013 #5
    So for the non-aqueous reaction, it would be something like, but not necessarily, 4Na + 2Al(OH)3 → 2NaAlO2 + 2NaOH + H2?
     
  7. Nov 11, 2013 #6

    Borek

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    I would check if there is no mixed oxide like Na2O·Al2O3 - which would be a most likely product then.

    Many salts can be described as mixed oxides, and they are present in many minerals (see for example http://en.wikipedia.org/wiki/Spinel_group).
     
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