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How does (AlK(SO4)2•12H2O) balance?

  1. Oct 5, 2011 #1
    So like say you have (AlK(SO4)2•12H2O)
    And then how does the reaction go?

    Is it like (AlK(SO4)2•12H2O) -> AlOH + H2SO4 or something like this?

    Was also wondering how like
    (NaCH3COO) would work with H2O as well...
    same for this
    (Cu(NO3)2•2.5H2O)

    These are acid-base, and I am pretty new to them, could I get some clarifications? Thanks
     
  2. jcsd
  3. Oct 6, 2011 #2

    Borek

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    Staff: Mentor

    What reaction?

    Please ask homework (and homework type) questions in appropriate forum.
     
  4. Oct 6, 2011 #3
    Sorry, and It's just reacting with plain o'll water.
     
  5. Oct 6, 2011 #4

    Borek

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    Staff: Mentor

    So you are asking about dissolution and dissociation.

    In most cases dissociation is just about separating ions. What ions do you see in the alum molecule?

    Don't worry about water of crystallization - it will just became part of the solution, there is already plenty of water around.
     
  6. Oct 6, 2011 #5
    Just Al3+ K- and SO4 2-
     
  7. Oct 7, 2011 #6

    Borek

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    No such thing as K-.

    So if you have Al3+, why do you try to put something like AlOH between products?

    Are you expected to know all the details of cation hydrolysis in such solutions?
     
  8. Oct 7, 2011 #7
    Yeah I believe so.
     
  9. Oct 7, 2011 #8

    Borek

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    Staff: Mentor

    That's a tricky thing - Al3+ creates a series of complexes, 4 at least (or rather I have equilibrium constants for 4 reactions). But these are things like Al3(OH)45+ - I don't think anybody by graduate students should worry about them.
     
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