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Chemistry

I am predicting acid-base equilibria using the Brondsted-Lowry predictions. I am asked to do "solid aluminum sulfate added to water".

So I know there is an Al^2+ ion and a SO4^-2 ion, but does water dissociate into H30^+ and OH^-? If so, then H30^+ and OH^- would be the strongest acid and base...and I would end up with a neutral reaction (shouldn't it be acidic?). If not, then I would only have a strong base (SO4^-2), and no acid to work with. Help!

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Stuck

  1. Also, if products are favoured, does that mean that the resulting solution is basic? And if reactants are favoured, the solution is acidic? Or is there no distinction?

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    Stuck

  2. Let's start over.
    Al2(SO4)3 + H2O ==> Al^+3 (you had +2) + SO4^-2. Not balanced.
    But it doesn't stop there.
    Al^+3 actually is not that but Al(H2O)6 with +3 charge in solution. It is a hydrated ion with a +3 charge. It is acidic because it ionizes. It's tough to write on the board because I can't do subscripts and superscripts. Here is what happens. One of the H2O molecules that is in the hydrated ion splits into H^+ and OH^-. The H^+ reacts with H2O to form the H3O^+. What is left is the Al(H2O)5(OH)^+2. Now that you know what happens I'll do my best to write the equation.
    Al(H2O)6(+3 charge) + H2O ==> Al(H2O)5(OH)^+2 + H3O^+.
    The solution is acidic because of the H3O^+. The products are favored. But that means the solution is acidic. Where is this knowledge useful. Some plants don't grow well in basic soil; they need acidic soil. If a person has soil that is basic, s/he adds a little Al2(SO4)3 to the soil (solid S will do the same thing but it's a different reaction) and it becomes acidic (pH 5.5 -6--something like that) and those acidic plants thrive. (The S works because it's a bacteria found in the soil that eats on it and converts it to an acid.).

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    DrBob222

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