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Acid/Base Equilibria (hard, i think).

  1. Apr 4, 2007 #1
    1. The problem statement, all variables and given/known data
    1 oz of 80% pure (NH4)2SO4, 1 gallon of water @ 25 degrees Celsius.
    1 gallon = 3.785 L, 1 oz = 28.4 grams.

    Suppose an avid gardener prepares a solution by putting 1 oz of 80% pure Ammonium Sulfate in a gallon of water at 25 degrees C.
    a) what is concentration of HSO4 and NH3 in solution at equlibrium.
    b) suppose you want to make sure pH is at 7.0. If you have pure NH3 and H2SO4 at hand, how many grams (of which one) would you have to add?

    2. Relevant equations
    (NH4)2SO4 dissolves in water to form 2NH4+ and SO4(2-)

    NH4 + H2O => H3O + NH3
    SO4 + H2O => OH + HSO4

    3. The attempt at a solution
    i already finished part a by first changing oz and gallons into moles/L => Molarity. then i calculated equilibrium amounts by initial, change, end calculations.

    my main problem lies in (b) as i have no clue how to approach this problem. can someone help?
  2. jcsd
  3. Apr 4, 2007 #2
    NH3 is a strong base while H2SO4 is strong acid. Since water has a leveling effect, almost 100% of a strong acid yields H3O and almost 100% of a strong base yields OH. This means that when you mix a strong base and a strong acid in equal quantities, the concentration of OH and H3O is almost the same. From Le Chatelier principle, some of the OH and H3O will react to produce H2O (inverse of autoionization). This said, the final concentration of OH and H3O is very close to 10^-7. So you need the same concentration (once mixed in) of each in order to get a Ph of 7.
  4. Apr 8, 2007 #3


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    Staff: Mentor

    It is not. I am afraid that's not the only reason that makes your discussion invalid, but the most obvious one.
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