Acid/Base Equilibria (hard)?

Question

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?

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

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

[b]3. The attempt at a solution[/b]
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?

Answer 1

pH is a logarithmic scale measuring the concentration of active H+ ions in a solution.
You can calculate it using molarity, the formula is: pH = -log H3O mol/L (log has the base 10 here).
So with given pH, you can calculate what molarity you need, and when you have that, you can calculate how many of which to add to achieve that molarity, the calculating method should be the reversal of a).

Answer 2

Forget the hydrolysis of the sulphate ion - it is negligible. This is a question about buffer solutions. You are being asked how much ammonia will be needed to form a (NH4)+/NH3 buffer solution of pH = 7.

[OH-] = Kb x [NH3]/[(NH4)+]

Set [OH-] equal to 10-7 and you're away!