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?
(NH4)2SO4 dissolves in water to form 2NH4+ and SO4(2-)
NH4 + H2O => H3O + NH3
SO4 + H2O => OH + HSO4
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?