Is it possible to calculate that X (known) Kg of SO2 = Y (solve) gallons of "typical" acid rain?

Question

I need to create a mathematical equation to use in the interpretation of Life Cycle Assessment results. The LCA output is kg of SO2 (sulfur dioxide). 



My thought is X number of gallons of Acid Rain (maybe it's Olympic swimming pools). Some assumptions need to be made: 1) the pH of the acid rain is 4.3 AND 2) 65% of the acidity is due to H2SO4 (the other 35% is HNO3).

Answer 1

You'll have to make quite a few assumptions. Here are a few of the issues:

1) SO₂ is the acid anhydride of sulfurous acid (H₂SO₃) which is a weak acid (doesn't acidify water nearly as much as sulfuric.

2) If you had SO₃, you could assumes that that will yield H₂SO₄ 1:1 in water by moles and that you will get 2 moles of H+ for every mole of sulfuric acid.

3) If you assume that only sulfuric acid produces the acid rain, you can solve for molarity (moles/liter) of H+ for pH of 4.3 ([H⁺] = 10^-pH) . You can then divide the moles of H⁺ derived from the SO₃ and divide by the molarity to find liters of acid rain solution.

4) You could figure out what part of the H⁺ comes from sulfuric acid. This seems arbitrary in that you assume nitric acid comes from somewhere in just the right proportion to create acid rain. You would end up with the same volume as in (3) divided by .65.

Not completely satisfying response, I'm sure. Take care.