Question

If the pH of a 1.00-in. rainfall over 1300miles2 is 3.20, how many kilograms of sulfuric...

If the pH of a 1.00-in. rainfall over 1300miles2 is 3.20, how many kilograms of sulfuric acid, H2SO4, are present, assuming that it is the only acid contributing to the pH?

The presence of SO2 in the atmosphere and the sulfuric acid that it produces result in the phenomenon of acid rain. Uncontaminated rainwater is naturally acidic and generally has a pH value of about 5.6. The pH of acid rain is about 4. This acidity has affected many lakes in northern Europe, the northern United States, and Canada, reducing fish populations and affecting other parts of the ecological network within the lakes and surrounding forests.

The pH of most natural waters containing living organisms is between 6.5 and 8.5, but freshwater pH values are far below 6.5 in many parts of the continental United States. At pH levels below 4.0, all vertebrates, most invertebrates, and many microorganisms are destroyed. More than 300 lakes in New York State contain no fish, and 140 lakes in Ontario, Canada, are devoid of life. The acid rain that appears to have killed the organisms in these lakes originates hundreds of kilometers upwind in the Ohio Valley and Great Lakes regions. Some of these regions are recovering as sulfur emissions from fossil fuel combustion decrease, in part because of the U.S. Clean Air Act of 1990, which required that power plants reduce their sulfur emissions by 80%.

Since
[H+] [SO4--] / [HSO4-] = Ka2
then at pH 3.2
[SO4--] / [HSO4-] = Ka2 / 10^-3.2 = 19
Since all of the H2SO4 is deprotonated, the total amount of sulfuric acid is
[SO4--] + [HSO4-] = 19 [HSO4-] + [HSO4-] = 20 [HSO4-]
The proportion of total sulfuric acid that's doubly deprotonated is
[SO4--] / 20 [HSO4-] = 19/20
and the proportion that's singly deprotonated is
[HSO4-] / 20 [HSO4-] = 1/20
The moles of hydrogen ion per mole of sulfuric acid at this pH is
(19/20)*2 + (1/20)*1 = 39/20
The amount of sulfuric acid in the rain then is
[H+] (1 mol sulfuric acid / 39/20 mol H+) = (20/39) 10^-3.2 mol/L
Multiply by the molecular weight of H2SO4 for grams acid per liter, then multiply by the number of liters in 1" x 1400 square miles for the total mass of acid.