If the pH of a 1.00-in. rainfall over 1300 miles2 is 3.30, how many kilograms of sulfuric acid, H2SO4, are present, assuming that it is the only acid contributing to the pH?
For sulfuric acid, Ka1 is very large and Ka2 is 0.012.
Volume of rainfall = Area X height
Height = 1 in = 0.254 dm
1mile2 = 2.59 X 10^8 dm^2
so 1300 miles^2 = 3367 X 10^8 dm^2
Volume = 0.254 X 3367 X 10^8 = 855.218 X 10^8 dm^3
H2SO4 ionizes as
H2SO4 --> H+ + HSO4-
Ka1 is very high so it completely dissociates , due to first dissociation [H+] = Concentraion of H2SO4
HSO4- --> H+ + SO4-2
We also consider it as high dissciation
so each mole of H2SO4 will give 2 moles of H+ (proton)
pH = 3.30 = -log[H+]
So [H+] = 0.000501 M = 0.000501 moles / L or moles / dm^3
Volume = 855.218 X 10^8 dm^3
so moles of H+ present in this volume = 0.000501 X 855.218 X 10^8 = 4.28 X 10^7 moles
So moles of H2SO4 = 4.28 X 10^7 / 2 = 2.14 X 10^7 moles
Molecular weight of H2SO4 = 98 g / mole
so mass of H2SO4 = Moles X molecular weight = 98 x 2.14 X 10^7 grams = 209.72 X 10^7 grams = 2.09 X 10^6 Kg
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