Question

If the pH of a 1.00-in. rainfall over 1700 miles2 is 3.70, 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.

Answer 1

pH = 3.70

total volume

V = 1 in * 1700 mi^2 = (1.57828*10^-5 miles)(1700 miles^2) = 0.02683076 mi^3

change to m3

0.02683076 mi^3 * (1609 m/ 1mi) ^3 = 111763786.45 m3

change to liters

V = 111763786.45*10^3 L = 111763786450 Liters

now..

[H+] = 10^-pH = 10^-3.7 = 0.0001995 M

for 1st ionization,

100% of H+ is due to H2SO4

H2SO4 = 2H+ + SO4-2

0.0001995 mol of H+ per Liter * 111763786450 liters = 22296875.3968 mol of [H+]

1 mol of [H+] = 1/2 mol of H2SO4 approx

so

22296875.3968 mol o fH+ = 1/2*(22296875.3968) = 11148437.6 mol of H2SO4

mass of H2SO4 = mol*MW = 11148437.6*98 = 1092546884.8 g = 1092546.9 kg of H2SO4 approx