Question

Acid rain was at one time an important point of contention between the United States and Canada. Much of this acid was the result of the emission of sulfur oxides by coal-fired electricity generating plants in southern Indiana and Ohio. These sulfur oxides, when dissolved in rainwater, formed sulfuric acid and hence “acid rain”. How many metric tonnes of Indiana coal, which averages 3.5% sulfur by weight, would yield the H2SO4 required to produce a 0.9 inch rainfall of pH 3.90 precipitation over a 104 square mile area?

Answer 1

First, find concnetration of H+ needed

[H+] = 10^-pH = 10^-3.90 = 0.0001258

now, find moles of H+ needed

mol = M*V = 0.0001258*V

V = A*h = 104 mi^2 * 0.9 in

change all to m

A = 2.694*10^8 m^2

h = 0.02286

V= (0.02286)(2.694*10^8 ) = 6158484 m^3

but we need liter so

V = 6158484000 L = 6.158*10^9 liter

now, calculate moles

mol = M*V = 0.0001258*V = (0.0001258)(6.158*10^9) = 774676.4 mol of H+

recall that we have 2 mol of H+ per 1 mol of H2SO4 so

774676.4 /2 = 387338.2 mol of H2SO4 are needed

Note that we need 1 mol of SO4 per 1 mol of H2SO4 so

387338.2 mol of S are required

calculate mass

mass = mol*MW = 387338.2*32.065 = 12419999.383 grams or = 12419.99 kg or = 12.42 ton of S

but only 3.5% of the coal is S so

12.42/0.035 = 354.86 ton of coal are needed to get a pH change to 3.90 to that amount of volume of rain