Question

A typical coal-fired electric generating plant will burn about 3 metric tons of coal per hour. Most of the coal burned in the United States contains 1 to 4 % by weight sulfur in the form of pyrite, which is oxidized as the coal burns: 4FeS2(s) + 11 O2(g) 2 Fe2O3(s) + 8 SO2(g) Once in the atmosphere, the SO2 is oxidized to SO3, which then reacts with water in the atmosphere to form sulfuric acid: SO3(g) + H2O(l) H2SO4(aq) If 68.0 metric tons of coal that contains 1.55 % by weight S is burned and all of the sulfuric acid that is formed rains down into a pond of dimensions 333 m x 235 m x 4.06 m, what is the pH of the pond? (OK to assume 2 mol H3O+ per mole H2SO4.) Hint: 1 metric ton = 1000 kg

Answer 1

68.0 metric tons of coal = 68000 kg coal.

mass of S = 68000 * 1.55 / 100 = 1054 kg. = 1054000 g S.

mole of S = 1054000 g / 32 g / mole = 32937.5 mole.

thus

mole of H2SO4 produce = 32937.5 mole

volume of the pond = 333 m x 235 m x 4.06 m = 317715.3 m^3 = 317715.3 * 1000 = 317715300 L

thus

[H2SO4] = 32937.5 mole / 317715300 L = 1.04 * 10^-4 M

thus

[H3O+] = 2 * 1.04 * 10^-4 = 2.08 * 10^-4 M

pH of the pond = - log [H3O+] = - log (2.08 * 10^-4) = 3.68 (answer)