Question

One defense against acid rain is the liming of lakes. Ground limestone is added to offset the effect of acid rain. If we assume that a lake drains 10 km2 and that that the annual precipitation in the entire drainage basin is 1 m (1 m3/m2 of surface), how much limestone, CaCO3, should be added to the lake to change the acid rainfall (pH4.5) to the equivalent of normal rainfall (pH 5.6)?

Answer 1

CaCO3 + 2H+ = Ca+2 + H2CO3

then

1 mol fo CaCO3 = 2 mol of H+

Vtotal:

V = 10 km2 * 1 m = (10km)(1000 m /km)^2 * 1m = 10^7 m * 1m = 10^7 m3 = 10^10 Liters

V = 10^10 L

if we need

pH1 = 4.5

pH2 = 5.6

[H+]1 = 10^-pH = 10^-4.5

[H+]2 = 10^-pH = 10^-5.6

d[H+] = 10^-4.5 - 10^-5.6 = 0.0000291108 M must be eliminated

mol of H+ = M*V = 0.0000291108*10^10 = 291108 mol fo H+

then

mol of CaCO3 = 1/2*mol of H = 1/2*291108 = 145554 mol of CaCO3

mass = mol*MW = 145554*100 =14555400 g = 14555.4 kg = 14.55 ton of CaCO3