Question

Sulfur dioxide concentration in MN are typically in the range of 20-40 ppb. How would this level of sulfur dioxide change the pH of pure rain water (no CO2 considered)?

Answer 1

concentration of SO2 = 20 - 40 ppb

1 ppb = 10^-3 ppm

1 ppm = 1 mg in 1 kg of solution

based on density of water,

1 ppm = 1 mg in 1 L of solution

1 ppm = (1*10^-3/18)moles / L of solution

1 ppm = 5.55 * 10^-5 moles/L

concentration of SO2 = (20 * 10^-3 * (5.55 * 10^-5 mol/L)) - (40 * 10^-3 * (5.55 * 10^-5 mol/L))

Concentration of SO2 = (1.11 * 10^-6)mol/L - (2.22 * 10^-6)mol/L

for pH of water = 7

concentration of H+ = 10^-7 mol/L

so the concentration of H+ ion increses in the pure water due to SO2.

pH = -log[H+]

pH = -log(1.11 * 10^-6)

pH = 5.955

pH = -log[H+]

pH = -log(2.22 * 10^-6)

pH = 5.654

so the pH range become 5.955 - 5.654

Change in pH = (7 - 5.955) - (7 - 5.654) = 1.045 - 1.346