SO2 forms H2SO3 in water:
Step 1
H2SO3(aq) <=> H^+(aq) + HSO3^-(aq)
K1 = [H+][HSO3^-]/[H2SO3] = 1.71x10^-2
Step 2
HSO3-(aq) <=> H^+(aq) + SO3^-2(aq)
K2 = [H+][SO3^-2]/[HSO3^-] = 6.0x10^-8
The above sequence shows that Step 1 is the primary source of hydrogen ions and that we can ignore Step 2.
If the pH = 4.1, [H+]=7.94x10^-5
Substituting into the K1 expression,
[7.94x10^-5][7.94x10^-5]/[C-7.94x10^-5] = 1.71x10^-2
Solve for the molar concentration of H2SO3, C. Multiply the value of C by the molar mass of SO2 to get grams of SO2 per liter. Convert the last value to milligrams/liter which is equivalent to parts per million (ppm).
Calculate the concentration of SO2(ppm) that must be reached in polluted air (in ppm) if the dissolved gas is to produce a pH of 4.1 in raindrops without any oxidation of the gas.
Hint: Sulfurous acid equilibria
how would i do this.
what equations would i use.?
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