Question

In the midst of the acid rain crisis in the 1980's, the average atmospheric SO₂ levels were of the order of 150ppbv. Calculate the pH of natural rainwater at sea level in equilitrium with SO₂(K_H=5.4M^-1atm, Ka1=1.7*10^-2, K_a2=6.4*10^-8) Assume that there are no other proton sources in the system. PPBV means ppb of volume.

Answer 1

pH of natural rainwater = 3.93

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When SO2 is in contact with H2O, following equilibrium establishes, H2O + SO2 <==> H2SO3

H2SO3 ionize as follows,

H2SO3 + H2O <==> H3O^+ + HSO3^-, Ka1 = 1.7 x 10^-2

HSO3^- + H2O <==> H3O^+ + SO3^2-, Ka2 = 6.4 x 10^-8

Concentration of atmospheric SO2 = 150 ppbv = 150 x 10^-9 atm Henry’s Law constant for SO2, KH = 5.4 M/atm

[H2SO3] = KH x PSO2, KH = Henry’s Law constant, PSO2 = partial pressure of SO2.

[H2SO3] = 5.4 M/atm x 150 x 10^-9 atm = 8.1 x 10^-7 M

For the ionization, since Ka2 is very-very less as compared to Ka1, the contribution from Ka2 can be excluded.

Ka1 = [H3O+][HSO3^-]/[H2SO3] = 1.7 x 10^-2

If x fraction of [H2SO3] get ionized then

[H3O+] = x

[HSO3^-] = x

[H2SO3] = 8.1 x 10^-7

[H2SO3] will not be changing, since atmospheric SO2 remain nearly constant.

i.e., 1.7 x 10^-2 = x.x/(8.1 x 10^-7)

x = 0.000117346

[H3O+] = 0.000117346

pH = -log[H3O+] = -log (0.000117346) = 3.93