Question

If the level of atmospheric CO2 were to double from its current concentration of 406 ppm, what would be the calculated pH of rainwater, assuming that H2CO3(carbonic acid) is the only source of acidity?

Answer 1

pH = -log[H⁺]

Then

CO₂(g) + H₂O(l) --> H₂CO₃(aq)

H₂CO₃(aq) --> H⁺ + HCO₃^-(Aq)

Ka1 = [H⁺][HCO₃^-]/[H₂CO₃]

[CO₂] = mol of CO₂ / Volume

406 ppm = 406 mg of CO2 / liter of air

Mole = mass/MW = 406/44 = 9.22 mmol = 9.22*10^-3 mol

P_CO2 = (9.22*10^-3)*P_total = 9.22*10^-3 atm

CO₂ in water = H*P_CO2

CO₂ in water = (3.4*10^-2)*9.22*10^-3 = 3.13*10^-4 M of CO₂ in water

Assume CO₂ = H₂CO₃

[H₂CO₃] = 3.13*10^-4 M initially

10^-6.34 = x*x/(3.13*10^-4 - x)

x = [H⁺] = 1.19*10^-3

pH = -log(x) = -log(1.19*10^-3)

= 2.92