Question

If the concentration of atmospheric CO2 were double from its current value of 390 ppm, what would be the calculated pH of rainwater (assuming that CO2 were the only acidic input)? With respect to rising CO2 levels, do we have to be concerned about enhanced acidity of rain in addition to potential climate warming?

Answer 1

Probable Concentration of CO2 = 390 ppm = 390 x 1ppm = 390 x (1 mg of CO2/1 Kg of water)

= 390 mg of CO2 / 1 kg water.

Mass of CO2 = 390 mg = 0.390 g.

Molar mass of CO2 = 44.0 g/mole.

# of moles of CO2 = Mass of CO2 / Molar mass of CO2 = 0.390 / 44.0 = 0.0089 moles.

I.e. there are 0.0089 moles in 1 Kg i.e. 1000 g i.e. 1L water (density of water is apprx. 1 g/mL)

Hence we have concentration of CO2 dissolved is

[CO2] = 0.0089 moles/L = 0.0089 M

CO2 dissolve and ionize in water following the equation,

CO2 + H2O H2CO3 H⁺ (aq) + HCO₃^- (aq) Ka = 4.5 x 10^-7.

ICE table

H2CO3 H⁺ (aq) + HCO₃^- (aq)

Initially 0.0089 0 0

Change -X +X +X

Eqm conc. (0.0089-X) X X

Then,

Ka = [H⁺][HCO₃^-]/[H2CO3] = 4.5 x 10^-7.

(X)(X) / (0.0089 - X) = 4.5 x 10^-7.

H2CO3 being a very weak acid, " small concentration assumption" holds true. so 0.0089 - X 0.0089 M.

X² / 0.0089 = 4.5 x 10^-7.

X² = 0.0089 x 4.5 x 10^-7.

X² = 4.01 x 10^-9.

X = 6.33 x 10^-5.

By ICE table.

[H+] = X = 6.33 x 10^-5.

pH = -log(H+) = -log(6.33 x 10^-5)

pH = 4.2

pH of the water will be 4.2.

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