Question

If the level of atmospheric CO₂ were to double from its current concentration of 407 ppm, what would be the calculated pH of rainwater, assuming that H₂CO₃(carbonic acid) is the only source of acidity?

Answer 1

There are two important processes going on here: the dissolution of CO₂ in water (governed by Henry's Law, with a constant of 0.034 M/bar at 298 K) and then its first dissociation (which is governed by the acid-base equilibrium with a K_a1* = 4.47×10⁻⁷, where the * means that it takes into account the fact that CO₂ is not completely hydrated as carbonic acid). We will consider the second dissociation of carbonic acid as negligible, since it has a very low K_a2 value (4.69×10⁻¹¹) and the concentration is very low.

If the concentration of CO₂ in air were to double, it's usual partial pressure of 0.3 Torr would become 0.6 Torr, which are equivalent to 8.00x10^-4 bar. With this in mind, we can calculate the concentration of CO₂ in rainwater using Henry's law:

And now we can calculate the concentration of H⁺ from the equilibrium:

The ICE table is:

	CO2	HCO3-	H+
initial	2.72x10-5	0	0
change	-x	+x	+x
equilibrium	2.72x10-5 - x	x	x

And the equilibrium constant would be:

This is a quadratic equation which can be solved to yield:

Clearly the first solution is the valid one, which means the concentration of H⁺ is 3.28x10^-6 M, which results in a pH value of: