Question

Calculate the volume (in mL) of CO₂ at 25 ^oC and 1.0 atm dissolved in a 10.0 L bucket of rainwater that has a pH of 4.5.

I know the answer is 580, I just don't know how to solve it. Please help!

Answer 1

The author considered that the pH of rainwater is only due to H2CO3 (dissolved CO2)!.

H₂CO₃(aq) ⇌ H⁺(aq) + HCO₃^–(aq)         Ka1= 4.3x10^-7 (ignore the second step).

At pH = 4.5     [H+] = 10^-4.5 = 3.2x10^-5

Calculate as usual:

[H⁺] = [HCO₃^–]

[H⁺] = (K_a . C_H2CO3)^1/2

(3.2x10^-5 )² / 4.3x10^-7 = C_H2CO3

C_H2CO3 = 2.32x10^-3 mol/L_water

2.4x10^-3 mol CO₂/L_water x 22.4 L CO₂ /mol x (298K/273K) = 0.058 L CO₂ /L water

58 mL CO₂ x 10 L = 580 mL CO₂

(the molar volume 22.4 L at std cond was corected for 298 K)