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 pH of 4.5. The answer is supposed to be 580 but I keep getting 568. Please Help!

Answer 1

The reaction taking place is:

CO₂ + H₂O ----> H₂CO₃

Assuming that 'x' M of carbonic acid is present in water.

Carbonic acid dissociates via the following reaction:

H₂CO₃ ----> H⁺ + 0 HCO₃^-

Initial x 0 0

Eqb. x-a a a

K_a = 4.3*10^-7 = ( [H⁺]*[HCO₃^-] ) / [H₂CO₃] = a²/(x-a)

Given: a = [H⁺] = 10^-pH = 10^-4.5 = 3.16*10^-5 M

Putting values we get:

x = 0.00235 M

Volume of solution taken = 10 L

So, moles of carbonic acid = 0.00235*10 = 0.0235 moles

In order to produce 0.0235 moles of acid, 0.0235 moles of Co2 must be present.

Now,

Using ideal gas equation:

PV = nRT

1*V = 0.0235*0.0821*298

Thus, V = 1.739 L