Question

Suppose the total pressure in a bottle of soda water before it is opened is 40 psig (http://hypertextbook.com/facts/2000/SeemaMeraj.shtml) and that the gas mixture is 75% CO2 and 25% other gases. The temperature is 10°C. The Henry’s Law constant is 0.104x104 atm.

a) If the system is in equilibrium, how much CO2 is dissolved in the soda? (For this problem,ignore reactions between the CO2 and the water.) I know the answer is about 6500 mg/L but I'm having trouble finding the steps to get there!

b) Calculate the new equilibrium concentration when the bottle is opened. The volumetric fraction of CO2 in the normal atmosphere is 358 ppmv. Knowing this, explain why the CO2 acts as it does when the bottle is opened.

Answer 1

the total pressure = 40psig

the concentration of CO2 in bottle = 75%

so the pressure of CO2 = 0.75 X 40 = 30 psig

1psi = 0.068 atm

30psi = 2.04 atm

1psi =2.72 atm

So 30 psi = 81.6 atm

We know that

The given henry's constant = K^px_H

K^px_H = p / x = .104 X 10^4 atm

x = Ca X Molar mass of H2O / Density of water

x = Pressure / K^px_H = 2.04 / 0.104 X 10^4 = 19.6 X 10^-4 mol / L

x = Ca X 18 / 1000 g / L

Ca = x X 1000 / 18 = 19.6 X 10^-4 X 1000 / 18 = 0.1088 mole / L

Concentration of CO2 = 0.1088 X 44 g / L = 4.787 g / L = 4787 mg / L