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.

PLEASE ANSWER BOTH PARTS OR I WILL NOT "THUMBS UP" YOUR ANSWER!

Answer 1

a) Given that the total pressure in a bottle of soda water is 40 psig = 2.721 atm

Percentage of CO₂ in the mixture is 75%.

Henry’s Law constant K_H = 0.104 x10⁴atm/M.

         According to Henry’s law:

         P = K_H × X

X = P/K_H
X = 2.721 atm/0.104 x10⁴atm
X = 0.0026

Mole fraction of CO₂, X = n_CO2/(n_H2O+ n_CO2)

         Value of mole fraction is very small so it is negligible as compared to 1.

         We get: X = n_CO2/n_H2O

Density of water = 1 g/ ml , 1000g per 1 L.

n_H2O = 1000/18 = 55.55 mol

Therefore, n_CO2 = 55.55 mol x0.0026 = 0.144 mol

Mass of 1 mole CO₂ = 44 g

Mass of 0.144 mole CO₂ = 0.144 x 44 g = 6.336 g = 6336 mg/L

b) 358 ppmv CO₂ = 358 x 10^-6 m³ CO₂/ 1 m³ air.

      n = PV/RT

Making sure that everything is in SI units:

      n = (101,325)(358 x 10^-6 )/(8.31441)(298.15) = 0.0146 mol/L

When you open the cap, the pressure inside is released and the CO₂ escapes out.