Suppose the total pressure in a bottle of soda water before it is opened is 40 p

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!

Explanation / Answer

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 KH = 0.104 x104atm/M.

         According to Henry’s law:

         P = KH × X

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

Mole fraction of CO2, X = nCO2/(nH2O+ nCO2)

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

         We get: X = nCO2/nH2O

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

nH2O = 1000/18 = 55.55 mol

Therefore, nCO2 = 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.

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