Question

What is the equilibrium concentration of carbon dioxide in water that is in contact with air at 25 °C and 2.72 atm. The mole fraction of CO₂ in air is 3.01×10^-4. The Henry's law constant for carbon dioxide is 4.48×10^-5 M/mmHg.

A.2.79×10^-5 M

B.3.67×10^-8 M

C.9.26×10^-2 M

D.0.622 M

E.6.72 M

Answer 1

Correct answer is : (A.) 2.79 x 10^-5 M

Explanation

Given : Total pressure = 2.72 atm

mole fraction of CO₂ = 3.01 x 10^-4

Solution :

Partial pressure of CO₂ in air = (Total pressure) * (mole fraction of CO₂ in air)

Partial pressure of CO₂ in air = (2.72 atm) * (3.01 x 10^-4)

Partial pressure of CO₂ in air = 8.19 x 10^-4 atm

Partial pressure of CO₂ in air = 8.19 x 10^-4 atm * (760 mmHg / 1 atm)

Partial pressure of CO₂ in air = 0.622 mmHg

According to Henry's law,

Solubility of gas = (K_H) * (partial pressure of gas)

where K_H is the Henry's law constant

Concentration of CO₂ in water = (K_H CO₂) * (Partial pressure of CO₂ in air)

Concentration of CO₂ in water = (4.48 x 10^-5 M/mmHg) * (0.622 mmHg)

Concentration of CO₂ in water = 2.79 x 10^-5 M