What is the equilibrium concentration of carbon dioxide in water that is in contact with air at 25 °C and 4.17 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.
To find the equilibrium concentration of carbon dioxide (CO2) in water, we can use Henry's law. Henry's law states that the concentration of a gas dissolved in a liquid is directly proportional to the partial pressure of the gas above the liquid.
First, let's convert the given pressure from atm to mmHg. 1 atm is equivalent to 760 mmHg. Therefore, 4.17 atm is equal to 3167.2 mmHg.
Next, we can calculate the partial pressure of CO2 using the mole fraction of CO2 in air and the given total pressure. The mole fraction of CO2 is the ratio of the moles of CO2 to the total moles of air.
Given:
Mole fraction of CO2 in air = 3.01×10^-4
Total pressure = 4.17 atm = 3167.2 mmHg
To find the partial pressure of CO2:
Partial pressure of CO2 = Mole fraction of CO2 in air * Total pressure
Partial pressure of CO2 = 3.01×10^-4 * 3167.2
Partial pressure of CO2 = 0.9525 mmHg
Now, we can use Henry's law to find the equilibrium concentration of CO2 in water. Henry's law constant relates the concentration of a gas in a liquid to its partial pressure.
Given:
Henry's law constant for CO2 = 4.48×10^-5 M/mmHg
Equilibrium concentration of CO2 in water = Henry's law constant * Partial pressure of CO2
Equilibrium concentration of CO2 in water = 4.48×10^-5 * 0.9525
Equilibrium concentration of CO2 in water = 4.255×10^-5 M
Therefore, the equilibrium concentration of CO2 in water that is in contact with air at 25 °C and 4.17 atm is approximately 4.255×10^-5 M.