What is the equilibrium concentration of carbon dioxide in water that is in contact with air at 25 °C and 2.55 atm. The mole fraction of CO2 in air is 3.01 x 10-4. The Henry's law constant for carbon dioxide is 4.48 x 10-5 M/mmHg.

0.583 M
8.68E-2 M
6.72 M
2.61E-5 M
3.44E-8 M

I would do this.

XCO2 = pCO2/Ptotal
pCO2 = Ptotal x XCO2 = 2.55 atm x (760 mm/1 atm) x 3.01E-4 = ?
Then C = pk
C = ? x 4.48E-5
Solve for C in M

To find the equilibrium concentration of carbon dioxide (CO2) in water, we can use Henry's Law. Henry's Law relates the concentration of a gas dissolved in a liquid to the partial pressure of the gas above the liquid.

Henry's Law equation is defined as:
C = k * P
where C is the concentration of the gas in the liquid, k is the Henry's Law constant, and P is the partial pressure of the gas.

In this case, we are given:
- Temperature (T) = 25 °C (which is 298 K)
- Partial pressure of CO2 in air (P) = 3.01 x 10^-4 atm
- Henry's Law constant for CO2 (k) = 4.48 x 10^-5 M/mmHg

First, we need to convert the partial pressure of CO2 from atm to mmHg:
Partial pressure of CO2 (P) = 3.01 x 10^-4 atm * 760 mmHg/atm = 0.2288 mmHg

Now we can calculate the equilibrium concentration of CO2 (C):
C = k * P
C = (4.48 x 10^-5 M/mmHg) * (0.2288 mmHg)
C ≈ 1.024 x 10^-5 M

Therefore, the equilibrium concentration of CO2 in water in contact with air at 25 °C and 2.55 atm is approximately 1.024 x 10^-5 M.
The answer is 3.44E-8 M.

To find the equilibrium concentration of carbon dioxide 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. The constant of proportionality is called the Henry's Law constant (K_H).

The equation for Henry's Law is:

C = K_H * P

where:
C = concentration of the gas in the liquid
K_H = Henry's Law constant
P = partial pressure of the gas above the liquid

In this case, we are given the following information:

Partial pressure of CO2 in air (P): 2.55 atm
Mole fraction of CO2 in air: 3.01 x 10^(-4)
Henry's Law constant for CO2 (K_H): 4.48 x 10^(-5) M/mmHg

To find the partial pressure of CO2 in the liquid, we need to convert the mole fraction to partial pressure using Dalton's Law of partial pressures.

Dalton's Law states that the total pressure of a mixture of gases is the sum of the partial pressures of the individual gases.

The equation for Dalton's Law is:

P_total = P1 + P2 + ... + Pn

where:
P_total = total pressure of the mixture
P1, P2, ... Pn = partial pressures of the individual gases

In this case, the total pressure (P_total) is given as 2.55 atm. Since the mole fraction of CO2 is given, we can calculate the partial pressure of CO2 using:

P_Co2 = P_total * mole fraction of CO2

Substituting the given values:

P_Co2 = 2.55 atm * 3.01 x 10^(-4)

Now that we have the partial pressure of CO2 (P_Co2), we can use Henry's Law to find the equilibrium concentration of CO2 in water.

C = K_H * P_Co2

Substituting the given values:

C = 4.48 x 10^(-5) M/mmHg * P_Co2

Note that the units of the Henry's Law constant need to be consistent with the units of the partial pressure. In this case, we have partial pressure in atm, so we don't need to convert the units.

Finally, to convert the concentration from mmHg to M (Molarity), we need to use the conversion factor:

1 mmHg = 0.133322 atm

So, the equilibrium concentration of CO2 in water can be calculated as:

C = 4.48 x 10^(-5) M/mmHg * P_Co2 * (1 mmHg / 0.133322 atm)

Simplifying the equation:

C = 4.48 x 10^(-5) M/mmHg * P_Co2 * 0.133322

Now, we can substitute the value of P_Co2 we calculated earlier and solve for C:

C = 4.48 x 10^(-5) M/mmHg * (2.55 atm * 3.01 x 10^(-4)) * 0.133322

C ≈ 8.68 x 10^(-2) M

Therefore, the equilibrium concentration of carbon dioxide in water in contact with air at the given conditions is approximately 8.68 x 10^(-2) M.

The correct answer is: 8.68E-2 M.