Question

The solubility of carbon dioxide in water is very low in air (1.05x10-5 M at 25 degrees C) because the partial pressure of carbon dioxide in air is only 0.00030 atm. What pressure of carbon dioxide is needed to dissolve 100.0 mg of carbon dioxide in 1.00 L of water?

a. 0.0649 atm
b. 2.86 atm
c. 28.6 atm
d. 64.9 atm

Answer 1

B) 2.86 atm

In order to calculate this, you have to use the Henry's law which is:

C= kP

C = concentration of a dissolved gas

k = Henry's Law constant

P = partial pressure of the gas

C = kP

1.05x10^-5 M = k(0.00030 atm)

When we solve for k, we get:

k = (1.05x10^-5 M) / (0.00030 atm)

k = 3.50x10^-2 M/atm

We can now apply Henry's rule to calculate the partial pressure of carbon dioxide that will result in a concentration of 100.0 mg/L (or 0.1 g/L) in water:

0.1 g/L = (3.50x10^-2 M/atm)P

When we convert units, we get:

0.1 g/L = (3.50x10^-2 mol/L)P

P = (0.1 g/L) / (3.50x10^-2 mol/L)

P = 2.86 atm