Question

Henry’s Law Practice
1. The solubility of a gas is 0.58 g/L at a pressure of 104 kPa. What is the solubility if the pressure increases to 250 kPa at constant temperature?


2. The solubility of a gas is 1.6 g/L at 760 mmHg. What is the solubility of the same gas at a pressure of 2.5 atm? Assume temperature to be constant.


3. A gas has a solubility of 2.45 g/L at a pressure of 0.750 atm. What pressure would be required to produce an aqueous solution containing 6.25 g/L of this gas at constant temperature?

Answer 1

1.

Solution.

Henry's law states that the solubility of a gas in liquids at a constant temperature is proportional to its pressure. We write the formula:

$$S = K \times P$$

The Henry's constant for each substance is individual, so we express it and find it through the solubility of the gas and its pressure.

$$K = \frac{S}{P}$$ $$K = 5.58 * 10^{-6}$$

And now we find the solubility of the gas at a different pressure.

S = 1.39 g/L

Answer:

S = 1.39 g/L

2.

Solution.

To begin with, we will translate the pressure from mm Hg of the column and from atmospheres to Pascals.

760 mmHg = 101324.72 Pa

2.5 atm = 253312.5 Pa

$$S = K \times P$$

$$K = \frac{S}{P}$$ $$K = 1.58 * 10^{-5}$$

Now we find the solubility of the gas at a different pressure.

S = 4.00 g/L

Answer:

S = 4.00 g/L

3.

Solution.

To begin with, we will translate the pressure from atmospheres to Pascals.

0.750 atm = 75993.75 Pa

$$S = K \times P$$

$$K = \frac{S}{P}$$

$$K = 3.22*10^{-5}$$

Now we find the gas pressure to get a solution with a concentration of 6.25 g / L.

$$P = \frac{S}{K}$$

P = 194.1 kPa

Answer:

P = 194.1 kPa