Question

The solubility of nitrogen in water is 8.21 × 10-4 mol/L at 0°C when the N2 pressure above water is 0.790 atm. Calculate the Henry’s law constant for N2 in units of mol/L · atm for Henry’s law in the form C = kP, where C is the gas concentration in mol/L. Calculate the solubility of N2 in water when the partial pressure of nitrogen above water is 1.10 atm at 0°C.

Answer 1

The Henry’s law constant for nitrogen is 0.001039 mol/L/atm

The solubility of Nitrogen gas in water at the given conditions is 0.001143 mol/L.

How to determine the solubility of a gas.

Henry's law constant k : C = kP

where

C is the gas concentration in mol/L and

P is the partial pressure of the gas.

Rearrange the equation to solve for k:

k = C/P

Given that

C = 0.000821 mol/L

P = 0.790 atm at 0°C,

Substitute these values to find k.

k = 0.000821 /0.790

k = 0.001039 mol/L/atm

Calculate the solubility C when the partial pressure of nitrogen above water is 1.10 atm.

C = kP

When

P = 1.10 atm.

Substitute the values to find the solubility.

C = 0.001039 * 1.10

C = 0.001143 mol/L

The solubility of Nitrogen gas in water at the given conditions is 0.001143 mol/L.