Introductory University Chemistry I

Solutions and Solubility

Henry's Law and the Solubility of Gases

Concentration of solutions was taken up as part of molar stoichiometry in earlier sectionsof introductory university chemistry. The concentrations used in this section will all be molar concentrations, because in homogeneous solutions active mass is the ratio of amount of substance to unit volume. Molar concentration will be indicated by c, with the solute in parenthesis following the symbol, or by placing the solute molecule or ion symbol in square brackets. Other concentration units, such as molality, are less commonly usedin aqueous equilibrium calculations.

Gases dissolve in liquids to form solutions. This dissolution is an equilibrium process for which an equilibrium constant can be written. For example, the equilibrium between oxygen gas and dissolved oxygen in water is O2(aq) <--> O2(g). The equilibrium constant for this equilibrium is K = p(O2)/c(O2). The form of the equilibrium constant shows that the concentration of a solute gas in a solution is directly proportional to the partial pressure of that gas above the solution. This statement, known as Henry's law, was first proposed in 1800 by J.W. Henry as an empirical law well before the development of our modern ideas of chemical equilibrium.

Stating the pressure-concentration ratio as an equation and use of the usual modern symbol for the Henry's law constant on a concentration basis gives the following form of Henry's law:

p = k'cc

In this form p is the partial pressure of the gas, c is its molar concentration, and k'c is the Henry's law constant on the molar concentration scale. Henry's law is found to be an accurate description of the behavior of gases dissolving in liquids when concentrations and partial pressures are reasonably low. As concentrations and partial pressures increase, deviations from Henry's law become noticeable. This behavior is very similar to the behavior of gases, which are found to deviate from the ideal gas law as pressures increase and temperatures decrease. For this reason, slutions which are found to obey Henry's law are sometimes called ideal dilute solutions.

Values of the Henry's law constants for many gases in many different solvents have been measured. The table below gives a few selected values of Henry's law constants for gases dissolved in water.


Table: Molar Henry's Law Constants for Aqueous Solutions at 25oC

Gas          Constant            Constant
             (Pa/(mol/dm3))      (atm/(mol/dm3))
He           282.7   x 10+6      2865.
O2            74.68  x 10+6       756.7
N2           155.    x 10+6      1600.
H2           121.2   x 10+6      1228.
CO2            2.937 x 10+6        29.76
NH3            5.69  x 10+6        56.9
Values in this table are calculated from tables of molar thermodynamic properties of pure substances and aqueous solutes
The inverse of the Henry's law constant, multiplied by the partial pressure of the gas above the solution, is the molar solubility of the gas. Thus oxygen at one atmosphere would have a molar solubility of (1/756.7)mol/dm3 or 1.32 mmol/dm3.
Example. The amount of oxygen dissolved in air-saturated water under normal atmospheric conditions at 25oC can be calculated as follows. Normal atmospheric conditions are 20.948 mole per cent oxygen, which makes the partial pressure of oxygen 0.20948 atm or 20.67 kPa. Using Henry's law, the concentration of oxygen is 0.20948 atm/(756.7 atm/(mol/dm3)) which is 2.768 x 10-4 mol/dm3 or 0.2768 mmol/dm3.
Example. We can estimate the amount of nitrogen that a diver must lose from his bloodstream (about 5 dm3) in rising from a depth of 100 m to the surface in order to avoid formation of nitrogen bubbles in his bloodstream, a painful and often fatal condition known as "the bends", as follows.

The density of water is about 1 kg/dm3 or 1000 kg/m3. A column of water 100 m high would have a mass of 100000 kg/m2 of its base which would exert an additional force of 980665 N/m2 or 980665 Pa. (The total pressure would be 980.665 kPa plus 101.325 kPa or 1081.990 kPa.) The pressure change of 980665 Pa would produce a concentration change of 980665/(155,000,000) = 6.33 mmol/dm3. The amount of nitrogen which must be lost is then (5)(6.33) = 31.65 mmol or, at room temperature and pressure, somewhat over 750 cm3 of nitrogen. This is enough nitrogen gas to create massive bubbles in the bloodstream. A depth of 100 m is considered a deep dive; considerable decompression time will be required in order to reach the surface safely.


The value of the Henry's law constant is found to be temperature dependent. The value generally increases with increasing temperature. As a consequence, the solubility of gases generally decreases with increasing temperature. One example of this can be seen when water is heated on a stove. The gas bubbles which appear on the sides of the pan well below the boiling point of water are bubbles of air, which is evolved when water which was air-saturated at lower temperatures is heated and the amount of air which it can contain (the molar solubility of air) decreases. Addition of boiled or distilled water to a fish tank will cause the fish to die of suffocation unless the water has been allowed to reaerate before addition.

The decrease in solubility of gases with increasing temperature is an example of the operation of Le Chatelier's principle. The heat or enthalpy change of the dissolution reaction of most gases is negative, which is to say the reaction is exothermic. As a consequence, increasing the temperature leads to gas evolution.


Example. For the dissolution of oxygen in water, O2(g) <--> O2(aq), the enthalpy change under standard conditions is (-11.7)-0 = -11.7 kJ/mole.

Study Problems

1. Calculate the amount of carbon dioxide dissolved in one litre of soda pop if the manufacturer uses a pressure of 2.4 atm of CO2 to carbonate the soda pop.

2. A diver is safer using a helium/oxygen mixture than a nitrogen/oxygen mixture when diving to longer depths or remaining submerged for longer periods of time. Using the Henry's law constant for nitrogen and assuming that the rate of gas loss from the bloodstream into the lungs is independent of the nature of the gas, how much faster could a diver using a helium/oxygen mixture ascend than if the diver were using a nitrogen/oxygen mixture such as ordinary compressed air?

3. Water is used as a scrubber to remove gases such as ammonia from industrial gas mixtures and from natural gas. How much ammonia could one cubic metre of water possibly remove from such a gas mixture? Upon what properties of the gas mixture does your answer depend?


[NEXT: Solubility of Substances in Water]
[PREVIOUS: Dissolution, Evolution, and Precipitation]
[MAIN HEAD] [COURSE OUTLINE]
Copyright 1995 James A. Plambeck (Jim.Plambeck@ualberta.ca). Updated November 3, 1995 jp