 # Henry's Law for Gas Solubility Chemistry Tutorial

## Key Concepts

• Henry's Law1 states that:
At constant temperature, the concentration of a gas in solution is proportional to the partial pressure of the gas above the solution.
• Henry's Law can be written as a relationship:

c ∝ P

where
c = concentration of gas in solution2
P = partial pressure in the gas phase3
• Henry's Law can also be written as an equation:

P = Kc

where K = the constant of proportionality = Henry's Law constant
• To determine the molar solubility of a gas:

 c = 1   K × P

where 1/K is the inverse of Henry's Law constant
c is the concentration of the gas in mol L-1 (molarity)
P is the partial pressure of the gas

No ads = no money for us = no free stuff for you!

## Deriving Henry's Law

Consider a cold can of cola on a hot summer day.
Before you open the can, there is carbon dioxide gas dissolved in the cola drink, and, carbon dioxide gas above the drink at a pressure slightly greater than atmospheric pressure.
As soon as you open the can you can hear the "hiss" as gas escapes from the can into the atmosphere in order to equalise the pressure above the cola in the can with atmospheric pressure.
Then you pour the cola into a cold glass.
And you see that bubbles of gas are still rising through the cola to its surface and then escaping into the atmosphere.
Why? Henry's Law!

A gas, such as carbon dioxide, can dissolve in a liquid, such as water, to form a solution.
The gas is the solute and the liquid is the solvent.
In a closed system, solute particles in the gas phase above the solution are in equibrium with solute gas particles in solution as shown by the following equation:

 general equation example solute(g) + solvent(l) ⇋ solution CO2(g) + H2O(l) ⇋ CO2(aq)

By Le Chatelier's Principle, increasing the pressure on the system will force the equilibrium position to shift to the right, favouring the formation of the solution by dissolving more of the gaseous solute in the solvent.
Similarly, a decrease in pressure will favour the formation of gas and solvent, that is, gas will leave the solution to increase the pressure above the solution.

For the cola in the closed can, the partial pressure of carbon dioxide above the solution is very high.
For the cola in the glass, the partial pressure of carbon dioxide above the solution is very low because the partial pressure of carbon dioxide in the atmosphere is very low.
Therefore, by Le Chatelier's principle, as the cola sits in the glass, carbon dioxide is continually escaping from the solution in order to increase the partial pressure of carbon dioxide in the atmosphere.

solute(g) + solvent(l) ⇋ solution

• Increasing the partial pressure of the gaseous solute increases the concentration of solution.
• Decreasing the partial pressure of the gaseous solute decreases the concentration of solution.

Now we can see that partial pressure of a gas is proportional to its concentration in solution

Which we could write as:

partial pressure ∝ concentration

or as:

P ∝ c

So, using a constant of proportionality, K, we can write an expression for Henry's Law:

P = Kc

K is known as the Henry's Law constant.

Consider the system in which oxygen gas dissolved in water is in equilibrium with oxygen gas above the water.

chemical equation: O2(g) + H2O(l) ⇋ O2(aq) P(O2(g)) = Kc(O2(aq))

Using the concentration of oxygen gas in water at various partial pressures at 25°C we obtain a graph like the one below:

 partial pressure / kPa Solubility of O2(g) in water at 25°Cconcentration/ mol L-1

We can see that K, Henry's Law constant, is the slope (gradient) of the line4.
The slope of the line ≈ 60 kPa ÷ 0.0008 mol L-1 = 75 000 kPa/mol L-1
Therefore, for an aqueous solution of oxygen gas at 25°C we can write:

P(O2(g)) = 75 000 × c(O2(g))

The value of K, Henry's Law constant, is dependent on the

• gas used
• solvent used
• temperature of the system
• units of concentration used
• units of pressure used

The table below gives the value of Henry's Law constant for various aqueous solutions of gases at 25°C given that pressure is measured in kPa and concentration in mol L-1:

Aqueous Solutions at 25°C
Gas K
( kPa / mol L-1 )
helium 282 700
nitrogen 155 000
hydrogen 121 200
oxygen 74 680
ammonia 5 690
carbon dioxide 2 937

The data in the table above have been plotted on the graph below:

 partial pressure / kPa Solubility of gases in water at 25°Cconcentration/ mol L-1

From the graph we we can see that a greater partial pressure is required to dissolve 0.0002 moles of helium gas compared to any of the other gases listed.

Henry's Law only applies when the gas and its solution are essentially ideal, that is,

• concentration of gas in solution must be low
• partial pressure above solution must be low
• solute does not interact strongly with the solvent5

The curves for He(g), H2(g) and N2(g) are all linear up to about 10,130 kPa (100 atm), that is, these gases are obeying Henry's Law, however, in this range O2(g) starts to deviate.

Do you know this?

Play the game now!

## Molar Solubility of Gases

We can rearrange Henry's Law as given above to write an expression for the concentration of a gas:

 Henry's Law: P = Kc divide both sides of equation by K: P   K = KcK Molar Solubility of Gas:(c in mol L-1) c = P   K = P × 1 K

The table below gives the value of the inverse of Henry's Law constant for various aqueous solutions of gases at 25°C given that pressure is measured in kPa and concentration in mol L-1:

Aqueous Solutions at 25°C
Gas 1/K
( mol L-1/ kPa )
molar solubility
(101.3 kPa)
helium 3.537 × 10-6 3.583 × 10-4
nitrogen 6.452 × 10-6 6.536 × 10-4
hydrogen 8.251 × 10-6 8.358 × 10-4
oxygen 1.33 × 10-5 1.356 × 10-3
ammonia 1.757 × 10-4 0.01780
carbon dioxide 3.405 × 10-4 0.03449

We can then use these values to compare the molar solubility of the gases in water at 25°C and the same partial pressure, for example, 101.3 kPa (or 1 atm):

molar solubility of gas at 101.3 kPa

= c(gas)
= P × 1/K
= 101.3 × 1/k

The graph below shows how the solubility of a gas increases as its partial pressure increases according to Henry's Law:

 concentration / mol L-1 Molar Solubility of gases in water at 25°Cpartial pressure/ kPa

Comparing different solutions of gas dissolved in water with the same partial pressure at the same temperature, we see that oxygen gas is more soluble than hydrogen, nitrogen or helium gases.

Water at sea level can dissolve more oxygen gas than water on top of a mountain.
This is because air pressure at sea level is greater than on top of the mountain, and therefore, the partial pressure of oxygen in air at sea level is greater than on top of the mountain.

Do you understand this?

Take the test now!

## Henry's Law Calculations in Air

Air is made up of a mixture of gases.
If a solution is exposed to air, then the concentration of gases in solution will be dependent on the partial pressure of the gases in the atmosphere.
If we use Henry's Law to calculate the concentration of a gas at atmospheric pressure in equilibrium with air, we need to know the partial pressure of the gas in the atmosphere.

Composition of the Atmosphere
(clean dry air at 25°C and 101.3 kPa)
gas concentration
/ ppm by volume
partial pressure
kPa
nitrogen 780 900 79.1
oxygen 209 400 21.2
argon 9 300 0.94
carbon dioxide 315 0.032
neon 18 0.0018
helium 5.2 0.00053
methane 1.1 0.00011
krypton 1.0 0.00010
nitrous oxide 0.5 0.00005
hydrogen 0.5 0.00005
xenon 0.08 0.000008
nitrogen dioxide 0.02 0.000002
ozone 0.02 0.000002
molar concentration of gas = 1/k × partial pressure

c(g) = 1/k × P(g)

c(N2) = 6.452 × 10-6 × 79.1 = 5.10 × 10-4 mol L-1

c(O2) = 1.339 × 10-5 × 21.2 = 2.84 × 10-4 mol L-1

c(CO2) = 3.405 × 10-4 × 0.032 = 1.09 × 10-5 mol L-1

c(He) = 3.537 × 10-6 × 0.00053 = 1.87 × 10-9 mol L-1

c(H2) = 0.00005 × 8.251 × 10-6 = 4.13 × 10-10 mol L-1

Even though O2(g) is more soluble in water than N2(g), because the partial pressure of nitrogen in the atmosphere is greater, the concentration of nitrogen in water is slightly greater than that of oxygen.

Similarly, H2(g) is more soluble in water than N2(g) or He(g), yet, because the partial pressure of H2(g) in the atmosphere is much less than N2(g) or He(g) its concentration in water is less than that of N2 or He.

Can you apply this?

Join AUS-e-TUTE!

Take the test now!

## Effect of Temperature on Solubility of Gases

As the temperature of an aqueous solution increases, a gas generally becomes less soluble in the water, as shown in the graph below:

 concentration(mole fraction) Solubility of gases in waterTemperature / °C

Pure water at room temperature is made up of water molecules that are hydrogen-bonded to each other. This creates an arrangement of water molecules containing 'holes'. Molecules of gas solute can occupy these 'holes', but because the intermolecular attraction between the water and gas molecules is less than the intermolecular attraction between water and water molecules, energy is released.

solute(g) + water(l) ⇋ solution(aq) + energy

By Le Chatelier's Principle if the system initially at equilibrium is then heated, the equilibrium position shifts to the left to counteract the effect of the additional heat. Therefore, gas comes out of solution and escapes into the gaseous phase, decreasing the concentration of gas in solution and increasing its partial pressure.

Similarly, if the system is initially at equilibrium and is then cooled, by Le Chatelier's Principle the equilibrium position shifts to the right to provide more energy to offset the reduction in energy. More gaseous solute dissolves in the water which increases the concentration of gas in solution.

The effect of increased temperature on gas solubility is obvious if you leave a cool glass of a carbonated drink (cola for instance) sitting on a table in a warm room. As the solution warms, carbon dioxide gas becomes less soluble and escapes from the solution.

The effect can also be seen if you leave some cool, fresh water in a glass on a table in a warm room, you will see small bubbles of gas appear and float to the surface which will then escape into the atmosphere. As the water in the glass warms, the dissolved atmospheric gases become less soluble and escape from the solution into the atmosphere.

The effect of temperature on gas solubility is extremely important.
Oxygen gas, O2(g), dissolved in water is essential for the survival of living things in water. Oxygen gas is more soluble in cold water than hot water. So, if hot water from a power plant is discharged into a river, the resulting decrease in oxygen gas in the water can lead to the death of fish.

Can you apply this?

Join AUS-e-TUTE!

Take the test now!

## Worked Example of a Henry's Law Question

Question: At 25°C the carbon dioxide gas pressure inside a 1.00 L bottle of cola is 253 kPa.
Determine the amount in moles of carbon dioxide dissolved in the aqueous cola solution.

(K(CO2(g)) = 2,937 kPa/mol L-1 at 25°C)

Solution:

(Based on the StoPGoPS approach to problem solving.)

1. What is the question asking you to do?

Calculate moles of carbon dissolved in solution
n(CO2(g)) = ? mol

2. What data (information) have you been given in the question?

Extract the data from the question:

T = temperature = 25°C
K(CO2(g)) = Henry's Law constant = 2,937 kPa/mol L-1 at 25°C
V = volume of solution = 1.00 L
P = CO2(g) pressure = 253 kPa
3. What is the relationship between what you know and what you need to find out?
Assume this is an ideal solution.

(A) Henry's Law: P = Kc
P = partial pressure of gas
c = concentration of gas in solution
K = Henry's Law constant in suitable units and at the same temperature

(B) c = n/V
c = concentration of gas in mol L-1
n = amount of gas dissolved in moles
V = volume of solution in litres

Substitute the equation for concentration (B) into Henry's Law (A):

P = K × n/V

Rearrange the equation to find n:

n = PV/K

4. Substitute in the values and solve for n:

 n = PV K = 253 × 1.00 2937 = 0.0861 mol

Work backwards:
If n(CO2(g)) = 0.0861 mol
and V(CO2(aq)) = 1.00 L
c(CO2(aq)) = n/V = 0.0861/1.00 = 0.0861 mol L-1
P = Kc
P = 2,937 × 0.0861 = 252.9 ≈ 253 kPa
Since the calculated pressure using our concentration is the same as that given in the question, we are confident our answer is plausible.
6. State your solution to the problem " moles of carbon dissolved in solution ":

n(CO2(g)) = 0.0861 mol

Can you apply this?

Join AUS-e-TUTE!

Take the test now!

## Worked Example of a Henry's Law Question

Question: A 1.00 L bottle of unopened cola has a carbon dioxide concentration of 8.6 × 10-2 mol L-1 with a carbon dioxide partial pressure of 2.5 atm at 25°C.
At 25°C, the atmosphere contains carbon dioxide at a partial pressure of 3.2 × 10-4 atm.
After the bottle of cola has been opened and allowed to come to equilibrium with the atmosphere at 25°C, what is the concentration of carbon dioxide gas in the cola solution in mol L-1?

Solution:

(Based on the StoPGoPS approach to problem solving.)

1. What is the question asking you to do?

Calculate equilibrium concentration of carbon dioxide gas
c(CO2(g)) = ? mol L-1

2. What data (information) have you been given in the question?

Extract the data from the question:

 initial(before opening) final(after opening) initial temperature = Tinitial = 25deg;C intial volume of solution = Vinitial = 1.00 L intial partial pressure of CO2(g) = P(CO2(g)initial) = 2.5 atm intial concentration of CO2(aq) = c(CO2(g)initial) = 8.6 × 10-2 mol L-1 final temperature = Tfinal = 25°C final volume of solution = Vfinal = 1.00 L final partial pressure of CO2(g) = P(CO2(g)final) = 3.2 × 10-4 atm final concentration of CO2(aq) = c(CO2(g)final) = ? mol L-1

3. What is the relationship between what you know and what you need to find out?

(A) Henry's Law: P = kc

 initial final P(CO2(g)initial) = Kinitialc(CO2(aq)initial) So, P(CO2(g)initial)/c(CO2(aq)initial) = Kinitial P(CO2(g)final) = Kfinalc(CO2(aq)final) So, P(CO2(g)final)/c2(aq)final) = Kfinal

(B) Since we are considering the same gas, CO2(g), in the same solution, CO2(aq), at the same temperature, 25°C,

Kinitial = Kfinal = K(CO2)

therefore:

P(CO2(g)initial)/c(CO2(aq)initial) = P(CO2(g)final)/c(CO2(aq)final)

rearranging the expression above gives us:

 c(CO2(aq)final) = c(CO2(aq)initial) × P(CO2(g)final)P(CO2(g)initial)

4. Substitute in the values and solve for CO2(g) concentration:

 c(CO2(aq)final) = c(CO2(aq)initial) × P(CO2(g)final)P(CO2(g)initial) = 8.6 × 10-2 × 3.2 × 10-42.5 = 1.1 × 10-5 mol L-1

work backwards
calculate K in atm/mol L-1 for c(final) = 1.1 × 10-5 mol L-1 and P(final) = 3.2 × 10-4 atm
K = P/c = 3.2 × 10-4/1.1 × 10-5 = 29.1 atm/mol L-1
use K to calculate c(initial) for a P(initial) of 2.5 atm
c = P/K = 2.5/29.1 = 0.086 = 8.6 × 10-2 mol L-1
Since the value we calculated for c(initial) is the same as that given in the question, we are confident our answer is plausible.
6. State your solution to the problem "equilibrium concentration of carbon dioxide gas ":

c(CO2(g)) = 1.1 × 10-5 mol L-1

Quick Question 1

An aqueous solution of hydrogen gas at a concentration of 7.21 × 10-5 mol L-1 is in equilibrium with its vapor.
Determine the partial pressure in kPa of its vapor if K = 121200 kPa/mol L-1.

P(H2(g)) = kPa

1. Henry's Law was formulated by William Henry in 1803 as an empirical law, long before Le Chatelier suggested that systems at equilibrium tend to compensate for the effects of perturbing influences in 1884.

2. What are the units for concentration? You could use molarity, mole fraction, w/w, parts per million, etc, BUT you must be consistent because these units will decide the units of the Henry's Law constant.

3. What are the units for pressure? You could use kilopascals, atmospheres, torr, etc, BUT you must be consistent because these units will decide the units of the Henry's Law constant.

4. The equation of a straight line is y = mx + b
where x and y are the coordinates of a point on the line (x,y), m is the slope or gradient of the line, and, b is the intercept on the y axis.
A straight line that goes through the origin has a y intercept of 0, so b = 0 and the equation of the line is y = mx
Henry's Law is P = Kc where the slope or gradient of the line is K and (c,P) represents a point on the line.

5. Carbon dioxide, CO2(g), does react to some extent with water to form carbonic acid, H2CO3(aq), but the equilibrium constant is very small, so Henry's Law can be used for very dilute solutions of carbon dioxide gas at reasonable pressures. (refer to polyprotic weak acids)
Similarly, although ammonia gas, NH3(g), reacts with water to some extent, its equilibrium constant is also very small, so we can use Henry's Law for low concentrations at reasonable pressures.(refer to base dissociation constants)