Henry’s Law

william henry of henry's law

Core Concepts

To understand the dynamics of gases dissolved in liquid, chemists use Henry’s Law. When you open a can of soda, you hear an example of Henry’s Law. When you open the can, the pressure of carbon dioxide decreases quickly. According to Henry’s law, the concentration of carbon dioxide in the drink will also decrease. The hissing sound you hear is the carbon dioxide escaping from the liquid. Learn more about Henry’s law below.

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Who was William Henry?

William Henry was an English chemist who studied in the early 19th century. He lived from 1775-1836. His work focused on the solubility of gasses and results in Henry’s Law being named after him.

william of Henry's Law
William Henry. 1775-1836. Source.

What is Henry’s Law?

Henry’s Law describes the relationship between the partial pressure of gas above a liquid and the gas dissolved in the liquid. In general, the higher the pressure then more gas will be dissolved in the liquid. This relationship is described by the following equation:

    \begin{gather*} {\text{Solubility} = \text{Henry's Constant * Partial Pressure of Gas}} \\ {C_{g}=kP_{g}}  \end{gather*}

 The constant in the equation is dependent on the gas being examined. The equation is discussed in mor detail below along with examples.

The law can be used even when multiple gasses are present. Atmospheric chemistry and environmental research commonly utilize Henry’s Law as they examine gasses passing back and forth between the atmosphere and our water systems.

Henry’s Law Constant

The constant (k} is determined through experimentation. The gas pressure above the liquid is carefully controlled, and then the concentration in the solution is measured. The constant is experimentally determined by dividing the concentration by the partial pressure at different points.

The units on the constant are normally \text{molm} ^{-3}\text{Pa} ^{-1} which is the same thing as \text{molL} ^{-1}\text{atm}^{-1}. In fraction form, these units would appear as the following:

    \begin{gather*} {\frac{\text{mol}}{\text{m}^{3}\text{Pa}} \text{or} \frac{\text{mol}}{\text{L}\text{atm}}} \end{gather*}

List of Henry’s Law Constants

Below is a list of Henry’s Law constants for some common gases in water. A more complete list can be found in this paper and is the source for the constants listed below. There is also a searchable database of constants based on the above paper at this link.

Gas in WaterHenry’s Law Constant (mol*m-3*Pa-1)Henry’s Law Constant (mol*L-1*atm-1)
Hydrogen Bromide (HBr)0.2424.3
Perchloric Acid (HClO4)9.9 * 1031.0 * 106
Hydrogen Fluoride (HF)1.3 * 1021.32 * 104
Nitric Acid (HNO3)2.1 * 1032.12 * 105
Hydrogen (H2)7.7 * 10-67.8 * 10-4
Deuterium (D2)7.9 * 10-68.0 * 10-4
Oxygen (O2)1.2 * 10-51.2 * 103
Bromine (Br2)7.2 * 10-30.73
Sulfur Dioxide (SO2)1.2 * 10-21.22
Methane (CH4)1.4 * 10-51.41 * 10-3
Benzene (C6H6)1.8 * 10-30.18
Henry’s Law Constants for Various Gases in Water

Limitations of Henry’s Law

There are several situations where Henry’s Law proves to be not accurate.

First, the system must be at equilibrium. If the system is not at equilibrium, then the concentration in the solution based on the law will be inaccurate. For a description of what equilibrium is, see this ChemTalk Article.

The equation also does not work well if the gas reacts with the solvent. The reaction means that Henry’s law does not apply to the situation.

And the third limitation is that the law breaks down at high concentrations and pressures.

Henry’s Law Example Problems

Question #1: What is the concentration of hydrogen in water when the partial pressure of hydrogen is 3 atm?

Answer and Solution #1: To answer this problem we will use the equation from the law. From the table above, we know that Henry’s constant for hydrogen in water is 7.8 * 10^{-4} \text{molL} ^{-1}\text{atm}^{-1}.

    \begin{gather*} {C_{g}=kP_{g}} \\ {\text{Concentration} = 7.8 * 10^{-4} \text{molL} ^{-1}\text{atm}^{-1} * 3\text{atm}} \\ \text{Concentration} = 2.34 * 10^{-3} \text{M} \end{gather*}

Remember to pay attention to units in these problems. Our partial pressure was given in atm so in the problem use Henry’s constant that uses the units of atm.

Question #2: Calculate the partial pressure of methane in the air if the concentration in a water sample has a concentration of 7.8 * 10^{-4}\text{M}.

Answer and Solution #2: In this problem we are solving for partial pressure. So we start by rearranging our equation.

    \begin{align*} {C_{g}&=kP_{g}\vphantom{\frac{1}{2}}} \\ {\frac{C_{g}}{k}&=P_{g}} \end{align*}

Then we plug in the values we know. The constant comes from a table of values. The one for methane in water can be found in the table above.

    \begin{align*} {P_{g} &= \frac{7.8 * 10^{-4}\text{molL}^{-1}}{1.41*10^{-3}\text{molL} ^{-1}\text{atm}^{-1}}} \\ {P_{g} &= 0.55 \text{atm}} \end{align*}

Henry’s Law Practice Problems

Problem 1

You have three experiment chambers involving gaseous layers above water. In the first chamber, you have 1\text{atm} of HBr above the water, in the second you have 1\text{atm} of H2, and in the third you have 1\text{atm} of HF. Which chamber has the highest concentration of gas dissolved in the water?

Problem 2

Above 1.00\text{L} of water, you have a 5.00\text{atm} gaseous layer of some unknown pure gas. Using fractional distillation, you extract 7.05*10^{-3}\text{mol} of liquid from the water layer. What is the identity of the gas?

Henry’s Law Practice Problem Solutions

1: HF (1.34*10^{4}\text{M})

2: CH4