# Henry's Law

Henry’s law states that the concentration of a dissolved gas in a liquid is directly proportional to the partial pressure of the gas above the liquid.

$C_g = kP_g$

Here, Cg is the concentration of the dissolved gas (in mol L–1), k is Henry’s constant (specific to the gas), and Pg is the partial pressure of the gas. In the figure above, the amount of dissolved carbon dioxide in water increases when increasing the pressure of carbon dioxide inside the container (left to right).

Henry’s constant can be determined from careful experimentation. A fictitious gas dissolved in water is found to have the following concentrations and partial pressures at a constant temperature:

• 0.100 m – 2.0 atm
• 0.125 m – 2.5 atm
• 0.150 m – 3.0 atm

By dividing the concentration (in m) by the partial pressure of the gas (in atm), the Henry’s constant for this gas at the measured temperature is

\begin{align*} k_{\mathrm{H}} &= \dfrac{C_{\mathrm{g}}}{P_{\mathrm{g}}} \\[1.5ex] &= \dfrac{0.100~m}{2.0~\mathrm{atm}} \\[1.5ex] &= 0.05~m~\mathrm{atm}^{-1} \end{align*}

## Henry’s Law Constant

Henry’s Law constants are generally reported in the scientific literature with SI units given as mol m–3 Pa–1. The conversion to mol L–1 atm–1 can be written as

$\dfrac{1~\mathrm{mol}}{\mathrm{m^3~Pa}} \left ( \dfrac{1~\mathrm{m^3}}{1000~\mathrm{L}}\right ) \left ( \dfrac{101,325~\mathrm{Pa}}{1~\mathrm{atm}} \right ) = 101.325~\mathrm{mol~L^{-1}~atm^{-1}}$

Therefore, the conversion factor from SI units to mol L–1 atm–1 is 101.325.

Example:

An authoritative database on Henry’s Law constants (see paper) reports kH (given as as Hcp) for carbon dioxide (CO2) as 3.3 × 10–4 mol m–3 Pa–1. We can convert this to mol L–1 atm–1 given as

$\dfrac{3.3\times 10^{-4}~\mathrm{mol}}{\mathrm{m^3~Pa}} \left( \dfrac{101.325~\mathrm{m^3~Pa}}{\mathrm{L~atm}} \right ) = 3.34\times 10^{-4}~\mathrm{mol~L^{-1}~atm^{-1}}$

## Practice Problems

### Problem 1

The Henry’s law constant for CO2 is 3.34 × 10–2 M atm–1 at 25 °C. What pressure (in atm) of CO2 is needed to maintain a CO2 concentration of 0.100 M in a can of lemon-lime soda?1

Solution

\begin{align*} C_{\mathrm{g}} &= k_{\mathrm{H}} P_{\mathrm{g}} \\[1.5ex] P_{\mathrm{g}} &= \dfrac{C_{\mathrm{g}}}{k_{\mathrm{H}}} = \dfrac{0.100~M}{3.34\times 10^{-2}~M~\mathrm{atm}^{-1}} \\[1.5ex] &= 2.99~\mathrm{atm} \end{align*}

### Problem 2

What is the maximum concentration (in M) of CO2 that can be dissolved in a can of lemon-lime soda if the maximum pressure that the can is able to withstand is 4.6 atm (at 25 °C)? k(CO2) = 3.34 × 10–2 M atm–1

Solution

\begin{align*} C_{\mathrm{g}} &= k_{\mathrm{H}} P_{\mathrm{g}} \\[1.5ex] &= \left ( 3.34\times 10^{-2}~M~\mathrm{atm}^{-1} \right ) \left ( 4.6~\mathrm{atm} \right )\\[1.5ex] &= 0.15~M \end{align*}

### Problem 3

What mass (in g) of CO2 is dissolved in 500. mL of lemon-lime soda that is under a pressure of 3.80 atm (at 25 °C)? k(CO2) = 3.34 × 10–2 M atm–1

Solution

\begin{align*} C_{\mathrm{g}} &= k_{\mathrm{H}} P_{\mathrm{g}} \\[1.5ex] &= \left ( 3.34\times 10^{-2}~M~\mathrm{atm}^{-1} \right ) \left ( 3.80~\mathrm{atm} \right )\\[1.5ex] &= 0.127~M \\[2.5ex] m_{\mathrm{CO_2}} &= M_{\mathrm{solute}} \times V_{\mathrm{solution}} \times \mathrm{~molar~mass} \\[1.5ex] &= \left ( \dfrac{0.12~\mathrm{mol~CO_2}}{\mathrm{L~soln}} \right ) \left ( 0.500~\mathrm{L~soln} \right ) \left ( \dfrac{44.01~\mathrm{g}}{\mathrm{mol}} \right ) \\[1.5ex] &= 2.79~\mathrm{g} \end{align*}

### Problem 4

At 20 °C, the concentration of O2 in water with a partial pressure of 1 atm is 1.38 × 10–3 M. What is the solubility (in M) of O2 when the partial pressure is 0.204 atm? k(O2) = 1.32 × 10–3 M atm–1

Solution

\begin{align*} C_{\mathrm{g}} &= k_{\mathrm{H}} P_{\mathrm{g}} \\[1.5ex] &= \left ( 1.32\times 10^{-3}~M~\mathrm{atm}^{-1} \right ) \left ( 0.204~\mathrm{atm} \right )\\[1.5ex] &= 2.69\times 10^{-4}~M \end{align*}

## Timed Assessments

### Henry’s Law

VIDEO

1. 7.06 atm
2. 0.156 M
3. 0.703 M atm–1

VIDEO