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.

Figure: 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

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

Practice Problems


Problem 1

The Henry’s law constant for CO2 is 3.4 \(\times\) 10–2 M atm–1 at 25 °C. What pressure (in atm) of CO2 is needed to maintain a CO2 concentration of 0.10 M in a can of lemon-lime soda?1

Solution

\[\begin{align*} C_{\mathrm{g}} &= k_{\mathrm{H}} P_{\mathrm{g}} \\[1.5ex] P_{\mathrm{g}} &= \dfrac{C_{\mathrm{g}}}{k_{\mathrm{H}}} = \dfrac{0.10~M}{3.4\times 10^{-2}~M~\mathrm{atm}^{-1}} \\[1.5ex] &= 2.9~\mathrm{atm} \end{align*}\]


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 \(^{\circ}\)C)? \(k_{\mathrm{CO_2}} = 3.2\times 10^{-2}\) M atm–1

Solution

\[\begin{align*} C_{\mathrm{g}} &= k_{\mathrm{H}} P_{\mathrm{g}} \\[1.5ex] &= \left ( 3.2\times 10^{-2}~M~\mathrm{atm}^{-1} \right ) \left ( 4.6~\mathrm{atm} \right )\\[1.5ex] &= 0.15~M \end{align*}\]


Problem 3

What mass (in g) of CO2 is dissolved in 500. mL of lemon-lime soda that is under a pressure of 3.8 atm (at 25 \(^{\circ}\)C)? \(k_{\mathrm{CO_2}} = 3.2\times 10^{-2}\) M atm–1

Solution

\[\begin{align*} C_{\mathrm{g}} &= k_{\mathrm{H}} P_{\mathrm{g}} \\[1.5ex] &= \left ( 3.2\times 10^{-2}~M~\mathrm{atm}^{-1} \right ) \left ( 3.8~\mathrm{atm} \right )\\[1.5ex] &= 0.12~M \\[2.5ex] m_{\mathrm{CO_2}} &= M_{\mathrm{solute}} \times V_{\mathrm{solution}} \times \mathrm{~molar~mass} \\[1.5ex] &= \left ( \dfrac{0.12~\mathrm{mol~CO_2}}{\mathrm{L~soln}} \right ) \left ( 0.500~\mathrm{L~soln} \right ) \left ( \dfrac{44.01~\mathrm{g}}{\mathrm{mol}} \right ) \\[1.5ex] &= 2.6~\mathrm{g} \end{align*}\]


Problem 4

At 20 °C, the concentration of O2 in water with a partial pressure of 1 atm is 1.38 \(\times\) 10–3 M. What is the solubility (in M) of O2 when the partial pressure is 0.204 atm? \(k_{\mathrm{O_2}} = 1.3\times 10^{-3}\) M atm–1

Solution

\[\begin{align*} C_{\mathrm{g}} &= k_{\mathrm{H}} P_{\mathrm{g}} \\[1.5ex] &= \left ( 1.3\times 10^{-3}~M~\mathrm{atm}^{-1} \right ) \left ( 0.204~\mathrm{atm} \right )\\[1.5ex] &= 2.6\times 10^{-4}~M \end{align*}\]


Timed Assessments


Henry’s Law

VIDEO

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


Henry’s Law - mass

VIDEO

Answers
  1. 2.54 g
  2. 18.7 g
  3. 102.7 g


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