Henry’s Law — Effect of Pressure on the Solubility of Gases
At constant temperature, the equilibrium solubility of a sparingly soluble gas in a liquid is
directly proportional to the partial pressure of that gas above the solution. In the
concentration–pressure form used by the calculator:
\[
C = k\,P_{\mathrm{gas}}
\]
Here \(C\) is the concentration of dissolved gas, \(P_{\mathrm{gas}}\) is the gas’s partial
pressure at the interface, and \(k\) is Henry’s constant. The unit of \(k\) follows the chosen
concentration unit:
mol·L⁻¹·atm⁻¹ (if \(C\) in mol·L⁻¹), mg·L⁻¹·atm⁻¹ (if \(C\) in mg·L⁻¹),
or mL·L⁻¹·atm⁻¹ (if \(C\) in mL·L⁻¹).
Symbols & Units
- \(C\) — solubility (e.g., \(\mathrm{mol\,L^{-1}}\), \(\mathrm{mg\,L^{-1}}\), or \(\mathrm{mL\,L^{-1}}\)).
- \(P_{\mathrm{gas}}\) — partial pressure of the gas (atm). The calculator converts kPa and bar to atm.
- \(k\) — Henry’s constant with units \([\text{concentration}]\,\mathrm{atm^{-1}}\).
- \(V_g\) — molar gas volume at the stated conditions (optional, \(\mathrm{L\,mol^{-1}}\)) used only to convert \(\mathrm{mL\,L^{-1}}\rightarrow \mathrm{mol\,L^{-1}}\):
\(\displaystyle C\ (\mathrm{mol\,L^{-1}})=\frac{C\ (\mathrm{mL\,L^{-1}})}{1000}\cdot\frac{1}{V_g}\).
How the Calculator Solves
- Evaluate the constant \(k\) from one reference pair.
\[
k = \frac{C_1}{P_1}
\]
Use the same concentration unit you intend to use for predictions.
The result has units \([\text{concentration}]\,\mathrm{atm^{-1}}\) (for example \(\mathrm{mol\,L^{-1}\,atm^{-1}}\)).
- Predict a new solubility at a different pressure.
\[
C_2 = k\,P_2
\]
- Or compute the required pressure for a target solubility.
\[
P_2 = \frac{C_2}{k}
\]
- (Optional) Convert \(\mathrm{mL\,L^{-1}}\) to \(\mathrm{mol\,L^{-1}}\).
If you provide \(V_g\) (e.g., \(22.414\ \mathrm{L\,mol^{-1}}\) at STP):
\[
C\ (\mathrm{mol\,L^{-1}})=\frac{C\ (\mathrm{mL\,L^{-1}})}{1000}\cdot\frac{1}{V_g}.
\]
Pressure Units
The tool accepts atm, kPa, or bar for \(P_1\) and \(P_2\) and internally converts to atm using
\(1\ \mathrm{atm}=101.325\ \mathrm{kPa}=1.01325\ \mathrm{bar}\). Always use the partial
pressure of the gas of interest (e.g., for \(\mathrm{O_2}\) in air near sea level,
\(P_{\mathrm{O_2}}\approx 0.2095\ \mathrm{atm}\)).
Worked Example
At \(0^\circ\mathrm{C}\) and \(P_1=1.00\ \mathrm{atm}\), the solubility of \(\mathrm{O_2(g)}\) in water is
\(C_1=48.9\ \mathrm{mL\,L^{-1}}\). Find the solubility at \(P_2=0.2095\ \mathrm{atm}\), and then express it in
\(\mathrm{mol\,L^{-1}}\) using \(V_g=22.4\ \mathrm{L\,mol^{-1}}\).
\[
k=\frac{C_1}{P_1}=\frac{48.9\ \mathrm{mL\,L^{-1}}}{1.00\ \mathrm{atm}}=48.9\ \mathrm{mL\,L^{-1}\,atm^{-1}}
\]
\[
C_2=k\,P_2=(48.9\ \mathrm{mL\,L^{-1}\,atm^{-1}})\cdot(0.2095\ \mathrm{atm})=10.25\ \mathrm{mL\,L^{-1}}
\]
\[
C_2\ (\mathrm{mol\,L^{-1}})=\frac{10.25\ \mathrm{mL\,L^{-1}}}{1000}\cdot\frac{1}{22.4\ \mathrm{L\,mol^{-1}}}
=4.57\times 10^{-4}\ \mathrm{mol\,L^{-1}}.
\]
Assumptions & Limitations
- Dilute solutions. Henry’s law holds best at low \(C\) where gas–solute interactions are ideal.
- Constant temperature. \(k\) is sensitive to \(T\); do not mix data sets at different temperatures.
- No chemical reaction. If the gas reacts (e.g., \(\mathrm{CO_2}\) forming \(\mathrm{HCO_3^-}\)), the simple linear relation may not apply.
- Partial pressure, not total pressure. Use \(P_{\mathrm{gas}}\) of the species of interest.
- Salting-out & solvent identity. Electrolytes and different solvents change \(k\) significantly.
Quality Checks Before Trusting a Result
- Units are consistent: concentration unit for \(C_1\), \(C_2\) matches the unit used to form \(k\).
- Pressures converted properly to atm before applying formulas.
- If converting mL·L⁻¹ to mol·L⁻¹, a plausible \(V_g\) (same \(T,P\)) is used.
Other common forms of Henry’s law exist, e.g., \(P_{\mathrm{gas}}=k_x\,x_{\mathrm{gas}}\) (mole-fraction form).
The calculator here uses the concentration–pressure form \(C=k\,P_{\mathrm{gas}}\) for clarity in laboratory contexts.
