Practice: Henry’s Law Constant for nitrogen in water is 1.67 × 10^{-4} M • atm^{–1}. If a closed canister contains 0.103 M nitrogen, what would be its pressure in atm?

To calculate *solubility* of a dissolves gas, **Henry's Law Constant** and *partial pressure* are used.

Concept #1: Henry's Law Formula

Example #1: Calculate the solubility of carbon dioxide gas, CO_{2}, when its Henry’s Law Constant is 8.20 x 10^{2} M/atm at 3.29 atm?

Concept #2: Henry’s Law (2 Point Form)

**The 2 Point Form of Henry's Law Formula** is used when dealing with 2 pressures and 2 solubilities of a given gas.

Example #2: At a pressure of 2.88 atm the solubility of dichloromethane, CH_{2}Cl_{2}, is 0.384 mg/L. If the solubility decreases to 0.225 mg/L, what is the new pressure?

Practice: Henry’s Law Constant for nitrogen in water is 1.67 × 10^{-4} M • atm^{–1}. If a closed canister contains 0.103 M nitrogen, what would be its pressure in atm?

Practice: At 0°C and 1.00 atm, as much as 0.84 g of O_{2} can dissolve in 1.0 L of water. At 0°C and 4.00 atm, how many grams of O_{2} dissolve in 1.0 L of water?

Practice: The atmospheric pressure in a lab is calculated as 1.3 atm. If oxygen gas contributes 62% of this atmospheric pressure, determine its mass (in g) dissolved at room temperature in 25 L of water. The Henry’s Law Constant for oxygen in water at this temperature is 5.3 × 10^{–5} M/atm.