Solubility of gases in liquids
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Henry's Law |
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The solubility of a gas in a liquid depends on temperature, the partial pressure of the gas over the liquid, the nature of the solvent and the nature of the gas. The most common solvent is water. Gas solubility is always limited by the equilibrium between the gas and a saturated solution of the gas. The dissolved gas will always follow Henry's law. The concentration of dissolved gas depends on the partial pressure of the gas. The partial pressure controls the number of gas molecule collisions with the surface of the solution. If the partial pressure is doubled the number of collisions with the surface will double. The increased number of collisions produce more dissolved gas. |
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The illustration
shows that if the pressure is doubled then the
concentration of dissolved gas will double.
Low pressure
equilibrium Double the pressure
equilibrium Low
concentration Double the
concentration
The dissolving process for gases is an equilibrium. The solubility of a gas depends directly on the gas pressure. The number of molecules leaving the gas phase to enter the solution equals the number of gas molecules leaving the solution. If the temperature stays constant increasing the pressure will increase the amount of dissolved gas.
The Henry's law constant "k" is different for every gas, temperature and solvent. The units on "k" depend on the units used for concentration and pressure. The value for k is the same for the same temperature, gas and solvent. This means the concentration to pressure ratio is the same when pressures change. The following equation can be used to relate pressure and concentration changes. |
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Example: |
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The concentration of dissolved oxygen is 0.44g / 100 mL solution. The partial pressure of oxygen is 150 mm Hg. What is the predicted concentration if the partial pressure for oxygen is 56 mm Hg? |
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Solution: |
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P1 = 150 mm Hg--------------------------------------------- |
C1 = 0.44 g O2 /100 mL solution |
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P2 = 56 mm Hg----------------------------------------------- |
C2 = ? |
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Online
Introductory Chemistry
Dr. Walt Volland all
rights reserved revised March 29, 2005