What is a solution? Solutions are homogeneous mixtures of two or more pure substances. A homogeneous mixture is a physical combination of two or more pure substances whose distribution throughout the mixture is uniform. What this means is that if we were to make a solution and take only a portion of the solution at random called an aliquot, the proportion of each pure substance in the aliquot would be the same as the proportion of that pure substance in the whole solution. We call these proportions to the whole solution the concentration. We are going to restrict our focus on this topic to two component solution systems with the minor component called the solute, and the major component called the solvent. Furthermore, we will restrict most of our comments to aqueous solutions where water is the solvent.

Within any solution there are electrical interactions called intermolecular forces between the solute and solvent molecules. In aqueous solutions the polar water molecules will be attracted to other molecules that are also polar. The water molecules will surround each solute molecule and form a "solvent cage" isolating each solute molecule. We say the solute molecules have been solvated. If the bonds holding the solute molecule together are weak enough, the weakest bond within the solute molecule could be broken by the electrical force of attraction the water molecules have for the solute molecules. If this should take place the solute molecule will undergo what is called ionization forming ions (fragments of the solute molecule) within the solution. In some cases the ionization process can be quite complete while in other cases the ionization process will occur in a very limited manner. Those solutions that do not undergo ionization have the solute molecules solvated within the solution and are called non-electrolytes. Such solutions will not conduct electrical current because of the absence of ions within the solution. Those solutions whose molecules undergo ionization during the solution process are called electrolytes and those solutions will conduct electrical current because of the presence of ions.

When solutions form the solvent molecules form solvent cages around the solute molecules. This solvation process involves a certain amount of thermal energy to be exchanged between the solution system and its immediate surrounding environment. Some solutions absorb energy as they are formed. The solution is said to have an endothermic Heat of Solution.

Most aqueous solutions involving liquid or solid solutes will have endothermic Heats of Solution. However, a few are exceptions to this statement.

Other solutions involving gaseous solutes in water will release thermal energy during the solution formation process. These solutions are said to have an exothermic Heat of Solution.

Solutions can be unsaturated, saturated, or supersaturated. Unsaturated solutions are those that are below the solubility limits of the solute in that solvent. Saturated solutions are those that are at the solubility limits. Supersaturated solutions are those solutions that are above the solubility limits. Supersaturated solutions are meta stable. Such solutions will have the excess solute crystalize out with any disturbance of the supersaturated solution. Providing a tiny crystal of the solute or scratching the sides of the container which introduce micro chips of glass into the supersaturated solution so the excess solute can crystalize upon its surface, will illicit the dramatic crystallization of the excess solute out of the solution restablishing a saturated solution.

The solubility of a particular solute in a solvent is the maximum amount of solute that will dissolve in a specified amount of solution or solvent. It represents the saturated level of the solution where no more solute will dissolve within the solution. This saturated condition creates a dynamic physical equilibrium between the solute and solvent and the solution:

Such dynamic equilibrium involve two processes, a forward process and a reverse process. When the rate of the forward process is equal to the rate of the reverse process, then the system is said to be in a dynamic equilibrium. Dynamic Equilibria will involve two opposing processes occuring simultaneously. We can prove this by taking a salt crystal and chipping off one end. Then we suspend the deformed cystal in a saturated solution of the salt and allow the saturated solution to be in contact with the deformed cystal for several weeks. When we come back we will have discovered that the deformed crystal will have been reformed. This could only happen if the solution process and its reverse process, the dissolution process, were occuring simultaneously.

There are three factors that explain why solubilities will vary even within the same solution.

1. Solute/Solvent interactions- The molecular size of the solute molecules will affect the solubility. The larger the solute molecules the more likely the solubility will diminish. In addition, the polar nature of the solute molecules compared to the solvent will also alter the solubility. For aqueous solutions where polar water is the solvent, the more polar the solute molecules are the more likely the solubility will be higher. Non-Polar Hydrocarbons have a very low solubility in polar water. So do such things as Carbon Tetrachloride, Oxygen gas, Nitrogen gas, and other non-polar molecular substances. On the other hand polar solutes like ionic salts and polar molecular substances will have higher solubilities in water. There is an old addage "Like dissolves in Like". If we were to use a non-polar solvent instead of polar water then the non-polar solutes above would have higher solubilities in that solvent.

2. Temperature- Generally speaking the water solubility of a liquid or solid will increase with increasing temperature. However, there are some exceptions to this. Some solutes like solid Ce2(SO4)3 will have a decreasing water solubility with increasing temperature. This depends upon the thermodynamics of the solution process. Since a solution that has reached the solubility limit is in equilibrium, the various laws of equilibrium apply to that system. There is a principle that will be more completely discussed in a future lessons called Le Chatlier's Principle. The principle says that when a stress is applied externally to an equilibrium, the equilibrium is disrupted temporarily and will shift in such a way as to undo the stress that had been applied. One such stress that can be applied is temperature change. According to the Principle, increasing the temperature of an equilibrium will always favor the endothermic process of an equilibrium since it is the endothermic process that can absorb the added energy resulting from the increase in temperature. That effectively counteracts the temperature increase. Since most liquid and solid solutes dissolved in water have the solution formation process endothermic, that would be favored when the temperature was increased resulting in an increase in the solubility limit. However some solutes like Ce₂(SO₄)₃ have an exothermic Heat of Solution. Since increasing the temperature will always favor the endothermic process the dissolution (solution breakdown) process endothermic) will be favored and the solubility will decrease. Gaseous solutes always have an exothermic Heat of Solution. Consequently, the solubility of all gases in water decrease with increasing temperature. That is why carbonated drinks that have Carbon Dioxide gas dissolved in them will become "flat" tasting when heated. The sparkle of the drink will have disappeared along with the Carbon Dioxide gas.

3. Pressure- Pressure changes above the solution do not affect the solubility limits of solids or liquids dissolved in water. However gaseous solutes are affected. If the pressure of the gas is increased above the gaseous solution then the solubility will be increased in a linear fashion. This was investigated by Henry and resulted in Henry's Law.

Henry noted that when you increased the pressure of a gas above a gaseous solution, the concentration of the gas in the solution would increase in a linear fashion.

where: k = Henry's constant for that gas

P = Partial Pressure of the gas above the solution

C = concentration of the gas in the solution

Twenty-seven grams of acetylene gas dissolves in 1 liter of acetone at 1 atm partial pressure of acetylene. If the partial pressure of acetylene is increased to 6 atm, what is the solubility of acetylene in acetone in grams / liter?

1. Identify the Initial Pressure

   1 atm pressure given

2. Identify the Initial Concentration in grams/liter:

   27 grams / 1 liter

3. Determine the Henry's Constant, k from the Henry's Law math statement:

   C = k P

   C = 27 grams/liter

   P = 1 atm

   27 grams / liter = k (1atm)

   k = 27 / 1 = 27 gram/ liter-atm

4. Identify the final pressure, 6 atm

5. Use the k determined above and the final pressure and determine the final cancentration using Henry's Law math formula:

   C = 27 grams / liter-atm X 6 atm = 162 grams/ liter

Sample Problem:

A liter of water dissolves 0.0404 grams of Oxygen gas when the partial pressure of the Oxygen gas is 1 atm. How many grams of Oxygen will dissolve in 1 liter of water when the partial pressure is 10 atm? Please work the problem and click here in order to check your results.

Further information about dilution and Molarity Problems are found at ChemTeam's Introduction to Solutions

R. H. Logan, Instructor of Chemistry, Dallas County Community College District, North Lake College.

Send Comments to R.H. Logan:

rhlogan@ix.netcom.com

All textual content copyrighted (c) 1997
R.H. Logan, Instructor of Chemistry, DCCCD
All Rights reserved

Sample Problem:

A liter of water dissolves 0.0404 grams of Oxygen gas when the partial pressure of the Oxygen gas is 1 atm. How many grams of Oxygen will dissolve in 1 liter of water when the partial pressure is 10 atm?

1. Identify the initial concentration, .0404 grams / 1 liter

2. Identify the initial pressure, 1 atm

3. Calculate the Henry's Law Constant, k from Henry's Law math statement:

   .0404 g / l = k (1atm)

   k = .0404 grams / liter / 1 atm = k = .0404 grams / liter atm

4. Identify the final pressure, 4 atm

   Plug k and the final pressure into Henry's Equation and solve for final Concentration:

   C = k P = .0404 grams / liter-atm (4 atm) = .1616 grams / liter

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   Revised: 10/21/98

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