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General Chemistry--Unit 2

04/25/09

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Colligative Properties


�Before You Begin:

To master this material you need to be able to perform unit conversion chemical calculations and calculate the concentrations of solutions in molarity, molality, and mole fraction. You should also be familiar with the properties of mixtures and the kinetic molecular theory description of liquids.

 


Colligative Properties

Some of the properties of a solution are dependent primarily on the identity of the solvent and the amount (rather than the identity) of the solute. These properties are known collectively as colligative properties.

Vapor Pressure Lowering and Raoult's Law

In solutions that have a non-volatile solute and a liquid solvent, the presence of the solute alters the behavior of the solvent. The vapor pressure of a pure solvent is always higher than the vapor pressure of a solution. We can use kinetic molecular theory to understand this effect. Vapor pressure is due to the evaporation of a small amount of the liquid. Remember the discussion on phase changes. If the liquid is part of a solution, the non-volatile solute molecules act to disrupt the evaporation process by attracting the solute molecules, since they have intermolecular forces at least as strong as the liquid molecules, they impede evaporation. The presence of the solute does not affect the condensation process, because, since it is non-volatile, it is not present in the vapor phase.

 

'My Dog Has Fleas' Analogy

The solvent molecules outnumber the solute, and they are spread uniformly throughout the solution, so how do they ‘block’ the surface and prevent evaporation? It is a common misconception to think of the solute as a film on the surface that forms an impenetrable barrier to prevent vaporization. WRONG.  Remember that substances dissolve because they have favorable intermolecular attractions with the solvent. Each solute molecule has a sphere of solute with which it is interacting. The solvent sphere, in turn, has another layer of solvent molecules with which they are interacting, and so on, like layers of an onion. Because the non-volatile solute will not enter the vapor phase, the solvent molecules interacting with the solute are less likely to vaporize than if the solute were not there.

To understand the effect of a non-volatile solute on the vapor pressure, imagine that the solute molecules are pet fleas. I keep my pet fleas in my yard which is surrounded by a chain link fence. The fleas are energetic and tiny, so the ones inside the yard can hop through the fence (this is vaporization). The fleas outside the fence are free to hop back inside (this is condensation) just as easily as other fleas hop out. After a while, the number of fleas inside the yard and the number outside the yard settles down and stays fairly constant, even though individual fleas are moving in and out (this is the vapor pressure). Suppose I get a pet dog and keep it in the yard with the fleas. The dog will disrupt the flow of fleas out of the yard, but not because he blocks up all the holes in the fence. While some of the fleas, unaware of the dog, continue to hop in and out as before, some of the fleas are busy interacting with the dog. The dog is too big to fit through the holes in the chain link fence (he is non-volatile), so he wanders randomly throughout the yard. Fleas outside the fence are just as likely to hop inside as before. Fleas inside the fence can still hop out, but fewer choose to do so because they prefer to pursue a dog-related lifestyle (these are part of the solvent shell surrounding the solute particles). Eventually, the number of fleas outside the fence will drop relative to the pre-dog conditions (the vapor pressure of the solution is lower than that of the pure solvent). Note that this process doesn’t depend on the type of dog I have. I could get a pet Chihuahua or Doberman or a cat or even an elephant. Any pet that fleas like (solute) that is too large to fit through the fence (non-volatile) will decrease the number of fleas outside the fence (lower the vapor pressure).

 

Raoult’s Law

To calculate the extent of the vapor pressure lowering we can use Raoult’s Law:

                                                               P=XP'

 

where P is the partial pressure of the component A, X is the mole fraction of component A, and P° is the partial pressure of component A.

An ideal solution is one that obeys Raoult’s law, but the behavior of a real solution follows that predicted by Raoult’s law only approximately. If the intermolecular attractions between the solute and solvent are very strong, the vapor pressure of the solution is even lower than what is predicted by Raoult’s Law. The solute molecules have strong enough attractions to the solvent to prevent evaporation to a greater degree. If the intermolecular attractions between solute and solvent are weaker than those of the pure solvent for itself, the vapor pressure will be lower than that predicted by Raoult’s Law. The vapor pressure will still be lower than that of the pure solvent because part of the surface is physically blocked by solute. However, the lower intermolecular forces allow the solvent molecules to evaporate more easily than they would if the solute had stronger attractions.

Vapor pressure lowering is not often observed in a freshman chemistry lab because it is difficult to accurately measure small pressure differences unless students have access to expensive equipment.

 

Boiling Point Elevation

The boiling point of a pure liquid is lower than the boiling point of a mixture at any given pressure. The presence of solute molecules hinders the liquid leaving the mixture, so it takes higher average kinetic energy and higher temperature. The mechanism of this effect is similar to that of vapor pressure lowering—the solute contributes to the attractive force holding the liquid phase together without taking part in the vaporization.

The extent of the increase in temperature of the boiling point follows the formula

                                                       delta T, boil=Kim
delta T, boil=Tmix-Tpure

 

where ΔTb is the boiling point increase in degrees Celsius, m is molality of the solution in mol/kg, Kb is a proportionality constant with units of °C kg/mol, and i is the van’t Hoff i-factor, which is does not have units. The boiling point elevation constant is unique for each solvent. The i-factor is a measure of the number of particles each solute particle dissociates into when it dissolves. For non-electrolytes, i = 1. For weak solutions of strong electrolytes, i is about equal to the number of ions in the formula unit.

Note that the identity of the solute is not important, only the amount of solute in the solution matters. The properties of the solvent determine the value for the boiling point elevation constant and the boiling temperature of the pure liquid.

The boiling point elevation may be observed qualitatively by measuring the temperature of salt water that has reached a rolling boil and comparing this to the normal boiling temperature of 100 °C. It is not practical to measure the boiling point elevation quantitatively, however, because the concentration of the solution increases as the solvent vaporizes, so it is very difficult to determine the molality.

 

Freezing Point Depression

The freezing point of a pure liquid is higher than the freezing point of a mixture. The presence of solute molecules hinders the formation of the solid phase. We can use kinetic molecular theory to explain this if we think of the freezing point as being the same as the melting point, just from a different perspective. When an ideal mixture freezes, the molecules of one of the components group together to form a crystal structure held together by intermolecular forces. A pure crystalline solid forms as the remaining liquid gets more and more concentrated with the solute impurities. The molecules need kinetic energy to separate them physically from the mixture, so the solidification process doesn’t start until the liquid is colder than the temperature required to freeze the pure liquid. As the liquid freezes, molecules must move through the solute to join the crystal. At the same time, some of the individual molecules of pure solid are melting and mixing with the liquid solution. It is easier for the pure solid to melt than for the solvent to move through the solution so that it can freeze (by easier we mean less restricted by attractions with foreign particles). At a temperature that would be low enough to freeze a pure liquid, the melting process is easier than the freezing process, so the mixture remains liquid. The mixture has to be cooled lower than the normal freezing point for the pure crystalline solid to form from the mixture.

 

This diagram is a cooling curve for a solution. The cooling curve is a graph with the temperature on the y axis and time on the x axis. The liquid solution cools rapidly, so the initial slope of the graph is steep. At the freezing point of the solution the slope changes to nearly horizontal. The solution gradually cools until it has all frozen. The slope changes again as the solid solution continues to cool rapidly. Compared to the cooling curve for a pure liquid, the freezing point of the solution is a few degrees lower. Also a pure liquid has zero slope at its freezing point.

 

The extent of the decrease in temperature of the freezing point follows the formula

                                                        delta T,freeze=Kim
delta T,freeze=Tpure-Tmix

 

where ΔTf is the freezing point decrease in degrees Celsius, m is molality of the solution in mol/kg, Kf is a proportionality constant with units of °C kg/mol, and i is the van’t Hoff i-factor, which is does not have units. The freezing point depression constant is unique for each solvent. The i-factor is a measure of the number of particles each solute particle dissociates into when it dissolves. For non-electrolytes, i = 1. For weak solutions of strong electrolytes, i is about equal to the number of ions in the formula unit.

Note that the identity of the solute is not important, only the amount of solute in the solution matters. The properties of the solvent determine the value for the boiling point elevation constant and the boiling temperature of the pure liquid.

Freshman chemistry students usually perform a freezing point depression experiment to illustrate colligative properties. The purpose of the experiment is often to determine the formula weight of an unknown solute, unfortunately. This confuses some students; the main thrust of colligative properties is that they depend on the identity of the solvent and the amount of the solute yet, in this type of experiment, we use the information to find a formula weight for the solute—it seems contradictory!  The explanation is that the experiment uses the freezing point depression to find the molality. If you know the mass of the solvent used to make a solution with a specific molality, you can find the moles of solute. Divide the mass of the solute by the number of moles to get the formula weight of the solute.

A real world example of freezing point depression is the use of antifreeze in a car radiator, salting roads to melt ice, and using rock salt when you make ice cream.


Concept Check: If used to melt ice, which would do a better job, sugar (a non-electrolyte), salt, or calcium chloride?

Answer: The calcium chloride would work better. Because the sugar is a non-electrolyte, it does not break apart into smaller but more numerous ions when it dissolves. The sodium chloride and calcium chloride are ionic compounds. These ionize in solution to form two and three ions per formula unit, respectively. Because calcium chloride forms more ions, and colligative properties depend on the number of solute particles, the calcium chloride will result in a larger freezing point depression.

Additional benefits can be seen if we investigate cost (sugar is the most expensive, salt is the least expensive), damage (salt harms concrete and brick), and heat (calcium chloride actually releases considerable heat when it dissolves as we will see in the unit about thermochemistry).



Concept Check: Automobile antifreeze is ethylene glycol, C2H6O2. It is a non-electrolyte. If a radiator contains 40.0% antifreeze and 60.0% water, by mass, what is the freezing point of the solution in the radiator? The normal freezing point for water is 0.0 °C and Kf is 1.86 °C mol/kg.

Answer: Find the molality of the solution. The mass of the solvent is 0.0600 kg and the formula weight of the solute is 62.066 g/mol.

                                          m=[40.0g/(62.066g/mol)]/0.0600kg=10.7mol/kg

 

 

 

Use the freezing point depression formula

                                 delta T,freeze=Kim=(1.86Cmol/kg)(1)(10.7kg/mol)=20.0C
delta T, freeze=0.0C-20.0C= -20.0C


Osmosis

Osmosis is the passage of certain molecules through a membrane. For example, water will flow through a cell membrane. Membranes are semi-permeable, because they permit molecules that are small enough to pass while serving as a barrier to other molecules. For example, animal cells are surrounded by cell membranes which keep the contents of the cell inside but allow water to flow in and out.

If solutions of differing concentrations are separated by a semi-permeable membrane, small molecules will flow through the membrane from the side with the most to the side with the least until the amounts on both sides of the membrane are equal. Frequently, water is able to pass through a semi-permeable membrane, since it is a small molecule, but solute molecules cannot. If this is the case, water will flow from the side with the lowest concentration (more water relative to solute) to the side with the highest concentration (less water relative to solute), diluting it until both sides are equal.

We can use kinetic molecular theory to explain this process. If we examine a semi-permeable membrane under magnification, we see that it is porous. Small molecules can fit inside the pores. Random molecular motions send small molecules from one side of the membrane to the other through the pores. Large molecules cannot fit through the pores, though. Not only do the large molecules stay on one side of the barrier, they hinder the movement of small molecules through, rather like they hinder liquid molecules from evaporating from a solution (remember the dog and fleas analogy). The small molecules can travel into the side with large solute more readily than they can travel out of the side with the solute.

diagram of osmosis

 

 

The force generated by the net flow of small molecules through a semi-permeable membrane will support a column of liquid; much like air pressure supports a column of mercury in a barometer. We can prevent the flow of small molecules by exerting pressure against the direction of the flow. The amount of force needed to equal the force due to the flow of small molecules is called the osmotic pressure, π.

The osmotic pressure for an ideal solution is analogous to the ideal gas law

                                                               piV=nRT

 

where T is temperature in Kelvin, V is volume, R is the gas constant and n is the number of moles of solute. This can be rearranged to

                                                          pi=(n/V)RT=MRT

 

where M is molarity of the solution, since n is number of moles of solute and V is the volume of the solution.

This diagram illustrates osmotic pressure. A U-shaped tube has a semi-permeable membrane across the bottom separating the tube into two halves. A high concentration solution is in one half. A low concentration solution is in the other half. Osmosis occurs; the small particles travel from low to high concentration across the membrane. The liquid level on the high conentration side is higher than the liquid level on the low concentration side. The osmotic pressure is equal to the force needed to support a column of liquid of this height.

 

 

 

 

 

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