Colligative Properties

Colligative properties are properties of a solution that depend mainly on the relative numbers of particles of solvent and solute molecules and not on the detailed properties of the molecules themselves. You could almost refer to these as statistical properties because they can be understood solely on the basis of counting the relative numbers of particles in a solution. We will derive equations for the colligative properties of ideal solutions. The equations we derive will be valid for ideal solutions and for real solutions in the limit of small concentrations. Nonideal solutions require that corrections be made to these ideal equations because in nonideal solutions the details of intermolecular interactions become important.

We will discuss four colligative properties:

1. Vapor pressure depression
2. Boiling point elevation
3. Melting point depression
4. Osmotic pressure
In all of the following discussion we will denote the solvent as component #1 and the solute as component #2. For example, the mole fraction of component 1 will be symbolized by X1, and so on.

1) Vapor Pressure Depression

Vapor pressure depression is the simplest of the colligative properties and the easiest to understand on the basis of a physical model. A given solvent has a vapor pressure which we usually denote by p1* . That means that at equilibrium the gas phase above the solvent has a solvent partial pressure of p1* . When you add a solute to the solvent to make a solution the partial pressure of the solvent in the gas phase decreases. We can show that this is true using Raoult's law. In this case we have X1 close to 1 and X2 small. Then Raoult's law applied to the vapor pressure (partial pressure in the gas phase) of the solvent is,

p1 = X1p1* = (1 − X2)p1*,                (1)
where we have taken advantage of the fact that in a two-component system, X1 + X2 = 1. Rearrange equation 1 to show the change in the vapor pressure of component 1,
    p1p1* = − X2 p1*.                (2)
We see that the change in the vapor pressure is negative which means that the vapor pressure decreased and it decreased by an amount that is proportional to the vapor pressure of the pure liquid and proportional to the fraction of solute molecules in the solution (that is, the ratio of the number of solute molecules to the total number of molecules).

Vapor pressure depression is relatively easy to understand on the basis of a physical model. At the surface of a liquid there is a competition between the kinetic energy of the molecules (thermal energy), which is trying to push the molecules off the surface into the gas phase, and the intermolecular forces, which are trying to keep the molecules on the surface. (We will see later that the kinetic energy depends on temperature and that the average kinetic energy is proportional to the Kelvin temperature. However, at any given temperature there is a distribution of kinetic energies with most of the molecules having a kinetic energy near the average, but with some higher and some lower. As you increase the temperature you get more molecules with the higher kinetic energy.)

Some of the molecules on the surface with sufficient kinetic energy to overcome the intermolecular forces will escape into the gas phase and contribute to the pressure of solvent molecules in the gas phase. As this pressure increases, there are more molecules in the gas phase colliding with the surface and sticking. At equilibrium the number of molecules leaving the surface just balances those returning to the surface. The measured pressure at this point is the vapor pressure. The solute molecules decrease the vapor pressure because some of the solvent molecules on the surface have now been replaced by solute molecules. There are fewer solvent molecules on the surface to escape and the vapor pressure goes down. You can calculate how much the pressure goes down by counting how many of the solvent molecules on the surface have been replaced by solute molecules.

We can see graphically how the vapor pressure of the solvent decreases with the addition of solute by looking at the vapor pressure diagram below. In this diagram the vapor pressure of the solute (component 1) is plotted against X1 at constant temperature using Raoult's law. The circled area represents the region of small X2 we are talking about, although for an ideal solution the straight line extends to X1 equals zero. The graph in the circled area would be approximately correct even for nonideal solutions because all solutions become ideal with respect to the solvent in the limit as X1 goes to unity.


    2) Boiling Point Elevation
It is not hard to understand qualitatively why the boiling point of a solution would be higher than the boiling point of the pure solvent. (We are assuming that the solute is a nonvolatile compound and does not make any significant contribution to the vapor pressure of the solution. If the solute is a volatile compound with a lower boiling point than the solvent the boiling point of the solution will actually go down - by Raoult's law.) We have learned that the vapor pressure of the solvent decreases in a solution. If our solvent was originally at the boiling point (p1 = 1 atm) adding a nonvolatile solute will lower the vapor pressure and the mixture will no longer be boiling. We can bring it back to boiling by increasing the temperature. Thus, the boiling point of the solution is higher that that of the pure solvent.

In the diagram below on the left side we have pure solvent at its boiling point. We know that the chemical potential of the liquid solvent must equal the chemical potential of the solvent in the vapor phase. That is,

                (3)
Thus, for the vaporization process,
liquid → vapor, >                (4)
we have,
                (5)

Approximating the vapor as an ideal gas and using the ideal solution equations for the solution we get.
                (6)
and
                (7)
where, as usual,  is the chemical potential of the pure solvent, is the standard state potential for the gas (vapor), and there is an implied 1 atm dividing p1 inside the logarithm. Then the difference in chemical potentials can be written,
                (8)
We started out, on the left of the diagram, with pure liquid solvent in equilibrium with its vapor at the normal boiling point. Let us now add a small amount of solute, component 2. Add enough to increase the mole fraction of 2 in the solution by dX2. We insist, however, that the solvent in the solution remain in equilibrium with the solvent vapor. We now ask the question what must the temperature change, dT, be to maintain equilibrium? The answer to this question is contained in the partial derivative,
                (9)
We can figure out how to calculate this derivative by using the cyclic relation
                (10)
The upper derivative on the right is easily obtained from our expression for Δμ1 above and the lower derivative on the left can be obtained by recognizing that,
                (11)
or, for the vaporization process,
                (12)
Then,
                (13)
We must simplify this expression, but we note first that the above expression requires no specification of the value of Δμ1 other than that it be held constant during the derivative process. However, we know that in order for the solvent to be in equilibrium with its vapor we must have not only Δμ1 = constant, but it must be held at the particular constant value of zero. That is, Δμ1 = 0. In this case,
                (14)
from which we conclude that,
                (15)
( We have added the subscript "vap" to remind ourselves that we are talking about the vaporization process. With these considerations Equation 13 simplifies to,
                (16)
We may need this form of the equation a little later on, but for now we are going to make some approximations. We assume that X2 is small so that 1 − X2 ≈ 1, then,
                (17)
We can set this up to integrate as follows
                (18)
If we are only going to integrate over a small range then our temperature is approximately constant at the boiling point, TBP, and we can write
                (19)
but ΔX2 = X2 − 0, since we started out at zero concentration of solute, so we get
                (20)
This equation actually solves our problem, but it is not in the form you saw in freshman chemistry. The freshman chemistry form had molality of the solute in place of the mole fraction of solute. To get the form you expect we must relate the mole fraction to the molality,
                (21)
where m2 is the molality of solute. That is, m2 is the number of moles in 1000 g of solvent. (Also, Mw1 is the formula weight of the solvent.) Since we are assuming a dilute solution the number of moles of solute is much smaller than the number of moles of solvent so we can drop the m2 in the denominator to get
                (22)
Plugging this expression for mole fraction into the formula we have above for the change in the boiling point we get
                (23)
The expression in square brackets is the boiling point elevation constant (or, if you like big words, the ebullioscopic constant).

If we are dealing with more concentrated solutions we can go back and integrate one of the earlier equations, like

                (24)
or
                (25)
The latter can be integrated from the boiling point of the pure liquid, TBP at X2 = 0 to some new boiling point, T, at concentration X2. Integration yields,
                (26)

3) Freezing point depression

Freezing point depression is done the same way we did boiling point elevation only this time we must keep the solvent in equilibrium with the pure solid rather than with the vapor. That means we consider the fusion process,

solid → liquid.                (27)
This time the chemical potential the solid, μ1s , which is the same as the chemical potential of the pure solid solvent, μ1s*, must be the same as the chemical potential of the solvent in the solution, which we have seen before,
                (28)
but this time
                (29)
and
                (30)
if the solid and solvent are in equilibrium. Under equilibrium conditions it is true, as before, that
                (31)
We ask the same question as in our discussion of boiling point elevation. If we add a small amount of solute, component 2, to the solvent to produce an increase in X2 of dX2, and we insist that the solid stay in equilibrium with the solvent, what must the temperature change be to maintain the equilibrium. This question is answered by the same type derivative we used in our discussion of boiling point elevation, namely,
               . (32)
This reduces to
                (33)
Notice that we have again used the fact that Δμ1 is not only constant, but constant at the value zero in order to be able to write
                (34)
Notice also that this time the derivative in Equation 33 is negative. (You might want to go back and see why we got a negative sign here when we had a positive sign in boiling point elevation.) Under the circumstance where the solution is very dilute, X2 << 1, this reduces to,
                (35)
and we can integrate through a short concentration range as before to get
                (36)
Again, this is not in the same form you saw in freshman chemistry. To get there we have to convert from mole fraction to molality as before. We get,
                (37)
The part in brackets is the freezing point depression constant or the "cryoscopic" constant.

If we want a larger concentration range we have to go back and integrate

                (38)
to get
                (39)
This expression will be of use later when we talk about solid-liquid phase diagrams.
 

4) Osmotic Pressure

The treatment of osmotic pressure is not identical to the treatments of boiling point elevation and freezing point depression we have just given. A cartoon of an osmotic pressure apparatus is shown below. A U-tube has a "semipermeable membrane inserted in the center of the cross piece as shown. (A semipermeable membrane will allow solvent to flow through it, but not solute.) The left arm of the U-tube is filled with pure solvent and the right arm contains solution. The chemical potential of the solvent is higher in the pure solvent than in the solution so that solvent will flow through the semipermeable membrane from the pure solvent side into the solution side unless it is restrained. (Another way to look at this is that the solution "would like to become more dilute" and it will do so if it can. Given the chance, material will flow from a region of higher chemical potential to a region of lower chemical potential.) To prevent the solvent from flowing we apply an external pressure, Π , to the right side. The pressure, Π , which exactly stops the flow is called the osmotic pressure.

In this experiment temperature is held constant. We will bring the solvent on the solution side into equilibrium with the pure solvent by the exertion of an external pressure on the solution. As we add solute we lower the chemical potential of the solvent in the solution so we must increase the pressure on the solution to bring the chemical potential back up. That is, as we make the solution we need to increase the external pressure to keep μ1l constant. Recall that

                (40)
The question we ask is: if we increase the concentration of solute by an amount dX2, how much does the external pressure have to change to keep the solution in equilibrium with the pure solvent? The derivative we want is,
                (41)
where we have used the fact that
               . (42)
Then Equation 41 simplifies to,
                (43)
Prepare to integrate this equation,
                (44)
which integrates to (calling the applied external pressure, or the osmotic pressure, Π),
                (45)
or, when the solution is dilute,
                (46)
But so,
.                (47)
This looks exactly like the ideal gas equation of state, except the pressure is the osmotic pressure, Π, and the system is a solution, not a gas.

This derivation of osmotic pressure assumed the solution is ideal. For ionic solutions, which are not ideal, you can get an approximate osmotic pressure by taking account of the dissociation of the compound into ions in solution. If the compound gives c ions per compound unit ("molecule") the osmotic pressure for dilute ionic solutions can be written,

                (48)
For more accurate work you can write an expansion similar to the virial expansion,
                (49)
or
                (50)
for ionic solutions.

Osmotic pressure is enormously important for life. For example, cell wall membranes, in many cases, are semipermeable membranes which will pass molecules of water but not solute molecules or ions. As a consequence the concentrations of material in the fluids inside and outside the cell must be carefully balanced so that they generate the same osmotic pressure. Otherwise, the difference in solute concentrations would generate a pressure differential across the cell wall and either collapse the cell or burst it. Fluids used in intravenous feeding, drug delivery, or just plain fluid replacement are called isotonic solutions because they are designed to have the same osmotic pressure as normal blood cells. The typical isotonic solution is 8.9 g NaCl per liter of water. If a blood cell is placed in a solution with a lower concentration than this (a hypotonic solution) fluid moves from the solution into the cell and cell bursts (called hemolysis). If a blood cell is placed in a solution with a higher concentration (a hypertonic solution) fluid moves from the cell to the surrounding solution and the cell shrinks and dies (called crenation).

It is instructive to calculate the osmotic pressure of an isotonic solution (8.9g NaCl per liter of water). This is the pressure that would be generated inside a blood cell that was placed in pure water. Do you suppose that the cell wall could withstand this pressure?

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Copyright 2004, W. R. Salzman
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Last updated 22 Oct 04
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