Colligative properties

Freezing point depression

There are a number of properties of a dilute solution that depend only on the number of particles and not on their kind. Colligative properties include the lowering of the vapor pressure of a solvent and elevation of the boiling temperature by the addition of a nonvolatile solute, the depression of the freezing point of a solution by a solute, and osmotic pressure.

At the freezing point of a solution, solid solvent is in equilibrium with the solvent in solution. As we have seen this equilibrium implies that the chemical potential of each phase is equal to the other

m₁^solid(T_fus) = m₁^soln(T_fus)

where subscript 1 denotes solvent and T_fus is the freezing point of the solution.

m₁^soln(T_fus) = m₁^* + RT ln a₁ = m₁^liq + RT ln a₁

where we have expressed the chemical potential of the pure liquid in a notation that will helpful for comparison with the solid.

m₁^solid = m₁^liq + RT ln a₁

Solving for a₁, we get

ln a₁ = (m₁^solid - m₁^liq)/RT = - D_fusG/RT

D_fusG = D_fusH - TD_fusS

ln a₁ = (m₁^solid - m₁^liq)/RT = - D_fusH/RT - D_fusS/R

For pure solvent a₁ = 1 and

ln 1 = - D_fusH/RT^* + D_fusS/R

For an activity a₂ of solute, ln a₁ = ln(1 - a₂) » - a₂

- a₂ = - D_fusH/RT + D_fusS/R

In the above we present the fusion temperature of pure solvent as T* and the fusion temperature of the solution as T. Noting that ln(1) = 0 we can eliminate D_fusS/R in the two equations.

D_fusS/R = D_fusH/RT^*

a₂ = D_fusH/RT - D_fusH/RT^*

a₂ = D_fusH/R(1/T - 1/T^*)

The above expression can be further approximated using

(1/T - 1/T^*) = (T^* - T)/TT^* » DT/T^*2

a₂ = D_fusH/RT^*2 DT

Note how this derivation differs from that in McQuarrie and Simon using the Gibbs-Helmholtz equation. The result, however, is the same. The above formula can be compared with the formula for freezing point depression

DT_fus = K_fm (m is molality)

For a dilute solution a₂ ® x₂ » M₁m/1000g kg^-1 for small values of m. Therefore,

K_f = RT^*2/D_fusH (M₁/1000g kg^-1)

is called the freezing point depression constant.

Boiling point elevation

We can determine the value of K_b for water.

The phenomenon of boiling point elevation can be derived in a completely analogous fashion. In both cases the physics behind the effect is a lowering of the chemical potential of the solution relative to the pure substance. This can be seen in the diagram below where we plot the chemical potential as a function of the temperature.

In this plot notice that the slope increases as the phase changes from solid to liquid and then to vapor. The slope is proportional to - S (since ¶m/¶T = -S) and the entropy increases in the same order. Notice that the violet line representing the chemical potential as a function of temperature is shifted down by the addition of solute. Mathematically this is due to

m₁^soln = m₁^liq + RT ln a₁

Because the chemical potential of the solid and vapor are not shifted by the addition of solute the intersection point (i.e. temperature of phase transition) goes down for fusion, but goes up for vaporization.

Osmotic pressure

Osmotic pressure arises from requirement that the chemical potential of a pure liquid and its solution must be the same if they are in contact through a semi-permeable membrane. Osmotic pressure is particularly applied to aqueous solutions where a semi-permeable membrane allows water to pass back and forth from pure water to the solution, but the solute cannot diffuse into the pure water. The point here is that the solute lowers the chemical potential on the solution side of the membrane and therefore there will be a tendency for water to move across the membrane to the solution side. Ultimately, there will be a balance of forces if a pressure builds up on the solution side of the membrane. This pressure can arise due to an increase in the hydrostatic pressure due to a rise in a column of solution or due to pressure inside a closed membrane. The easiest to visualize is a column of water. Recall that the pressure at the bottom of a column of a fluid is given by P = r g h. If water flows into the solution the height of the column of solution increases and the hydrostatic pressure also increases. At some point the chemical potential due to the concentration difference is exactly opposed to the chemical potential due to the pressure difference. We express this as

m₁^*(T,P) = m₁^soln(T,P+P,a₁)

The chemical potential of the solution is

m₁^soln(T,P) = m₁^*(T,P) + RT ln a₁

m₁^*(T,P) = m₁^soln(T,P+P,a₁) = m₁^*(T,P+P) + RT ln a₁

Recall that ¶m/¶T = V_m (subscript for molar volume) so

Thus,

Assuming V_m does not vary with applied pressure, we can write

PV_m + RT ln a₁ = 0

Since a₁ ® x₁ for a dilute solution and ln x₁ = ln(1-x₂) » -x₂ we have that

PV_m - RT x₂ = 0

which be expressed as

PV = n₂RT.

The above expression bears a surprising similarity to the ideal gas law. Keep in mind, however, that P is the osmotic pressure and n₂ is the number of moles of solute.

Thus, we can compute the osmotic pressure from

P = n₂RT/V

or

P = cRT

where c is the molarity, n₂/V, of the solution.

This equation is called the van't Hoff equation for osmotic pressure. The osmotic pressure can be used to determine the molecular masses of solutes, particularly solutes with large molecular masses such as polymers and proteins.

Example: determination of the molecular mass of a polymer

It is found that 10.9 grams of polymer dissolved in enough water to make 1 L of total solution. The solution has an osmotic pressure of 10 torr at 20 ^oC.

Solution: The molarity of the solution is given by

c = P/RT = (10 torr/760 torr/atm)/(0.082 L-atm/mole-K)(293 K)

= 5.47 x 10^-4 mol/L

In 1 L total solution then we have 5.47 x 10^-4 mol weighing 10.9 grams or 10.9 grams/5.47 x 10^-4 mol which corresponds to a molecular mass of 20,000 grams/mole.

Ideal solubility

The colligative properties can also be used to calculate the ideal solubility of a substance. If a solid solute is left in contact with a solvent, it dissolves until the solution is saturated. Saturation is a state of equilibrium, with the undissolved solute in equilibrium with the dissolved solute. To express this state of equilibrium we use the chemical potential for a solid solute in equilibrium with that solute in solution.

m^solid = m^liquid,* + RT ln x₂

Analogous to the discussion above for freezing point depression we obtain

ln x₂ = (m^solid - m^liquid)/RT = - D_fusG/RT

D_fusG = D_fusH - TD_fusS

ln x₂ = - D_fusH/RT + D_fusS/R

For the pure substance (i.e. pure solute)

D_fusH/RT_fus^* = D_fusS/R

Therefore,

ln x₂ = D_fusH/R[1/T_fus^* - 1/T]

Notice that this expression relates the mole fraction of solute that will dissolve at a given temperature to its D_fusH. The ideal solubility does not depend on the solvent, but only on the properties of the solute. This is analogous to the other colligative properties of freezing point depression or boiling point elevation in that those phenomena depend only on the properties of the solvent and not on the solute.

Example: Predict the solubility of iodine in water at 293 K given that its melting point is 387 K and its enthalpy of fusion is 15.5 kJ/mol. Give your answer in units of molality.

Solution: The mole fraction of I₂ is

x₂ = exp{ D_fusH/R[1/T_fus^* - 1/T] }

= exp{15,500/8.314[1/387 - 1/293]}

= 0.213

To convert to molality we use the relation

m = (0.213)/(1-0.213)(1000/126) = 2.15 molal