Colligative Properties

Colligative properties: Definition Those properties that depend on the number of particles (molecules or ions) of the solute rather than on their physical and chemical properties.

Solutions will have different properties than pure solvent.

Description. The colligative properties (depend upon number of particles) of solutions are:

vapor pressure lowering
boiling point elevation
freezing point depression
osmotic pressure

Osmotic pressure has the greatest direct relevance to pharmaceutics, because it is the property that largely determines the physiological acceptability of parenteral, ophthalmic, and nasal solutions.

However, osmotic pressure is extremely difficult to measure, so other colligative properties are determined and then related to osmotic pressure, since all colligative properties are interrelated.

Diffusion:
Substance tend to move from higher concentration to equilibrium (equal concentration).

1.Vapor Pressure Depression

The pressure brought by the vapor in equilibrium with its liquid is called the vapor pressure.

It increases upon increasing the temperature. The vapor pressure of a pure liquid depends on the rate of escape of the molecules from the surface. If the liquid is mixed with another substance, its concentration is decreased and the rate of escape is lowered. When a nonvolatile solute (i.e., a drug) is dissolved in a liquid solvent, the vapor pressure of the solvent is lowered because the drug does not contribute directly to the vapor pressure.

Draw solvent molecules on a surface of the solvent escaping into the vapor. Replace some of them with solute molecules, which have little, if any, vapor pressure of their own.

Solvent = A and Solute = B

P_A = X_AP^o_A

P_A = (1-X_B) P^o_A

P_A = P^o_A - X_B P^o_A

P^o_A - P_A = X_B P^o_A

êP = X_B P^o_A

P^o_A = Vapor pressure of solvent
P_A = Vapor pressure of solution
X_A = mole fraction of solvent
X_B = mole fraction of solute

Example:
0.5 mole of sucrose in 1000g of water at 20 C. Vapor pressure of water at 20 C is 17.54 mmHg.
Calculate the vapor pressure of solution.

moles of water = 1000/18.02 = 55.5 mole
moles of sucrose = 0.5 mole

X_B = 0.5/56 = 0.0089

êP = 0.0089 x 17.54 = 0.156 mmHg

Vapor pressure of solution = 17.54 - 0.156 = 17.38 mmHg

In pharmaceutical solutions, the solute is usually non-volatile and it does not contribute directly to the vapor pressure of the solution. However, its presence decreases the concentration of the solvent and its escape tendency, i.e., the vapor pressure of the solution is lower than that of the pure solvent. The vapor pressure lowering is proportional to the number of solute molecular particles or ions. The effect of a solute on the vapor pressure may be determined in dilute solutions by applying the Raoult's Law (Eq. 1).

p_a = p_a^ox_a

where: p_a is the partial vapor pressure of the solvent in the solution; p_a^o is the vapor pressure of the pure solvent; and x_a is the mole fraction of a.

Since X_a + X_b = 1 (X_b is the mole fraction of the solute), the above equation can be rewritten as

p_a = p_a^o(1-x_b) showing that the relative vapor pressure lowering of the solution is equal to the mole fraction of the solute(2).

This concept can be applied to calculate the vapor pressure for an aerosol propellant since mixtures of liquefied gas propellants can be considered as solutions. Mixtures of propellants result in reduced concentrations of any one propellant in the surface, thus in a reduction in the rate of escape and vapor pressure lowering of each component.

Example:
In a two component aerosol propellant system (A and B), the partial vapor pressure of A is:

p_a = p_a^ox_a or p_a = p_a^on_a / (n_a + n_b)

the partial vapor pressure of B is:

p_b = p_b^ox_b or p_b = p_b^o n_b / (n_b + n_a)

the total vapor pressure (P) is:

P = p_a + p_b

In a blend of two propellants (A/B = 30:70[g])where pure propellant A (mol. wt 120.93) has vapor pressure of 84.9 psi and propellant B (mol. wt 137.38) has vapor pressure 13.4 psi respectively the partial vapor pressure for A is:

The total vapor pressure of propellants in the above aerosol then equals:

P = p_a + p_b = 27.80 + 9.01 = 36.81 psi or,

P = 36.81 - 14.7 = 22.11 psi

Applying the same principle of Raoult's law one can also calculate the volume of two propellants (e.g., propane and isobutane) required to achieve a certain vapor pressure suitable for a propellant².

Application: Meter dose Inhalers

2. Boiling Point increase

The boiling point of a liquid is the temperature at which the vapor pressure of the liquid is equal to the external pressure of 760 mm Hg.

Since the vapor pressure of a solvent is lowered when a nonvolatile solute is added, the result is that the solution must be heated to a higher temperature than the pure solvent to reach the same vapor pressure. The boiling point of a solution is thus elevated in comparison to the boiling point of the pure solvent.

By using the mathematical relation between vapor pressure and temperature and the Raoult's Law, an equation is derived for the boiling point elevation of a solution:

where êT_b is the elevation of the boiling point, m is the molality of the solution and K_b is the proportionality constant, which can be also defined as the boiling point elevation for one molal (m) dilute solute solution i.e., K_b = êT_b/m. The values of K_b are different for different solvents; for water it is 0.515 C.(3)

From the above equation one can calculate the concentration of the solute in a solution by measuring the boiling point elevation and knowing the K_b.

Example:

Calculate the concentration of dextrose (mol. wt. 180) in 1000 g of water if the boiling point elevation of the solution is 0.284 C.:

ê T_b = K_bm or 0.284 = 0.512 m
then
m = 0.555 or 99.84 (=100)g of dextrose in 1000 g of water.

Example 1:

A 0.2 molal (mole/kg) aqueous solution of a drug gave a boiling point elevation of 0.103 C êT_b). Calculate the approximate molal elevation constant (K_b) for the solvent water.

Example 2:
An aqueous solution of drug (MW = 342) is prepared by dissolving 0.5 g drug in 100 g water. Calculate the boiling point elevation of this solution (êT_b). K_b = 0.515 deg. kg/mole

Example 3:
Calculate the K_b of water if the heat of evaporation of water (êHv) is 539.7 cal./g. MW = 18.02; B.P. = 373.1 K; R = 1.987 cal/mole deg.

Calculation of MW_B

Example:
A solution containing 10 g of drug dissolved in 100 g of water has a boiling point of 100.149 C. What is the molecular weight of the drug? K_b = 0.513 deg. kg/mol = 0.513 deg./molal

3. Freezing Point Depression

The freezing point of a liquid is the temperature at which the solid and the liquid phases are in equilibrium at one atmosphere.

The freezing point of a solution is the temperature at which the solid phase of the solvent and the liquid phase of the solution are in equilibrium at one atmosphere.

By using the thermodynamic equations for the effect of the temperature on the vapor pressure of the solid and liquid phases and Raoult's Law, the following equation is obtained:

where êT_f is the lowering of the freezing point of a solvent in a solution, m is the molality of the solute and K_f is the molal lowering of the freezing point. The K_f value for water is 1.858 C.

From êT_f equation the concentration of the solute in a solution can be calculated by measuring the freezing point depression of the solution and knowing the K_f of the solvent. The above two equations are valid only for very dilute solutions.

Example

What is the freezing point of a solution containing 3.42 g of drug and 500 g of water? MW of the drug = 342. K_f in dilute solution = 1.86 deg. kg/mol = 1.86 deg. molal

Example:

Calculate the concentration of dextrose (mol. wt 180) in 100 g of water if the freezing point depression of the solution is 150 C.:

êT_f = K_fm or 0.52 = 1.86 m
then
m = 0.280 or 50.4 (50)g of dextrose in 1000 g of water

Back to Home Page