Colligative Properties of Solution

Colligative properties of solution are properties of solution that depend on the amount of solute present in the solution and not on the nature of solute particles.  These are the vapor-pressure lowering, boiling-point elevation, freezing-point depression, and osmotic pressure.

Vapor-pressure Lowering

Some liquids like water contains molecules,  these molecules when absorb enough energy will eventually form into vapor.  The pressure exerted by the vapor is called vapor pressure.  Try boiling water in whistling tea kettle, you will notice that when it boils the tea kettle whistle, this is because of the pressure caused by the vapor.   What will happen to the vapor pressure when solute is added to the solvent?  Since solution contains already solute that is attached to the solvent, the vapor pressure of the solution is lower compared to the pure solvent.  The relationship is expressed  by Rault's Law, which states that the partial pressure  of a solution, P_A, equals the product of the mole fraction  of the solvent in the solution, X_A, times the vapor pressure of the pure solvent, P^o_A.

P_A  =  X_A P^o_A

For example, the vapor pressure of pure water at 20^oC is 17.5 torr.  What if glucose (C6H12O6) is added to water.  Let us say the mole fraction of water, X_H2O = 0.800 and mole fraction of glucose, X_C6H12O6 = 0.200.  What is the vapor pressure of  the solution?  Calculating the problem:

P_H2O = X_H2O P^o_H2O

           = (0.800)(17.5 torr) 

           = 14.0 torr

Based from the calculation above, there is a decrease in vapor pressure from 17.5 torr of pure solvent to 14 torr of vapor pressure of the solution.

Sample Problem:
Calculate the vapor pressure of a solution made by dissolving 218 g of glucose (molar mass = 180 g/mol) in 460 mL of water at 30^oC.  What is the vapor pressure lowering?  The vapor pressure of water at 30^oC is 31.82 mm Hg.  Assume the density of the solution is 1.00 g/mL.

Solution:

First, calculate the number of moles of glucose and water.

Then,  the mole fraction of water, X_1(water) is

For the vapor pressure of the solution, we have,

P_A  =  X_A P^o_A
Psolution = (0.955) (31.82 mm Hg)
                = 30.4 mm Hg


Therefore the vapor pressure lowering is 31.82 mm Hg - 30.4 mm Hg =  1.4 mm Hg

Boiling Point Elevation

Boiling point  is the temperature at which the vapor pressure equalizes with the atmospheric pressure.  Since the vapor pressure of the solution is lowered than its pure solvent, therefore the boiling point also of the solution is affected with the presence of the solute.  Below is the phase diagram illustrating the boiling point elevation and freezing point depression of aqueous solution:

Based from the graph above, the boiling point of solution is higher than that of the boiling point of pure water. The increase in boiling point is due to the solute present in the solution.  The normal boiling point of pure liquid  is the temperature  at which its vapor pressure equals 1 atm.  Since vapor pressure of the solution is lowered , it requires  higher temperature to attain vapor pressure of 1 atm. Thus, the boiling of the solution is higher than that of the pure liquid.  

The increase in boiling point relative to that of the pure solvent, is directly proportional to the number of solute particles per mole of solvent molecules, meaning directly proportional to the concentration expressed in molality.  Thus,

∆T_b = K_bm

where ∆T_b is boiling point elevation, m is molality of the solution, and K_b is the molal boiling-point elevation constant.  The unit of K_b is ^oC/m.

Below is a table showing the Molal Boiling-Point Elevation and Freezing-Point Depression Constants: 

Freezing-Point Depression

Freezing-point of a solution is the temperature at which  the first crystals of pure solvent begin to form equilibrium with the solution.  Based from the graph above, the freezing point of the solution is lower than that of the pure liquid.  This is because freezing involves a transition from disordered to ordered state wherein, energy must be removed from the system in order the process to occur.  Because solution has greater disorder than the solvent, more energy is needed to be removed from it in order to create order the same as that of the pure solvent. 

The freezing point depression, ∆T_f,  is directly proportional to the molality of the solute:

∆T_f = K_f m

where,  m is the concentration of the solute in molality, K_f is the molal freezing-point depression constant.  The unit also is ^oC/m.  

Sample Problem:

Automotive antifreeze consists of ethylene glycol (C₂H₆O₂), a nonvolatile nonelectrolyte.  Calculate the boiling point and freezing point of a 25.0 mass % solution of ethylene glycol in water.

Solution:  

The given in the problem is 25 % by mass of ethylene glycol solution.  Let us assume an amount of solution let say 1000 g of the solution.  Since the solution is 25 % by mass, this means that 250 g of ethylene glycol is in 750 g of water.  Using this quantities we can calculate the concentration of the solution in molality:

After calculating the molality, boiling point elevation and freezing point depression can now be calculated:

∆T_b = K_bm = (0.51 ^oC/m)(5.37 m) = 2.7 ^oC

∆T_f = K_f m = (1.86 ^oC/m)(5.37 m) = 10.0 ^oC

boiling point = (normal boiling point of solvent) + ∆T_b     
                      =  100 ^oC +  2.7 ^oC =  102.7 ^oC

Freezing point = (normal freezing point of solvent + ∆T_b
                        =  0.0 ^oC  - 10.0 ^oC =  -10.0 ^oC

Osmotic Pressure

Osmosis is the net movement of solvent molecules through semipermeable membrane from dilute solution to a concentrated solution.  Semipermeable membrane is a membrane that allows the passage of solvent molecules but block the passage of solute molecules.  Let us look at the figure below:

The figure above,  a is a container containing the same amount of pure solvent and solution  separated by a semi permeable membrane.  The left part contains the pure solvent while the right part contains the solution.  In container b, osmosis occurs, the solution part rises, this is because the solvent in the left compartment passes through the semipermeable membrane .  Osmotic pressure is equal to the hydrostatic pressure exerted by the column of fluid in the right tube at equilibrium.           

Osmotic pressure 𝝅 of the solution is the pressure required to stop osmosis.

The osmotic pressure of a solution is given by:

𝝅 = MRT  

where M is the molarity of the solution, R is the gas constant (0.0821 L.atm/K.mol) and T is the absolute temperature.  The osmotic pressure, 𝝅, is expressed in atmospheres.

Osmotic pressure, the same with other colligative properties, is directly proportional to the concentration of solution.  This means that all colligative properties of solution are dependent on the number of particles of solute present in the solution.  If two solutions have the same concentration and osmotic pressure, they are said to be isotonic.  If two solutions have different osmotic pressure, the more concentrated solution is said to be hypertonic, and the more dilute solution is said to be hypotonic.  

Osmosis plays a vital role in living systems.  Example, the membrane of red blood cells are semipermeable.  When red blood cells is placed in a solution that is hypertonic  relative to the intracellular solution (the solution within the cells) causes water to move out of the cell.  This causes the cell to shrivel, a process called crenation.  Placing the cell in a solution that is hypotonic relative to the intracellular fluid causes water to move into the cell.  This causes the cell to rupture, a process called hemolysis.  Some patients who need body fluids or nutrients replaced but cannot be done orally  are given solutions by intravenous (IV) infusion, which feeds nutrients directly into the veins.  To prevent crenation or hemolysis of red blood cells , the IV solutions must be isotonic with the intracellular fluids of the cells.

There are many interesting examples of osmosis, one of these is watering of plants.  Water moves from soil into plant roots and into the upper portions of the plants.   But if you will use seawater in watering the plants, the reverse will occur.  The water from the plant will flow down into the roots into the soil, this will cause the dehydration of plant.  This is because in osmosis the water moves from less concentrated to the more concentrated solution.

Sample Problem:

The average osmotic pressure of blood is 7.7 atm at 25^oC.  What concentration of glucose (C₆H₁₂O₆) will be isotonic with blood?

Solution:
The given above are the osmotic pressure and the temperature.  Temperature must be converted to K before calculating.

      𝝅 = MRT