# 13.3 Colligative Properties ## What is a Colligative Property?

A colligative property is a property of a solution that changes proportionally as a solute is added, and this change generally occurs with any solute and is not dependent upon the solute's identity.  The key is that the solute is distinct from the solvent and disrupts the pattern of intermolecular forces normally present in the pure solvent, and the more solute particles the greater the disruption.

There are four commonly studied colligative properties:

• Freezing Point Depression
• Boiling Point Elevation
• Vapor Pressure Depression
• Osmotic Pressure

As you add increasing concentrations of solute to water its freezing point decreases, its boiling point increases, and the vapor pressure of water above the solution decreases.

## What is the Van't Hoff Factor?

The van't Hoff Factor (i) has no units and is simply the number of ions a solute dissociates into per formula unit.  Since the change in a colligative property is proportional to the number of solute particles present in solution, then a solute that dissociates into multiple ions will result in a larger change and the van't Hoff factor allows us to factor that in.  If a solute is a nonelectrolyte (includes most molecular compounds) and doesn't dissociate into ions it has a van't Hoff factor of 1, whereas electrolytes will have values greater than 1.

For electrolytes, i > 1

For nonelectrolytes, i = 1

The van't Hoff factors for some typical compounds are shown below:

NaCl → Na+ + Cl-                  i = 2 CaCl2 → Ca2+ + 2Cl-            i = 3 Al(NO3)3 → Al3+ + 3NO3-      i = 4

For strong electrolytes having polyatomic ions do not make the mistake of trying to have the polyatomic ions dissociate into individual atoms. Try a couple yourself.  What is the van't Hoff factor for each of the following?

CH3OH
i = 1
• CH3OH is a nonelectrolyte
Ba(OH)2
i = 3
• Ba2+
• 2OH-
• 3 total ions

## Freezing Point Depression

The addition of solute to a solution results in the lowering of the freezing point (a.k.a. melting point), the occurrence of which we refer to as Freezing Point Depression.  When a liquid cools the molecules slow down until the point at which they no longer have enough kinetic energy to overcome the intermolecular forces that 'freeze' the molecules into a solid.  This is the freezing point.

The addition of solute adds molecules or ions that are distinct from the solvent molecules.  Upon freezing these solute particles will put irregularities into the crystal structure which make it more difficult to form when compared to the pure solvent.  The result is that the molecules must be moving even more slowly to 'freeze' into a solid which now occurs at a lower temperature than for the pure solvent, and the greater the concentration of solute particles, the lower the freezing point.

Mathematically, the change in the freezing point is calculated as follows:

ΔTF = -iKFm

i = the van't Hoff Factor

KF = the Freezing Point Constant (a.k.a. Cryoscopic Constant)

m = the molality of the solute

### Freezing Point Depression Calculation

What is the freezing point of 1m NaCl in H2O?

ΔTF = -iKFm

ΔTF = -(2)(1.86°C/m)(1m)

ΔTF = -3.72°C

TF = 0°C - 3.72°C

TF = -3.72°C

Constants for H2O
TF = 0°C KF = 1.86°C/m
TB = 100°C KB = 0.51°C/m
PH2O, 25°C = 24torr

## Boiling Point Elevation

The addition of solute to a solution results in the increase of the boiling point, the occurrence of which we refer to as Boiling Point Elevation.  As a liquid is heated its energy increases ('chemical potential' is the proper technical term but I'll substitute 'energy' for simplicity here).  When the energy of the liquid equals the energy of its gaseous form the two phases are said to be in equilibrium.  This is the boiling temperature.

When a solute spontaneously dissolves into a solvent, it lowers the energy of the solution.  As a result of this lowering of its energy, the solution must now be heated to an even higher temperature to have its energy equal to that of the gaseous form thus explaining the increase in the boiling point.

Mathematically, the change in the boiling point is calculated as follows:

ΔTB = iKBm

i = the van't Hoff Factor

KB = the Boiling Point Constant (a.k.a. Ebullioscopic Constant)

m = the molality of the solute

### Boiling Point Elevation Calculation

What is the boiling point of 1m NaCl in H2O?

ΔTB = iKBm

ΔTB = (2)(0.51°C/m)(1m)

ΔTB = 1.02°C

TB = 100°C + 1.02°C

TB = 101.02°C

Constants for H2O
TF = 0°C KF = 1.86°C/m
TB = 100°C KB = 0.51°C/m
PH2O, 25°C = 24torr

Which of the following aqueous solutions would have the highest boiling point and highest freezing point? (answered in the video lesson)

1m NaCl
1m CH3OH
1m Ba(OH)2
1m Al(NO3)3

Which of the following aqueous solutions would have the highest boiling point and highest freezing point? (answered in the video lesson)

2m NaCl
1.5m CH3OH
1.2m Ba(OH)2
0.9m Al(NO3)3

## Vapor Pressure Depression (Raoult's Law)

Mathematically, the vapor pressure of the solvent (A) above a solution is:

PA = χAPA*

PA = the partial pressure of A (solvent) above the solution

χA = the mole fraction of A (solvent)

PA * = the partial pressure of pure A (solvent)

### Raoult's Law Calculation

What is the vapor pressure of water at 25°C above a solution containing 162g H2O and 32g CH3OH?

(162g H2O)(1mol/18g) = 9mol H2O

(32g CH3OH)(1mol/32g) = 1mol CH3OH

χA = (9mol H2O)/(9mol H2O + 1mol CH3OH) = 9/10

PA = χAPA*

PA = (9/10)(24torr)

PA = 21.6torr

Constants for H2O
TF = 0°C KF = 1.86°C/m
TB = 100°C KB = 0.51°C/m
PH2O, 25°C = 24torr

## Osmotic Pressure

Mathematically, the osmotic pressure is calculated as follows:

π = iMRT

π = the osmotic pressure

i = the van't Hoff factor

M = the molarity of the solute

R = the universal gas constant (R = 0.08206 L.atm/mol.K)

T = the temperature (in Kelvin)

### Osmotic Pressure Calculation

What is the osmotic pressure at 25°C of the fluid in the following diagram?

π = iMRT

π = (2)(1.0M)(0.08206 L.atm/mol.K)(298K)

π = 48.9atm ~ 49atm 