Kf and Kb

Kf and Kb are the freezing point depression constant and boiling point elevation constant. When a solute is added to a solvent, the boiling point of the solution is always greater than the boiling point of the pure solvent. Adding a solute also lowers the freezing point.
Solvent Formula Melting Point (oC) Boiling Point(oC) Kf(oC/m) Kb(oC/m)
Benzene C6H6 5.455 80.2 5.065 2.61
Ethanol C2H5OH -- 78.3 -- 1.07
Water H20 0.000 100.000 1.858 0.521


To determine the amount of change in boiling point you will need this equation:
delta Tb = Kb * m
  • delta Tb change in boiling point
  • Kb boiling point elevation constant
  • m molality of solution


Then delta Tb is added to the normal boiling point of the pure solvent. Note that the identity of the solute is not important, just its concentration (expressed in molality). Therefore, boiling point elevation is a colligative property.

To determine the freezing point of a solution, you need to calculate the decrease in freezing point caused by the addition of a solute to the solvent. Use the equation:

delta Tf = Kf * m

  • delta Tf change in freezing point
  • Kf freezing point depression constant
  • m molality of solution

Then the delta Tf is subtracted from the normal freezing point of the pure solvent. Freezing point depression is also a colligative property.

example:

Calculate the boiling point and freezing point of a solution of .30 g of glycerol (C3H8O3) in 20.0 g of water.

moles glycerol = (.30 g) (1 mole / 92 g) = .0033 moles molality of solution = .0033 moles / .020 kg = .16 m

delta Tb = (.521 oC/m)(.16 m) = .083 oC

Boiling point = 100.00 + .083 = 100.083 oC

delta Tf = (1.858 oC/m)(.16m) = .30 oC

Freezing point = 0.00 oC - .30 oC = -.30 oC

In an ionic solution, the total concentration of ions is important. Therefore, another factor (i) is included in the equations.

delta Tb = Kb * m * i

delta Tf = Kf * m * i

"i" is the number of ions from each formula unit. In the previous example, if NaCl had been the solute your change in boiling point and freezing point would have needed to be multiplied by 2. (i=2 because NaCl consists of a Na+ ion and a Cl- ion.