Chapter 7: Unit 7. Colligative Properties of Solutions

Colligative Properties of Solutions

The colligative properties of the solution are physical properties that depend on the amount of solute particles in the solution.

Adding solute particles to the pure solvent will lead to a decrease in the freezing point and an increase in the boiling point of the solvent. Also adding the solute particles to pure solvent will lead to a decrease in the vapor pressure of the solvent.

Let us look at each colligative property by itself:

1. Freezing point depression (decrease):

When a nonvolatile solute is added to a pure solvent, the pure solvent freezing point is lowered because of the solute particles are preventing the pure solvent molecules to organize and hence less temperature is needed before the solvent molecules become organized to freeze.

Antifreeze ethylene glycol is added to car radiator to lower the freezing point of the radiator water solution. Ethylene glycol has higher boiling point of water.

Change in Freezing point:

The addition of the solute particles will lower the freezing point. This change can be calculated using the formula below:

∆T_F = T_F(pure solvent) – T_F(solution) = (i) x K_F x molality

Where:

T_F(pure solvent) = Freezing point of the pure solvent

T_F(solution) = Freezing point of the solution

I = Number of the particles present and it is called the van’t Hoff factor

K_F = Molal freezing point constant in (kg ^oC)/mol

m = molality in mol solute / kg solvent  = [ (mass /molar mass) solute ] / kg solvent

(molar mass) = [ (i) x K_F x mass solute] / [ (∆T_F) x (kg solvent) ]

The molal freezing point constant K_F is given in a table out of the reference:

http://www.kbcc.cuny.edu/academicDepartments/PHYSCI/PL/chm11/Documents/FreezingPoint.pdf

Where mp is the melting point which is equal the freezing point of the material.

The value of I depends on the status of the dissociation of compound in water. For example, nonelectrolytes do not dissociate and hence I = 1

1 mole of sugar C₆H₁₂O₆(s)  + H₂O(l)  à    1 mole of sugar C₆H₁₂O₆(aq)  [I = 1 mole of particles]

1 mole of NaCl(s) + H₂O(l)  à 1 mole Na⁺(aq)  + 1 mole Cl^–(aq)  [I = 2 moles of particles]

1 mole of Ca₃(PO₄)₂(s)  à 3 Ca²⁺(aq)  +  2 PO₄^3-(aq)    [I = 5 moles of particles]

Example:

80.50 grams of naphthalene are dissolved in 5546.5 grams of benzene. What will be the change in the freezing point of benzene? K_F of benzene from the table above is 5.12 (kg ^oC)/mol. Freezing point of benzene is 5.5 ^oC.

Solution:

Solute = naphthalene C₁₀H₈ with the molar mass of 128.2 g / mol

Solvent = benzene

Step 1: the molality should be calculated.

Moles of naphthalene = (80.50 g) / (128.2 g / mol) = 0.6279 moles

molality = 0.6279 moles / 5.5465 kg = 0.1132 m = 0.1132 mol / kg

Step 2: Freezing point depression formula is used.

∆T_F = T_F(pure solvent) – T_F(solution) = (i) x K_F x molality

∆T_F = 5.5 ^oC  – T_F(solution) = 1 x [5.12 (kg ^oC)/mol] x [0.1132 mol / kg]

1 mol naphthalene C₁₀H₈(s)  + benzene(l)   à         1 mol naphthalene C₁₀H₈(l)  [I = 1 mole of particles]

5.5 ^oC  – T_F(solution) = 1 x [5.12 (kg ^oC)/mol] x [0.1132 mol / kg]

5.5 ^oC  – T_F(solution) = 0.580 ⁰C

T_F(solution) = 5.5 ^oC  – 0.580 ⁰C = 4.9 ^oC

A You Tube video illustrates the Freezing Point Lowering calculations:https://www.youtube.com/embed/06Buf6N2Yp4?start=13&feature=oembed

1. Boiling point elevation (increase):

When a nonvolatile solute is added to a pure solvent, the pure solvent boiling point is increase. This is because the addition of a nonvolatile solute to a pure solvent will lower of vapor pressure of the pure solvent. The vapor pressure of pure solvent should reach atmospheric pressure before it begins to boil and since the vapor pressure is lowered by this addition, a higher temperature needed than the boiling point of original pure solvent boiling point to bring the solution to boiling.

Change in Boiling Point:

The addition of the solute particles will increase the boiling point. This change can be calculated using the formula below:

∆T_B = T_B(solution) – T_B(pure solvent)  = (i) x K_B x molality

Where:

T_B(pure solvent) = Boiling point of the pure solvent

T_B(solution) = Boiling point of the solution

I = Number of the particles present and it is called the van’t Hoff factor

K_B = Molal boiling point constant in (kg ^oC)/mol

m = molality in mol / kg

The molal boiling point constant K_B is given in a table out of the reference:

https://chem.libretexts.org/Textbook_Maps/General_Chemistry/Map%3A_General_Chemistry_(Petrucci_et_al.)/13%3A_Solutions_and_their_Physical_Properties/13.08%3A_Freezing-Point_Depression_and_Boiling-Point_Elevation_of_Nonelectrolyte_Solutions

Example:

25.50 grams of acetic acid CH₃COOH are dissolved in 2550.0 grams water H₂O. What will be the boiling point of the resulting solution?

Solution:

Solute = acetic acid CH₃COOH, molar mass = 60.00 g/mol and K_B = 0.51 (Kg °C/mol)

Solvent water H₂O

Step 1: the molality should be calculated.

Amount of solvent should converted into kilograms = 2550.0 gram = 2.5500 kg

Moles of acetic acids = (25.50 g) / (60.00 g/mol) = 0.4250 moles

molality = 0.425 moles acetic acid / 2.5500 kg water = 0.1667 mol/kg

Step 2: Boiling point elevation formula is used.

∆T_B = T_B(solution) – T_B(pure solvent)  = (i) x K_B x molality

CH₃COOH(aq)   +   H₂O(l)   à  H₃O⁺(aq)   + CH₃COO^–(aq)   [1 mole acetic acid will have 2 moles of particles and hence I = 2]

∆T_B = T_B(solution) – 100.0 ^oC = (2) x (0.51 kg ^oC/mol) x (0.1667 mol/kg) = 0.17 ^oC

T_B(solution) = ∆T_B  +  100.0 ^oC = 0.17 ^oC + 100.0 ^oC = 100.17 ^oC = 100.2 ^oC

A You Tube video illustrates the Boiling Point Elevation calculations:https://www.youtube.com/embed/QX4efwPcGt4?feature=oembed