Colligative Properties Flashcards

Colligative Properties

• Colligative properties are properties of solutions that depend on the number of solute particles, not their identities.
• Solutes that dissociate produce more particles than those that don't.

Van't Hoff factor (i)

• The Van't Hoff factor (i) is the number of moles of particles in solution resulting from each mole of dissolved solute.
• Example:
  • CaCl_2 (s) \rightarrow Ca^{2+}(aq) + 2Cl^-(aq)
  • For CaCl_2, i = 3 (1 calcium ion and 2 chloride ions)
  • Urea (a covalent compound): i = 1

Effects of Increasing Dissolved Particles

• As the number of dissolved particles increases:
  • Freezing point decreases.
  • Vapor pressure decreases.
  • Boiling point increases.
  • Osmotic pressure increases.

Freezing Point Depression and Boiling Point Elevation

Constants

• K_f: Freezing point depression constant.
• K_b: Boiling point elevation constant.
• Presence of solute expands the liquid range of a solvent.

Freezing Point Depression

• Formula: Tf = Tf^0 - \Delta T_f
  • T_f = New freezing point
  • T_f^0 = Freezing point of pure solvent
  • \Delta Tf = i \cdot m \cdot Kf
    • i = Van't Hoff factor
    • m = molal concentration (moles of solute / kg of solvent)
    • K_f = constant that depends on the solvent
• The freezing point decreases relative to the pure solvent.

Boiling Point Elevation

• Formula: Tb = Tb^0 + \Delta T_b
  • T_b = New boiling point
  • T_b^0 = Boiling point of pure solvent
  • \Delta Tb = i \cdot m \cdot Kb
    • i = Van't Hoff factor
    • m = molal concentration (moles of solute / kg of solvent)
    • K_b = constant that depends on the solvent
• The boiling point increases relative to the pure solvent.

Example Calculation: Freezing Point Depression

• Problem: Calculate the mass of calcium chloride (FW = 110.98 g/mol) you must add to 5.0 L of water to depress its freezing point to -5.0 °C (assume d = 1.0 kg/L).
• Solution:
  • \Delta T_f = 5.0 °C (since the freezing point is depressed from 0°C to -5.0°C)
  • \Delta Tf = i \cdot m \cdot Kf
  • For CaCl_2, i = 3
  • K_f for water = 1.86 °C/m
  • 5.0 = 3 \cdot m \cdot 1.86
  • m = \frac{5.0}{3 \cdot 1.86} = 0.894 mol/kg
  • Volume of water = 5.0 L
  • Density of water = 1.0 kg/L
  • Mass of water = 5.0 L * 1.0 kg/L = 5.0 kg
  • Moles of CaCl_2 needed = 0.894 mol/kg * 5.0 kg = 4.47 mol
  • Mass of CaCl_2 needed = 4.47 mol * 110.98 g/mol = 496.98 g ≈ 4.97 x 10^2 g or 0.50 kg

Raoult's Law for Vapor Pressures

• Pi = xi \cdot P_i^0

  • P_i = Vapor pressure of component i above the solution
  • x_i = Mole fraction of i in the solution
  • P_i^0 = Vapor pressure of pure component i under current conditions
  • P{total} = \sum Pi

• Volatile compounds have significant vapor pressures.

• Nonvolatile compounds do not have significant vapor pressures.

• If all solutes are non-volatile: P{solution} = X{solvent} \cdot P_{solvent}

• The mole fraction (x) must take into account the Van't Hoff factor:

  • Moles of dissolved particles, not just moles of compounds.

• Volatile solutes tend to have i = 1.