Colligative Properties Overview

Reminder

• Complete the following before the deadline:
  • Submit Unit 10 labs (3 labs).
  • Submit worksheets 1 and 2 of Unit 11.
• Check the calendar on Canvas for deadlines.

Chapter 15: Solutions

Colligative Properties

• Definition: Colligative properties describe how a solute affects the physical properties of a solvent.
• Key Types: Concentrate on vapor pressure, freezing point, and boiling point changes.
• Dependence: The effect magnitude depends on the number of solute particles, not their identity.
• Examples:
  • C₁₂H₂₂O₁₁ (s) → C₁₂H₂₂O₁₁ (aq): 1 particle.
  • NaCl (s) → Na⁺ (aq) + Cl⁻ (aq): 2 particles.
  • Na₂SO₄ (s) → 2 Na⁺ (aq) + SO₄²⁻ (aq): 3 particles.
  • AlCl₃ (s) → Al³⁺ (aq) + 3 Cl⁻ (aq): 4 particles.

Effects of Colligative Properties

• Vapor Pressure:

  • Lowered due to solute interactions with solvent particles, making it harder for solvent to vaporize.
  • Fewer vapor particles lead to lower vapor pressure.

• Boiling Point:

  • Increases because a decreased vapor pressure requires a higher temperature to reach boiling.
  • Note: boiling point relates to vapor pressure, not temperature.

• Freezing Point:

  • Decreases because solute particles hinder solvent crystal formation necessary for freezing.

Measuring Colligative Properties

• Temperature Changes:
  • Change in Freezing Point (∆Tᶠ) and Change in Boiling Point (∆Tᵇ) depend on solute particle number (moles, not concentration).
  • Use van’t Hoff factor (i), indicating the number of particles per formula unit for ionic solutes.
  • For ionic compounds, i equals the number of ions per formula unit.
  • For covalent compounds, i always equals 1.

Mathematical Expressions for Colligative Properties

• Boiling Point Elevation:
  $\Delta T<em>b = i imes K</em>b imes m$

  • Where:
  • $i$ = van’t Hoff factor.
  • $K_b$ = boiling point constant (unique to solvent).
  • $m$ = molality (moles solute / kg solvent).

• Freezing Point Depression:
  $\Delta T<em>f = i imes K</em>f imes m$

  • Where:
  • $i$ = van’t Hoff factor.
  • $K_f$ = freezing point constant (unique to solvent).
  • $m$ = molality (moles solute / kg solvent).

Specific Constants for Water

• Freezing Point Depression: For water, $K_f = -1.86° C/(mol/kg)$. Each mole of solute lowers it by 1.86°C.
• Boiling Point Elevation: For water, $K_b = 0.512° C/(mol/kg)$. Each mole of solute raises it by 0.512°C.

Calculating Final Temperatures

• Final Freezing Point:
  $T<em>f = T</em>{f, ext{normal}} - \Delta T_f$
• Final Boiling Point:
  $T<em>b = T</em>{b, ext{normal}} + \Delta T_b$

Example Calculation

• Given: 86.50 g of ammonium sulfate in 500.0 g of water.

• Molar Mass Calculation:

  • Ammonium sulfate: $ext{Weight} = 86.50 g, ext{Molar Mass of (NH₄)₂SO₄} = 132.17 g/mol$.
  • Moles of ammonium sulfate:
    $ext{Moles} = \frac{86.50}{132.17} = 0.6545$

• Ionic Dissociation:

  • ext{NH}42 ext{SO}4 ightarrow 2 ext{NH}4^+ + ext{SO}_4^{2-} (3 particles, so $i = 3$)

• Molality Calculation:
  $m = \frac{0.6545 ext{ moles}}{0.5000 ext{ kg}} = 1.309$

• Boiling Point Elevation:
  $\Delta T<em>b = (3)(0.512° C)(1.309) = 2.0100° C$$T</em>b = 100.0000° C + 2.0100° C = 102.0100° C$

• Freezing Point Depression:
  $\Delta T<em>f = (3)(-1.86° C)(1.309) = -7.3040° C$$T</em>f = 0.000° C - 7.3040° C = -7.3040° C$

Significant Figures

• Freezing points and boiling points are exact numbers for calculations.
• Example:
  • For $\Delta T<em>f = -1.753° C$, thus $T</em>f = 0.000° C - 1.753° C = -1.753° C$.
  • For $\Delta T<em>b = 0.5327° C$, thus $T</em>b = 100.0000° C + 0.5327° C = 100.5327° C$.