CH 13 (2/20)(PG 6-11)

Solute-Solvent Interactions

Solid-Liquid Interactions

• Ionic Solids:

  • May be soluble in polar water, but some are insoluble.

  • Expected to be insoluble in nonpolar solvents.

  • Wide ranges of solubility in water.

  • Solubility Examples:

  • NaCl: 36 g / 100 mL

  • AgCl: 0.0001 g / 100 mL

  • BaSO4: 0.0003 g / 100 mL

  • NaC2H3O2: 119 g / 100 mL

Alcohol Solubility in Water

• Solubility Values for Alcohols in mol / 100 g H2O:

  • CH3OH: 0.11

  • C2H5OH: 0.030

  • C3H7OH: 0.0058

• Solubility of Specific Compounds:

  • Acetic Acid:

  • Draw the Lewis structure for acetic acid.

  • Is acetic acid polar or non-polar?

  • Is acetic acid soluble in water?

  • Mineral Oil:

  • Composed of CH3(CH2)nCH3 molecules where n is about 10

  • Are molecules in mineral oil polar or non-polar?

  • Is mineral oil soluble in water?

Additional Solid-Liquid Interactions

• Molecular Solids:

  • Polar sugar is soluble in polar water but insoluble in nonpolar gasoline.

• Nonpolar Paraffin Wax:

  • A mixture of CH3(CH2)nCH3 molecules (n ~ 18-38) that is insoluble in polar water and soluble in nonpolar CCl4.

Enthalpy Changes in Solution

• Enthalpy (H):

  • ΔH equals the heat gained or lost by a system under constant pressure.

• Energetics of the Solution Process (ΔsolutionH):

  • Example: Dissolution of KF (s) to K+(aq) and F−(aq):

  • Energy must be supplied to separate the ions, overcoming attractive forces (endothermic).

  • Energy is evolved when individual ions dissolve in water, stabilizing by solvation (exothermic).

• Induced Chemical Equations:

  • Step 1: KF(s) → K+(g) + F−(g) (−ΔlatticeH)

  • Step 2: K+(g) + F−(g) → K+(aq) + F−(aq) (ΔhydrationH)

  • Overall Reaction: KF(s) → K+(aq) + F−(aq)

  • ΔsolutionH = −ΔlatticeH + ΔhydrationH

  • Example Calculation: ΔsolutionH = 821 kJ/mol + (-837 kJ/mol) = -16 kJ/mol

• Significance of ΔsolutionH:

  • Solution enthalpy is small, being the difference between two large quantities, which may show a positive or negative sign.

Temperature Effects on Solubility

• Solids in Liquids:

  • Solubility of solids in water often increases with temperature but may also decrease in some cases.

Pressure Effects on Solubility

• General Observations:

  • No significant effect of pressure on the solubility of liquids and solids.

  • For gases: Higher gas pressure leads to higher solubility.

• Henry’s Law:

  • Solubility of a gas is directly proportional to the gas pressure:

  • Sg = kHPg

    • where Sg = solubility of gas (usually mol/kg)

    • Pg = partial pressure of the gas over the solution

    • kH = Henry’s Law constant

Gas Mixtures & Partial Pressures

Dalton’s Law

• The total pressure of a mixture of gases equals the sum of the pressures that each gas would exert if present alone:

  • Ptotal = PA + PB + PC…

• Mathematical Representation:

  • PAV = nART

  • Ptotal = (nA + nB + nC)(RT/V)

  • Pressure of individual gases: PA = (nA/ntotal)(n totalRT/V)

Colligative Properties Overview

• Definition:

  • Properties depending on the number of solute particles, independent of the type of solute.

• Examples of Colligative Properties:

  • Vapor pressure lowering

  • Boiling point elevation

  • Freezing point depression

  • Osmotic pressure

Vapor Pressure Lowering

• Raoult’s Law:

  • Psolvent = Xsolvent * Pº solvent

  • Psolvent: Vapor pressure of the solvent over the solution

  • Pº solvent: Equilibrium vapor pressure of pure solvent

  • Xsolvent: Mole fraction of the solvent

  • Definition of Ideal Solution: A solution obeying Raoult's law.

Boiling Point Elevation & Freezing Point Depression

• Dependence On:

  • Molal concentration (m) of solute particles

  • Freezing point depression constant (Kfp)

  • Boiling point elevation constant (Kbp) characteristic of solvent

• Constants for Water**:

  • Kbp(H2O) = 0.51 °C kg/mol

  • Kfp(H2O) = −1.86 °C kg/mol

Identifying Changes in Temperature

• Boiling Point Elevation:

  • ΔTbp = Kbp * m

• Freezing Point Depression:

  • ΔTfp = Kfp * m

• Applying Relationships to Solvent**:

  • Tbp = Tbp° + ΔTbp

  • Tfp = Tfp° + ΔTfp

Colligative Properties of Electrolytes

• Van't Hoff Factor (i):

  • Each colligative property adjustment needs to incorporate how many particles the solute dissociates into.

• Calculating Freezing Point Changes:

  • Example Calculation for Saturated Solutions of NaCl and CaCl2 at 0 °C:

  • NaCl solubility: 35.7 g per 100.0 g of water

  • CaCl2 solubility: 59.5 g per 100.0 g of water

  • Kfp for H2O = −1.86 °C kg/mol

• Molecular Contribution to Solutions:

  • 1 mol of sucrose = 1 mol of particles

  • 1 mol of NaCl = 2 mol of particles (1 mol Na+ and 1 mol Cl−)

  • 1 mol of CaCl2 = 3 mol of particles