Solutions Lecture Notes

Chapter 14 Lecture Notes: Solutions

Solutions and their Components

• Solution:

  • Definition: A homogeneous mixture of two or more substances.

  • Components:

  • Solvent: The majority component of the solution.

  • Solute: The minority component of the solution.

Types of Solutions and Solubility

• Aqueous Solution: A solution where water acts as the solvent.

  • Examples include:

  • Solid in water: Saltwater (NaCl).

  • Liquid in water: Alcohol (ethanol).

  • Gas in water: Soda (CO₂).

• Solubility:

  • Definition: The amount of solute that can dissolve in a given amount of solvent.

  • Note: The phrase “likes dissolve likes” applies. However, entropy also influences the formation of solutions.

  • Entropy:

  • Definition: A measure of the disorder or dispersion of energy of a system.

  • Phenomenon: Entropy drives mixtures to occur, such as gases mixing due to a lack of attraction or repulsion.

Intermolecular Forces (IMFs) and Solution Formation

• To predict the formation of a solution, consider three types of interactions:

  1. Solute-Solute attractions (IMFs)

  2. Solvent-Solvent interactions (IMFs)

  3. Solute-Solvent interactions (IMFs)

  • If solute's attraction is stronger than solvent's, a solution does not form. Conversely, if the solvent's attraction is stronger, a solution will not occur.

  • When IMFs are similar for both solute and solvent, a solution tends to form.

• Example: Dissolving hexane (C₆H₁₄) in water does not occur, as hexane is nonpolar and water is polar.

Energetics of Solution Formation

• The dissolution process may either

  • Release heat (exothermic): Heat is produced when solute and solvent mix.

  • Absorb heat (endothermic): Heat is consumed when solute separates.

Steps in the Process of Dissolution:

1. Breaking solute interactions (endothermic): Solute molecules must overcome their attractive forces to separate.

2. Breaking solvent interactions (endothermic): Solvent particles must likewise separate to accommodate solute particles.

3. Mixing solute and solvent (exothermic): This process releases heat and corresponds to a decrease in potential energy.

Lattice Energy

• Definition: The energy released when one mole of ionic solid is formed from gaseous ions.

  • Influences the energy changes during solution formation.

Enthalpy of Hydration

• Enthalpy of hydration can be expressed as:
  \Delta H{sol} = \Delta H{solute} + \Delta H{solvent} + \Delta H{mix}

• If the process is exothermic, total heat change is negative (\Delta H < 0). If it is endothermic, the change is positive (\Delta H > 0). If there is no heat exchange, it can be approximated as zero (\Delta H_{sol} = 0).

Solution Equilibrium and Factors Affecting Solubility

• Saturated Solution: No more solute can dissolve in the solvent, leading to equilibrium.

• Unsaturated Solution: More solute can still dissolve in the solvent.

Temperature Effects on Solubility

• Many solids dissolve more effectively at increased temperatures due to enhanced molecular movements that facilitate breaking solute-solvent interactions.

• Exceptions: Solubility of gases decreases in higher temperatures due to increased kinetic energy causing gas molecules to escape the solvent. Cold water retains more dissolved gas due to:

  • Smaller intermolecular spaces in colder water.

Pressure Effects on Solubility of Gases

• Henry’s Law: Relates the solubility of a gas to its partial pressure above the liquid.

  • Equation: S = KH imes P{gas}

  • S: Solubility of the gas (mol/L)

  • K_H: Henry’s constant, unique to the specific gas

  • P_{gas}: Partial pressure of the gas (atm)

• Example Problem:

  • To find the required nitrogen gas pressure to maintain a concentration of 0.28 M:
    0.28 = (0.00061) P{gas} P{gas} = \frac{0.28}{0.00061} = 459.02 \text{ atm}

Solution Concentrations

Mole Concentration (Molarity)

• Symbol: M

• Units: mol/L

• Formula:
  M = \frac{n{solute}}{V{solution}}
  where n{solute}: moles of solute and V{solution}: volume of solution in liters.

• Notes: Molarity is temperature-dependent due to thermal expansion affecting the volume of solution.

Molality

• Symbol: m

• Units: mol/kg

• Formula:
  m = \frac{n{solute}}{m{solvent}}
  where m_{solvent} is the mass of the solvent in kg.

Mass Percent

• Symbol: mass%

• Units: unitless

• Formula:
  mass\% = \frac{mass{solute}}{mass{solute} + mass_{solvent}}\times 100\,\%

Parts Per Million (ppm)

• Formulas: ppm = \frac{mass{solute}}{mass{solution}}\times 10^6

  • Where mass of the solution = mass solute + mass solvent.

• Parts Per Billion (ppb):

  • ppb = \frac{mass{solute}}{mass{solution}} \times 10^9

Mole Fraction

• Symbol: χ

• Formula:
  \chi{solute} = \frac{n{solute}}{n{solute} + n{solvent}}

Colligative Properties

• Definition: Properties that depend on the number of solute particles in a solution, not their identity.

• Key Colligative Properties:

  1. Freezing Point Depression

  2. Boiling Point Elevation

  3. Osmotic Pressure

Freezing Point Depression

• Formula: \Delta Tf = kf m

  • Where:

  • k_f: freezing point depression constant

  • m: molality of the solution.

Boiling Point Elevation

• Formula: \Delta Tb = kb m

  • Where:

  • k_b: boiling point elevation constant

  • m: molality of the solution.

Osmotic Pressure

• Formula:
  \Pi = iMRT

  • Where:

  • \Pi: osmotic pressure

  • i: van’t Hoff factor (number of particles the solute breaks into)

  • M: molarity of the solution

  • R: ideal gas constant (0.08206\, L\cdot atm / (K\cdot mol))

  • T: temperature in Kelvin.

• Osmosis:

  • Defined as the movement of solvent from an area of lower solute concentration to an area of higher solute concentration across a semi-permeable membrane.

Example Calculations

Freezing Point Depression Calculation Example

• For a 2.25 m aqueous sucrose solution: \Delta Tf = kf m = 1.86(2.25) = 4.19°C

  • Thus, freezing point = -4.19°C.

Boiling Point Elevation Calculation Example

• For a 1.23 m aqueous sodium chloride solution with a van’t Hoff factor of 1.9: \Delta T_b = 1.9(1.23)(0.512) = 1.20°C

  • Thus, boiling point = 101.2°C.

Molar Mass Calculation Example

• For a solution involving osmotic pressure: \Pi = 3.33 = M (0.08206)(301.15)

  • Thus, the calculated molar mass of protein would approximate: M = 1.77 \times 10^{-4} \text{ mol/L}.

Freezing Point Calculation for Potassium Sulfate

• For 29.1 g of K₂SO₄ in 398 g of water, freezing point depression:
  \Delta T_f = 2.6(1.86)(0.420) = 2.03°C

• Thus, freezing point = -2.03°C.

Molar Mass Calculation for Non-Ionic Compound

• For a non-ionic compound with a resulting freezing point depression of -6.4°C: 6.4 = (1.86)(m)

  • Molar mass approximated through: m = 3.44 = \frac{101.5}{x}, yielding values around 81.2\, g/mol.