Notes on Solutions

Chapter 13: Solutions

13.5 Solution Concentration

• Solutions have variable composition; hence, to adequately describe a solution, one must specify the components and their relative amounts.

• The terms dilute and concentrated serve as qualitative descriptors of the solute amount in a solution.

• Concentration is defined as the amount of solute present in a specified amount of solution.

Concentration Units

Table 13.5: Solution Concentration Terms

i. Solution Concentration: Molarity

• Molarity (M or mol/L) is measured as the number of moles of solute per 1 liter of solution.

• Example of Molarity: For a sugar solution with a concentration of 2.0 M:

  • 1 L of solution contains 2.0 mol of sugar.

  • 2 L of solution contains 4.0 mol of sugar.

  • 0.5 L of solution contains 1.0 mol of sugar.

ii. Solution Concentration: Molality

• Molality (m or mol/kg) is defined as the number of moles of solute per 1 kilogram of solvent.

• Molality depends on the mass of solvent, not the solution, and does not vary with temperature, since it is based on mass instead of volume.

iii. Parts Solute in Parts Solution

• Parts can be evaluated by either mass or volume, with corresponding measurements:

  • By mass: in grams, kilograms, or pounds.

  • By volume: in milliliters, liters, or gallons.

• Percentage = amount of solute in every 100 parts of solution.

  • Example: A solution that is 0.9% by mass contains 0.9 g of solute in every 100 g of solution.

Parts per Million (ppm)

• ppm can be defined in two ways:

  • Grams of solute per 1,000,000 g of solution.

  • Milligrams (mg) of solute per 1000 g of solution.

• Note that 1 L of water at room temperature is approximately equal to 1000 g or 1 kg of water.

• For dilute aqueous solutions, the weight of the solution can be approximated by that of the water.

• ppm by mass formula:
  $ppm = \frac{mass \ of \ solute}{total \ mass \ of \ solution} \times 10^6$

Parts per Billion (ppb)

• ppb can be defined similarly to ppm:

  • Milligrams of solute per 1,000,000 g of solution.

  • Micrograms (μg) of solute per 1000 g of solution.

• ppb by mass formula:
  $ppb = \frac{mass \ of \ solute}{total \ mass \ of \ solution} \times 10^9$

iv. Solution Concentrations: Mole Fraction (χa)

• A mole fraction is defined as the ratio of the moles of one component to the total moles of all components in the solution.

  • Total of all mole fractions in a solution equals 1:
    $χ_a = \frac{moles \ of \ substance \ a}{moles \ of \ substance \ a + moles \ of \ substance \ b}$

  • Thus, $χ<em>a + χ</em>b + … = 1$

• Mole fractions are dimensionless (no units).

• Mole Percentage formula:
  $mol \% = mole \ fraction (χ_a) \times 100\%$

Converting Concentration Units

1. Write the given concentration as a ratio.

2. Separate the numerator and denominator into solute and solution parts.

3. Convert the solute part into the required unit.

4. Convert the solution part into the required unit.

5. Use the definitions to calculate the new concentration units.

Examples

Example 7: Molality of Aqueous Solution

• Problem: Calculate the molality (m) of an aqueous solution that is 10.5% glucose (C6H12O6) by mass.

Example 8: Molarity Calculation

• Problem: Given a 0.448-m aqueous solution of K2CrO4 with a density of 1.063 g/mL, find the molarity (M) of the solution. The molar mass of K2CrO4 is 194.2 g/mol.

Example 9: Molarity and Molality Calculation

• Problem: If the mole fraction of ethylene glycol (MM=62.07 g/mol) in an aqueous solution is 0.24 and the density of the solution is 1.06 g/mL, calculate the molarity and molality of the solution.

Example 10: Multiple Calculations

• Problem: An aqueous solution with a density of 1.23 g/mL contains 30% H2SO4 by mass.

  • Calculate:
    a) Molarity of H2SO4 in the solution
    b) Molality of H2SO4 in the solution
    c) Mole fraction of H2SO4 in the solution

13.6 Colligative Properties

• Colligative properties are defined as physical properties of a solution whose values depend solely on the number of solute particles in the solution, rather than the type of particles.

• The types of solute particles can include atoms, charged ions, or neutral molecular species.

• Colligative properties vary based on the number of dissolved particles; the behavior of non-electrolytes differs from that of electrolytes due to the number particles formed upon dissolution.

Importance of Electrolyte Types

• Understanding whether a solute is a strong electrolyte, a weak electrolyte, or a nonelectrolyte is critically important in the context of colligative properties:

  • Strong electrolyte example: 1 mole NaCl in H2O produces 2 moles of solute particles (1 mole Na+ and 1 mole Cl–).

  • Weak electrolyte example: 1 mole HF in H2O produces less than 2 moles of solute particles (H+ and F–).

  • Nonelectrolyte example: 1 mole CH3OH in H2O produces 1 mole of solute particles (CH3OH molecules).

Types of Colligative Properties

A. Vapor Pressure Lowering
B. Boiling Point Elevation
C. Freezing Point Depression
D. Osmotic Pressure

A. Vapor Pressure of Solutions

• Vapor pressure of a liquid is the pressure of a vapor above its liquid when both are in equilibrium.

• Liquids exhibiting high vapor pressures are termed volatile.

• The vapor pressure of a solvent within a solution is lower than that of the pure solvent due to the replacement of solvent molecules by solute particles.

• Adding a non-volatile solute reduces the rate of vaporization, decreasing the vapor concentration above the solution.

• This creates a dynamic equilibrium with fewer vapor molecules, leading to a lower vapor pressure compared to the pure solvent.

Vapor Pressure Lowering Formula

• The reduction in vapor pressure of a solvent can be quantified using:
  $\Delta P = P^{\circ}<em>{solvent} - P</em>{solution} = \chi<em>{solute} \cdot P^{\circ}</em>{solvent}$

• Where:

  • $\Delta P$ is the vapor pressure lowering.

  • $P^{\circ}_{solvent}$ is the vapor pressure of the pure solvent.

  • $P_{solution}$ is the vapor pressure of the solvent in the solution.

Raoult’s Law for Vapor Pressure

• In a solution, the vapor pressure of a volatile solvent is equal to its normal vapor pressure (i.e., $P^{\circ}<em>{solvent}$) multiplied by its mole fraction (i.e., $\chi</em>{solvent}$):
  $P<em>{solvent}^{solution} = \chi</em>{solvent} \cdot P^{\circ}_{solvent}$

• As the mole fraction is inherently less than 1, the vapor pressure of the solvent in solution will always be lower than in the pure state.

Raoult’s Law with Nonvolatile Solute

• For a solution with a nonvolatile solute, we use:
  $P<em>{solution} = \chi</em>{solvent} \cdot P^{\circ}_{solvent}$

Raoult’s Law with Volatile Solute

• When both solvent and solute exert vapor pressures:

  • $P<em>{solute} = \chi</em>{solute} \cdot P^{\circ}_{solute}$

  • $P<em>{solvent} = \chi</em>{solvent} \cdot P^{\circ}_{solvent}$

• The total vapor pressure of the solution is modeled by:
  $P<em>{solution} = \chi</em>{A} \cdot P^{\circ}<em>{A} + \chi</em>{B} \cdot P^{\circ}_{B}$

• Note: $\chi<em>A + \chi</em>B = 1$.

• Importantly, the composition of A and B in the liquid phase differs from their composition in the vapor phase.

Ideal Solutions

• In an ideal solution, both solute and solvent can evaporate, leading to both types being present in the vapor phase.

• In ideal solutions, the interactions between solute and solvent are equal to the sum of broken interactions between solute-solute and solvent-solvent.

• This effect effectively dilutes the solvent.

• Ideal solutions strictly follow Raoult’s law at all concentrations, and the total vapor pressure can be calculated as:
  $P<em>{total} = P</em>{solute} + P_{solvent}$

Deviations from Raoult’s Law: Non-Ideal Solutions

• Non-ideal solutions result from solute-solvent interactions being either stronger or weaker than the broken interactions:

  • When solute-solvent interactions are stronger, the vapor pressure of the solution will be lower than predicted (negative deviation).

  • When solute-solvent interactions are weaker, the vapor pressure will be higher than expected (positive deviation).

Summary of Ideal vs. Non-Ideal Solutions

• In ideal solutions, solute-solvent interactions equal broken interactions.

• If differences arise, resulting in stronger solute-solvent interactions, vapor pressure is lower than predicted (negative deviation).

• Conversely, weaker interactions than the broken ones yield a higher vapor pressure (positive deviation).

Examples Calculating Vapor Pressure and Total Pressure

• Example 11: Calculate the vapor pressure of a solution made by dissolving 350 g of glucose (C6H12O6) in 550 mL of water at 35°C. The vapor pressure of water at this temperature is 42.0 mm Hg.

• Example 11a: A solution made by mixing 2.00 mol of benzene and 2.00 mol of toluene at 25 °C. Determine:
  a) The partial pressure of each component
  b) The total vapor pressure of the solution
  c) The mole fraction of each component in vapor.

• Example 11b: For a solution with a mole fraction of toluene at 0.750, calculate the mole fraction of benzene in the vapor above this ideal solution.