Chapter 13: Properties of Solutions – General Chemistry Study Notes

Properties of Solutions

Solution Definitions

Solutions are homogeneous mixtures of two or more substances, where the components are present as atoms, molecules, or ions. The solvent is the substance present in the largest amount, while the solute is present in lesser amounts. Both the solvent and solute can be solids, liquids, or gases.

• Solvent: Substance in the largest amount in a solution.

• Solute: Substance(s) present in lesser amounts.

• Solution: Homogeneous mixture of solute(s) and solvent.

Examples: Salt water (water as solvent, salt as solute), steel (iron as solvent, carbon as solute).

Types of Solutions

Solutions can be classified based on the physical states of their components:

Solvent	Solute	Example
Solid	Solid	Steel (iron and carbon)
Liquid	Solid	Salt water (water and NaCl)
Liquid	Liquid	Alcoholic beverage (water and ethanol)
Gas	Gas	Air (oxygen and nitrogen)

Water as a Solvent

Water is the most common solvent, forming aqueous solutions. It is a polar molecule and forms hydrogen bonds between the hydrogen atom in one molecule and the oxygen atom in another.

• Polarity: Water has a partial negative charge on oxygen and partial positive charge on hydrogen.

• Hydrogen bonding: Strong intermolecular force that affects solubility and solution properties.

Classification by Amount of Solute Dissolved

Solutions are classified based on the amount of solute dissolved at a specific temperature:

• Unsaturated solution: Contains less solute than the solvent can dissolve.

• Saturated solution: Contains the maximum amount of solute that can dissolve; excess solute remains undissolved.

• Supersaturated solution: Contains more solute than the maximum amount; unstable and can precipitate excess solute.

Solubility

Solubility is the amount of solute that dissolves in a given amount of solvent, typically expressed as grams solute per 100 mL solution.

• Unsaturated solutions: Can dissolve more solute.

• Saturated solutions: Have undissolved solute at the bottom.

Practice: Saturated vs. Unsaturated

• A spoonful of salt added to boiling water dissolves: Unsaturated

• A layer of sugar forms on the bottom of a glass of tea as ice is added: Saturated

• At 40°C, solubility of KBr is 80g/100g H2O:

  • 60g KBr in 100g water: Unsaturated

  • 200g KBr in 200g water: Saturated

  • 25g KBr in 50g water: Unsaturated

Factors Affecting Solubility

Solvation is the process where solute molecules are separated and surrounded by solvent molecules. Solubility depends on:

• Solute-solute interactions

• Solvent-solvent interactions

• Solute-solvent interactions

Steps in the Dissolving Process

The dissolving process involves three steps, each with associated enthalpy changes:

1. Separating solute particles ($\Delta H_1 > 0$)

2. Separating solvent particles ($\Delta H_2 > 0$)

3. Solute-solvent interactions ($\Delta H_3 < 0$)

Total enthalpy change: $\Delta H_{soln} = \Delta H_1 + \Delta H_2 + \Delta H_3$

Enthalpy of Solution

If solute-solvent interactions are dominant, $\Delta H_{soln}$ is negative (exothermic). If not, $\Delta H_{soln}$ can be positive (endothermic).

Why Water and Oil Do Not Mix

• Oil is nonpolar; water is polar.

• Energy required to break water’s hydrogen bonds ($\Delta H_2$) is large and positive.

• Interactions between oil and water ($\Delta H_3$) are negligible.

• Overall, $\Delta H_{soln}$ is large and positive, so mixing does not occur.

Intermolecular Forces and Solubility

Solubility is influenced by the type and magnitude of intermolecular forces (IMFs). Substances with similar IMFs are likely to be soluble in each other.

• "Like dissolves like": Polar solvents dissolve polar solutes; nonpolar solvents dissolve nonpolar solutes.

• Miscible: Two liquids completely soluble in each other.

• Immiscible: Two liquids not soluble in each other.

Examples of Solubility

Solute	Solubility Greater In
Br2	Benzene (nonpolar)
NaI	Water (polar)
CCl4	Benzene (nonpolar)
CH2O	Water (polar)
C3H8	Benzene (nonpolar)
HCl	Water (polar)

Solution Concentrations

Ways to Express Concentration

• Mass/mass percent (% m/m)

• Mass/volume percent (% m/v)

• Volume/volume percent (% v/v)

• Parts per million (ppm) and parts per billion (ppb)

• Molarity (M)

• Molality (m)

• Mole fraction (X)

Mass/Mass Percent

Expresses the mass of solute per mass of solution, multiplied by 100.

• Formula: $\% (m/m) = \frac{\text{solute mass}}{\text{solution mass}} \times 100$

• Example: 8.00 g KCl in 50.00 g solution: $\frac{8.00}{50.00} \times 100 = 16.0\%$

Volume/Volume Percent

Used for liquid-liquid solutions.

• Formula: $\% (v/v) = \frac{\text{solute volume}}{\text{solution volume}} \times 100$

• Example: 40% alcohol by volume in liquor.

Mass/Volume Percent

Common in medical and biological contexts.

• Formula: $\% (m/v) = \frac{\text{grams of solute}}{\text{mL of solution}} \times 100$

• Example: D5W is a 5% (m/v) glucose solution.

Parts Per Million (ppm) and Parts Per Billion (ppb)

Used for very dilute solutions.

• Formula: $\text{ppm} = \frac{\text{mass of solute (g)}}{\text{mass of solution (g)}} \times 10^6$

• Formula: $\text{ppb} = \frac{\text{mass of solute (g)}}{\text{mass of solution (g)}} \times 10^9$

• Example: 9.5 x 10-5 g fluoride in 100 g water = 0.95 ppm

Molarity (M)

Molarity is the number of moles of solute per liter of solution.

• Formula: $M = \frac{\text{moles of solute}}{\text{liters of solution}}$

• To find moles: $\text{moles} = M \times V$

• To find volume: $V = \frac{\text{moles}}{M}$

Molality (m)

Molality is the number of moles of solute per kilogram of solvent.

• Formula: $m = \frac{\text{moles of solute}}{\text{kilograms of solvent}}$

• Example: 1.45 g sucrose in 30.0 mL water ($m = 0.141$ mol/kg)

Mole Fraction (X)

Mole fraction is the ratio of moles of one component to the total moles in solution.

• Formula: $X_A = \frac{n_A}{n_A + n_B}$

• Example: 32.0 g methanol in 90.0 g water ($X_{CH_3OH} = 0.167$)

Practice Problem: Concentration Calculations

Given: 170.1 g glucose (C6H12O6, 180.2 g/mol) in 1 L solution (density = 1.062 g/mL)

• Molality: $m = \frac{0.9440 \text{ mol}}{0.892 \text{ kg water}} = 1.06$ mol/kg

• Percent by mass: $\frac{170.1}{1062} \times 100 = 16.02\%$

• Parts per million: $\frac{170.1}{1062} \times 10^6 = 160,210$ ppm