Lecture 24

Chapter 11: Solutions

I. Concentration Units (11.1)

Understanding how to express the concentration of solutions is fundamental in chemistry. Several units are commonly used, each with specific applications and calculation methods.

• Percent by Mass: The mass of solute divided by the total mass of the solution, multiplied by 100%.

  • Formula: $\text{Percent by mass} = \frac{\text{mass of solute}}{\text{mass of solution}} \times 100\%$

  • Example: What is the percent by mass of a NaCl solution in water made up of 0.5 g NaCl in 55.2 g of water? Solution: Total mass of solution = 0.5 g + 55.2 g = 55.7 g $\text{Percent by mass} = \frac{0.5}{55.7} \times 100\% = 0.90\%$

• Mole Fraction: The ratio of the number of moles of a component to the total number of moles in the solution.

  • Formula: $X_A = \frac{n_A}{n_{\text{total}}}$ where $n_A$ is moles of component A, and $n_{\text{total}}$ is total moles of all components.

• Molality (m): The number of moles of solute per kilogram of solvent.

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

  • Example: What is the molality of a solution containing 25 g of NaCl in 200 mL of H2O? Solution: Moles NaCl = $\frac{25}{58.44} = 0.428$ mol Mass of water = 200 mL × 1 g/mL = 200 g = 0.200 kg $m = \frac{0.428}{0.200} = 2.14$ m

  • Practice Example: How many g of NaCl should be added to 50 mL of H2O to make a 0.47 m solution? Solution: Mass of water = 50 g = 0.050 kg Moles NaCl = $0.47 \times 0.050 = 0.0235$ mol Mass NaCl = $0.0235 \times 58.44 = 1.37$ g

II. Solution Dilution (11.1)

Solution dilution involves reducing the concentration of a solution by adding more solvent. The total number of moles of solute remains constant during dilution.

• Key Principle: The total number of moles of solute remains the same.

• Formula: $M_1V_1 = M_2V_2$ where $M_1$ and $V_1$ are the initial molarity and volume, and $M_2$ and $V_2$ are the final molarity and volume.

• Example: How would you prepare 250 mL of a 0.15 M NaCl solution from a 5.0 M stock solution? Solution: $M_1 = 5.0$ M, $V_1 = ?$, $M_2 = 0.15$ M, $V_2 = 250$ mL = 0.250 L $5.0 \times V_1 = 0.15 \times 0.250$ $V_1 = \frac{0.15 \times 0.250}{5.0} = 0.0075$ L = 7.5 mL Measure 7.5 mL of stock solution and dilute to 250 mL with water.

III. The Nature of Dissolved Species: Reactions in Aqueous Solution (11.2, 11.3)

When substances dissolve in water, they may undergo chemical reactions. These reactions are classified based on the nature of the reactants and products.

a) General Types of Reactions

• Single Displacement (Replacement) Reactions: An element replaces another in a compound.

  • General Form: $AB + C \rightarrow AC + B$

  • Examples:

    • Metal + Acid (water) → H2 is displaced (e.g., Zn + 2HCl → ZnCl2 + H2)

    • Metal + Metal Ion → Replace metal ion (e.g., Fe + CuSO4 → FeSO4 + Cu)

    • Halogen + Halide Ion → Replace halide ion (e.g., Cl2 + 2KBr → 2KCl + Br2)

• Double Displacement (Metathesis) Reactions: Ions in two compounds exchange partners.

  • General Form: $AB + CD \rightarrow AD + CB$

  • Driving Forces: Formation of a precipitate, gas, or weak electrolyte (such as water).

  • Types:

    • Precipitation Reaction: An insoluble product forms from soluble reactants. Example: $AgNO_3(aq) + NaCl(aq) \rightarrow AgCl(s) + NaNO_3(aq)$

    • Neutralization Reaction: Acid reacts with base to form water and a salt. Example: $HCl(aq) + NaOH(aq) \rightarrow NaCl(aq) + H_2O(l)$

    • Gas Formation: Carbonate or sulfite reacts with acid to form CO2 or SO2 gas. Example: $Na_2CO_3(aq) + 2HCl(aq) \rightarrow 2NaCl(aq) + H_2O(l) + CO_2(g)$

  • Salt: An ionic compound containing a cation other than H+ and an anion other than OH- or O2-.

b) Electrolytes vs Non-Electrolytes

• Electrolyte: A solute that produces ions in water, allowing the solution to conduct electricity.

• Non-electrolyte: A substance that does not produce ions in solution and does not conduct electricity.

• Strong Electrolyte: Completely dissociates into ions in water (e.g., NaCl, HCl, NaOH).

• Weak Electrolyte: Partially dissociates in water (e.g., acetic acid, NH3).

• Note: Strong acids, strong bases, and most soluble salts are strong electrolytes. Weak acids and bases are weak electrolytes.

c) Ionization and Equilibrium

• Ionization: The process by which a molecule forms ions in solution.

• Equilibrium: Represented by a double arrow ($\rightleftharpoons$), indicating that the forward and reverse reactions occur at the same rate, resulting in no net change in concentrations.

• Example: Acetic acid in water: $CH_3COOH(aq) \rightleftharpoons CH_3COO^-(aq) + H^+(aq)$

d) Solubility of Ionic Compounds

• Solubility: The maximum amount of solute that can dissolve in a given amount of solvent at a specific temperature.

• Classification: Substances are classified as soluble, slightly soluble, or insoluble.

• General Rule: "Like dissolves like" – polar solvents dissolve polar/ionic solutes; nonpolar solvents dissolve nonpolar solutes.

*Additional info: Some definitions and examples were expanded for clarity and completeness. Practice problems were solved using standard chemical calculations. The classification of electrolytes and solubility was inferred from standard general chemistry content.*