2/25/25 CCP Chemistry Lecture (Unit 2.3, Chap 5)

Overview

• Discussing Chapter 5, focusing on Molarity and its applications in chemical calculations.

• Reference to earlier chapters: Stoichiometry from Chapter 4 and basics from Chapters 3 and 4.

Molarity

• Definition: Molarity (M) is the number of moles of solute per liter of solution.

  • Commonly used by chemists to measure concentration.

  • In contrast, household products often use percentage concentrations (e.g., vinegar is 5% acetic acid).

• Conversion Examples:

  • Sodium Hydroxide:

    • 40 grams of NaOH in 200 mL of solution.

    • Convert grams to moles using molar mass:

      • Molar Mass of NaOH: 40 g/mol → 0.1 moles.

    • Convert 200 mL to liters: 200 mL = 0.2 L.

    • Molarity Calculation: 0.1 moles / 0.2 L = 0.5 M.

      • Capital M is pronounced "molar".

Comparison of Molar Mass and Molarity

• Molar Mass: Uses atomic weights to convert between grams and moles for pure substances.

• Molarity: Allows conversion between liters and moles in solutions.

  • Essential to know the molarity when working with solutions in stoichiometry.

Example Calculations

1. Potassium Bromide:

   • Question: How many grams of KBr needed for a 4 L of a 2 M solution?

   • Calculation steps:

     • 4 L × 2 moles/L = 8 moles of KBr.

     • Molar Mass of KBr: 119 g/mol → 8 moles × 119 g/mol = 952 g.

2. Chloride from Aluminum Chloride:

   • Question: How many moles of chloride are in 12 L of 0.5 M aluminum chloride?

   • Calculation:

     • 12 L × 0.5 moles/L = 6 moles AlCl3.

     • 1 mole AlCl3 produces 3 moles Cl → 6 moles × 3 = 18 moles Cl.

3. Stoichiometry with Reactions:

   • Reaction Between Silver Nitrate and Sodium Chloride:

   • Calculate grams of silver chloride from 0.5 L of 1.25 M silver nitrate.

   • Steps:

     • 0.5 L × 1.25 moles/L = 0.625 moles AgNO3.

     • Stoichiometry indicates a 1:1 ratio with AgCl → 0.625 moles AgCl produced.

     • Molar Mass of AgCl: 143.35 g/mol → 0.625 moles × 143.35 g/mol = 89.6 g.

4. Neutralization Reaction:

   • Required volume of sulfuric acid to neutralize NaOH:

     • Given: 50 mL of 0.65 M NaOH with 1.5 M sulfuric acid balance equation.

     • Steps:

       • 50 mL NaOH = 0.050 L → 0.050 L × 0.65 moles/L = 0.0325 moles NaOH.

       • Reaction shows 2:1 ratio → 0.0325 moles NaOH yields 0.01625 moles H2SO4.

       • Convert back to volume: 0.01625 moles × (1 L / 1.5 M) = 0.01083 L = 10.83 mL.

Dilutions

• Concept of Dilution: Adding solvent (usually water) to concentrate solutions to achieve a lower concentration.

• Equation for Dilutions: M1V1 = M2V2.

  • Example:

    • Given: 16 M nitric acid, to make 800 mL of 0.5 M.

    • Calculate: 16 M × V1 = 0.5 M × 0.800 L → V1 = (0.5 × 0.800) / 16 = 25 mL.

• Ensure units are consistent during calculations.

Final Considerations

• Important to keep track of units throughout all calculations.

• Understanding of stoichiometry problems expands as more conversion factors (like molarity) are introduced.

• Review notes from Chapter 4 for better understanding of past concepts.