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.
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".
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.
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.
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.
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.
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.
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.
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.