Lecture Notes: Dilution Problems and Solutions

Key Equation

• M1V1 = M2V2
  • M: Molarity/Concentration
  • V: Volume (must use same units for V1 and V2, either milliliters or liters)

Dilution Concepts

• Decreasing Concentration: Add solvent (water)
  • Example: 200 mL of 2M Sodium Chloride to 400 mL reduces concentration to 1M
• Increasing Concentration: Add solute or remove solvent
  • Methods:
    • Add more solute (e.g., sodium chloride)
    • Evaporate solvent (e.g., water)

Problem-Solving Examples

Problem 1

• Situation: 0.8M HCl solution diluted to 500 mL, new concentration 0.20M
  • Original Volume (V1): Use M1V1 = M2V2
    • M1 = 0.8, M2 = 0.2, V2 = 500
    • V1 = 125 mL
  • Water Added: V2 - V1 = 500 - 125 = 375 mL

Problem 2

• Situation: 300 mL KCl solution concentration increased from 0.25M to 0.75M
  • New Volume (V2): Use M1V1 = M2V2
    • M1 = 0.25, V1 = 300, M2 = 0.75
    • V2 = 100 mL
  • Water Removed: V1 - V2 = 300 - 100 = -200 mL (indicating removal)

Problem 3

• Situation: 350 mL water added to 225 mL of 4.5M NaOH
  • New Volume (V2): V1 + added water = 225 + 350 = 575 mL
  • New Concentration (M2): Use M1V1 = M2V2
    • M1 = 4.5, V1 = 225
    • M2 = 1.761M

Problem 4

• Situation: Mixing 60 mL of 0.8M NaCl with 80 mL of 0.5M KBr
  • New Volume: 60 + 80 = 140 mL
  • New Concentrations:
    • NaCl: M2 = (0.8 * 60) / 140 = 0.343M
    • KBr: M2 = (0.5 * 80) / 140 = 0.286M

Weighted Average Problem

Problem 5

• Situation: 150 mL of 0.6M CaCl2 mixed with 300 mL of 0.9M CaCl2
  • Combined Volume (V3): 150 + 300 = 450 mL
  • New Concentration (M3): (0.6150 + 0.9300) / 450 = 0.8M
  • Concentration will lean towards the solution with the greater volume.