Lecture Notes on Dilution Problems

Introduction

Topic: Solving dilution problems in chemistry.
Method Used: M1V1 = M2V2 formula, where:
- M1 is the initial concentration (molarity)
- V1 is the initial volume
- M2 is the final concentration
- V2 is the final volume

Problem Statement: 0.850 L of a 5.00 M solution of NaCl is diluted to 1.80 L. Find the concentration of the diluted solution.
Solution Steps:
1. Use the equation M1V1 = M2V2.
2. Solve for M2 (final concentration):
  - Rearrange to M2 = (M1 * V1) / V2.
3. Plug in values:
  - M1 = 5.00 M
  - V1 = 0.850 L
  - V2 = 1.80 L
  - Calculate M2 = (5 * 0.850) / 1.80 = 2.36 M
Conclusion: The diluted concentration is 2.36 M, which is less than the initial concentration, indicating correct dilution.

Problem Statement: What volume of 0.12 M HCl can be prepared from 11 mL of a 0.45 M HCl stock solution?
Solution Steps:
1. Use the equation M1V1 = M2V2.
2. Solve for V2 (desired volume):
  - Rearrange to V2 = (M1 * V1) / M2.
3. Plug in values:
  - M1 = 0.45 M
  - V1 = 11 mL
  - M2 = 0.12 M
  - Calculate V2 = (0.45 * 11) / 0.12 = 41 mL
Conclusion: The final volume required is 41 mL, which checks out as greater than the initial volume, indicating correct dilution.

Problem Statement: What volume of 1.59 M NaOH is required to prepare a 5.00 L solution of 0.1 M NaOH?
Solution Steps:
1. Use the equation M1V1 = M2V2.
2. Solve for V1 (volume of stock solution required):
  - Rearrange to V1 = (M2 * V2) / M1.
3. Plug in values:
  - M2 = 0.1 M
  - V2 = 5.00 L
  - M1 = 1.59 M
  - Calculate V1 = (0.1 * 5) / 1.59 = 0.314 L
Conclusion: The volume of stock solution needed is 0.314 L, which is much smaller than the final volume needed, indicating correct dilution.

Key Takeaway: The dilution formula M1V1 = M2V2 is crucial for solving these types of problems in chemistry.
Additional Resource: For more detailed explanations, refer to the tutorial on solutions and dilution for further guidance.