Table of Contents
Initial volume (V₁) = 100 mL
Final concentration (C₂) = 0.5 M
Final volume (V₂) = ?
Applying the formula:
2 M × 100 mL = 0.5 M × V₂
V₂ = (2 M × 100 mL) / 0.5 M = 400 mL
Since the initial solution was 100 mL, the amount of water to add is:
400 mL – 100 mL = 300 mL
Practice Problem 2: Preparing a Diluted Solution
Problem: How much of a 5 M stock solution of potassium permanganate (KMnO₄) is needed to prepare 250 mL of a 1 M solution?
Solution: Using C₁V₁ = C₂V₂:
Initial concentration (C₁) = 5 M
Final concentration (C₂) = 1 M
Final volume (V₂) = 250 mL
Calculating V₁:
V₁ = (C₂ × V₂) / C₁ = (1 M × 250 mL) / 5 M = 50 mL
Therefore, 50 mL of the 5 M stock solution is needed, and the remaining volume (200 mL) should be made up with water.
Additional Practice Tips
When solving dilution problems:
- Identify the known quantities: initial concentration and volume, and the desired concentration or volume.
- Use the dilution formula carefully, ensuring units are consistent.
- Double-check your calculations to avoid common mistakes like incorrect unit conversions.
Practice with various problems to become comfortable with different scenarios involving dilutions and concentrations. This will improve your problem-solving skills and deepen your understanding of solution chemistry.
Understanding how to solve word problems involving dilutions and concentrations is essential for students studying chemistry, pharmacy, and related fields. These problems often involve calculating the amount of a substance needed to prepare a solution of a desired concentration or determining the concentration of a solution after dilution.
Key Concepts in Dilutions and Concentrations
Before tackling practice problems, it is important to understand some fundamental concepts:
- Concentration: The amount of solute in a given volume of solution, often expressed as molarity (M), percentage, or parts per million (ppm).
- Dilution: The process of reducing the concentration of a solute in a solution, usually by adding solvent.
- Dilution formula: C₁V₁ = C₂V₂, where C and V represent concentration and volume, respectively.
Sample Word Problem 1: Diluting a Solution
Problem: A scientist has 100 mL of a 2 M solution of sodium chloride (NaCl). How much water should be added to dilute the solution to a concentration of 0.5 M?
Solution: Using the dilution formula C₁V₁ = C₂V₂:
Initial concentration (C₁) = 2 M
Initial volume (V₁) = 100 mL
Final concentration (C₂) = 0.5 M
Final volume (V₂) = ?
Applying the formula:
2 M × 100 mL = 0.5 M × V₂
V₂ = (2 M × 100 mL) / 0.5 M = 400 mL
Since the initial solution was 100 mL, the amount of water to add is:
400 mL – 100 mL = 300 mL
Practice Problem 2: Preparing a Diluted Solution
Problem: How much of a 5 M stock solution of potassium permanganate (KMnO₄) is needed to prepare 250 mL of a 1 M solution?
Solution: Using C₁V₁ = C₂V₂:
Initial concentration (C₁) = 5 M
Final concentration (C₂) = 1 M
Final volume (V₂) = 250 mL
Calculating V₁:
V₁ = (C₂ × V₂) / C₁ = (1 M × 250 mL) / 5 M = 50 mL
Therefore, 50 mL of the 5 M stock solution is needed, and the remaining volume (200 mL) should be made up with water.
Additional Practice Tips
When solving dilution problems:
- Identify the known quantities: initial concentration and volume, and the desired concentration or volume.
- Use the dilution formula carefully, ensuring units are consistent.
- Double-check your calculations to avoid common mistakes like incorrect unit conversions.
Practice with various problems to become comfortable with different scenarios involving dilutions and concentrations. This will improve your problem-solving skills and deepen your understanding of solution chemistry.