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Solution:
Using the formula: C1 × V1 = C2 × V2
3 M × V1 = 1.5 M × 100 mL
V1 = (1.5 M × 100 mL) / 3 M = 50 mL
Mix 50 mL of the 3 M stock solution with 50 mL of water to get 100 mL of 1.5 M solution.
Scenario 3: Serial Dilutions
Serial dilutions involve multiple steps to reach a very dilute concentration. Suppose you start with 10 mL of a 10 M solution and perform two serial dilutions, each by a factor of 10. What is the final concentration?
Solution:
After the first dilution:
V1 = 1 mL of 10 M solution + 9 mL water = 10 mL total
Concentration after first dilution: 10 M × (1 mL / 10 mL) = 1 M
Second dilution:
V1 = 1 mL of 1 M solution + 9 mL water = 10 mL total
Final concentration: 1 M × (1 mL / 10 mL) = 0.1 M
Practice Exercises
- Dilute 100 mL of a 5 M solution to 1 M. How much water do you add?
- You have a 0.2 M solution and need 250 mL of a 0.05 M solution. How much stock solution is required?
- Perform a serial dilution starting with 5 mL of a 20 M solution, each step diluting by a factor of 10. What is the concentration after three steps?
Summary
Practicing various dilution scenarios enhances understanding and confidence. Remember to always identify your initial concentration and volume, desired final concentration and volume, and apply the formula carefully. Serial dilutions are useful for creating very dilute solutions, especially in laboratory experiments.
Mastering the concept of dilution is essential for students and professionals working in fields like chemistry, biology, and medicine. Practicing with multiple scenarios helps solidify understanding and improves problem-solving skills. This article provides various dilution scenarios along with detailed solutions to aid in mastering this important skill.
Understanding Dilution
Dilution involves reducing the concentration of a solute in a solution, usually by adding more solvent. The key formula used is:
C1 × V1 = C2 × V2
Scenario 1: Basic Dilution
Suppose you have 50 mL of a 2 M (molar) solution. How much water must you add to dilute it to a 0.5 M solution?
Solution:
Using the formula: C1 × V1 = C2 × V2
2 M × 50 mL = 0.5 M × V2
V2 = (2 M × 50 mL) / 0.5 M = 200 mL
Therefore, total volume after dilution should be 200 mL. Since you started with 50 mL, add 150 mL of water.
Scenario 2: Diluting to a Desired Concentration
You need 100 mL of a 1.5 M solution. You have a stock solution of 3 M. How much of the stock solution do you need?
Solution:
Using the formula: C1 × V1 = C2 × V2
3 M × V1 = 1.5 M × 100 mL
V1 = (1.5 M × 100 mL) / 3 M = 50 mL
Mix 50 mL of the 3 M stock solution with 50 mL of water to get 100 mL of 1.5 M solution.
Scenario 3: Serial Dilutions
Serial dilutions involve multiple steps to reach a very dilute concentration. Suppose you start with 10 mL of a 10 M solution and perform two serial dilutions, each by a factor of 10. What is the final concentration?
Solution:
After the first dilution:
V1 = 1 mL of 10 M solution + 9 mL water = 10 mL total
Concentration after first dilution: 10 M × (1 mL / 10 mL) = 1 M
Second dilution:
V1 = 1 mL of 1 M solution + 9 mL water = 10 mL total
Final concentration: 1 M × (1 mL / 10 mL) = 0.1 M
Practice Exercises
- Dilute 100 mL of a 5 M solution to 1 M. How much water do you add?
- You have a 0.2 M solution and need 250 mL of a 0.05 M solution. How much stock solution is required?
- Perform a serial dilution starting with 5 mL of a 20 M solution, each step diluting by a factor of 10. What is the concentration after three steps?
Summary
Practicing various dilution scenarios enhances understanding and confidence. Remember to always identify your initial concentration and volume, desired final concentration and volume, and apply the formula carefully. Serial dilutions are useful for creating very dilute solutions, especially in laboratory experiments.