Calculate solution dilutions using the M₁V₁ = M₂V₂ equation
Unit: mol/L (M)
Unit: mL or L (consistent)
Unit: mol/L (M)
Unit: mL or L (match V₁)
Dilution Principle: M₁V₁ = M₂V₂ (moles of solute remain constant)
Dilution reduces the concentration of a solution by adding more solvent. The amount of solute remains constant.
M₁V₁ = M₂V₂
1:10 Dilution
1 part stock + 9 parts solvent = 10 parts total
10x Dilution
Final volume is 10× the initial volume
10-fold Dilution
Concentration reduced by factor of 10
Problem:
You have a 2.0 M stock solution of NaCl. How would you prepare 500 mL of 0.5 M NaCl solution?
Solution:
Given: M₁ = 2.0 M, M₂ = 0.5 M, V₂ = 500 mL
Find: V₁ (volume of stock needed)
M₁V₁ = M₂V₂
(2.0 M)(V₁) = (0.5 M)(500 mL)
V₁ = (0.5 × 500) / 2.0 = 125 mL
Preparation:
1. Measure 125 mL of 2.0 M stock solution
2. Add water to bring total volume to 500 mL
3. Mix thoroughly
Problem:
Prepare a 1:1000 dilution of a bacterial culture using serial 1:10 dilutions.
Solution:
Three consecutive 1:10 dilutions:
1st: 1 mL culture + 9 mL media = 1:10
2nd: 1 mL from 1st + 9 mL media = 1:100
3rd: 1 mL from 2nd + 9 mL media = 1:1000
Total dilution = 10 × 10 × 10 = 1000
Advantage: Serial dilutions avoid handling very small volumes that are difficult to measure accurately.
| Dilution Notation | Dilution Factor | Example (1 M stock) | How to Prepare |
|---|---|---|---|
| 1:2 | 2× | 0.5 M | 1 part stock + 1 part solvent |
| 1:5 | 5× | 0.2 M | 1 part stock + 4 parts solvent |
| 1:10 | 10× | 0.1 M | 1 part stock + 9 parts solvent |
| 1:100 | 100× | 0.01 M | 1 part stock + 99 parts solvent |
| 1:1000 | 1000× | 0.001 M | 1 part stock + 999 parts solvent |