Solution Preparation Formula
Calculate mass and volume requirements for preparing solutions of precise concentration
Key Formulas
Molarity (M)
M = n / V
M = moles solute / liters solution
Units: mol/L (M)
Mass from Molarity
mass = M × V × MM
M (mol/L) × V (L) × MM (g/mol) = mass (g)
Dilution Formula
M₁V₁ = M₂V₂
Initial concentration × initial volume = final concentration × final volume
Preparation Procedures
From Solid Solute
- Calculate mass: mass = M × V × MM
- Weigh solute: Use analytical balance
- Dissolve: Add some solvent, stir until dissolved
- Transfer: Pour into volumetric flask
- Rinse: Rinse weighing container, add to flask
- Fill to mark: Add solvent until meniscus at calibration line
- Mix: Invert flask several times
By Dilution (from Stock Solution)
- Calculate volume: V₁ = M₂V₂ / M₁
- Measure stock: Use pipette or graduated cylinder
- Transfer: Pour into volumetric flask
- Add solvent: Fill to mark (not before!)
- Mix thoroughly: Invert multiple times
From Concentrated Acid
⚠️ SAFETY: Always add acid to water, never water to acid!
- Calculate volume of concentrated acid needed
- Add ~50% of final volume of water to flask
- Slowly add concentrated acid while stirring
- Allow to cool (dilution is exothermic!)
- Fill to mark with water
- Mix thoroughly
Worked Examples
Example 1: Preparing from Solid
Problem: Prepare 500 mL of 0.100 M NaCl solution
Solution:
Given:
M = 0.100 M
V = 500 mL = 0.500 L
MM(NaCl) = 58.44 g/mol
Calculate mass:
mass = M × V × MM
mass = 0.100 × 0.500 × 58.44
mass = 2.92 g
Weigh 2.92 g NaCl, dissolve in water, dilute to 500 mL
Example 2: Dilution
Problem: Prepare 250 mL of 0.50 M HCl from 6.0 M stock
Solution:
Use M₁V₁ = M₂V₂
M₁ = 6.0 M (stock)
M₂ = 0.50 M (desired)
V₂ = 250 mL (final volume)
V₁ = M₂V₂ / M₁
V₁ = (0.50)(250) / 6.0
V₁ = 20.8 mL
Measure 20.8 mL of 6.0 M HCl, dilute to 250 mL
(Add acid to ~200 mL water, then dilute to 250 mL)
Example 3: Hydrated Salt
Problem: Prepare 100 mL of 0.200 M CuSO₄ from CuSO₄·5H₂O
Solution:
MM(CuSO₄·5H₂O) = 249.68 g/mol
M = 0.200 M
V = 0.100 L
mass = M × V × MM
mass = 0.200 × 0.100 × 249.68
mass = 5.00 g
Use 5.00 g CuSO₄·5H₂O (not anhydrous CuSO₄!)
Example 4: Serial Dilution
Problem: Prepare 1.0 mM solution from 1.0 M stock
Solution:
Direct dilution: 1000× dilution needed
For 100 mL: need 0.1 mL stock (too small to measure accurately!)
Better approach - two steps:
Step 1: 1.0 M → 0.010 M (100× dilution)
1.0 mL stock + 99 mL water = 100 mL of 0.010 M
Step 2: 0.010 M → 0.001 M (10× dilution)
10 mL of 0.010 M + 90 mL water = 100 mL of 1.0 mM
Serial dilution gives better accuracy for large dilutions
Common Mistakes
Adding Solvent to Solute to Reach Volume
WRONG! Final volume includes solute volume. Must dilute TO the mark.
Using mL in Molarity Formula
M = mol/L requires liters! Convert mL to L first (÷1000)
Ignoring Water of Hydration
Use formula mass of hydrated form (e.g., CuSO₄·5H₂O not CuSO₄)
Adding Water to Concentrated Acid
DANGEROUS! Always add acid to water slowly. "Do like you oughta, add acid to water"
Use Volumetric Glassware
For accurate solutions, use volumetric flasks and pipettes, not beakers or Erlenmeyer flasks