Quantitative Practical Skills: Calculations From Titration Data (Copy)

Quantitative Practical Skills

Calculations From Titration Data

Purpose of Titration Calculations in ATP

• Titration calculations are designed to test:
  • Numerical accuracy
  • Understanding of mole relationships
  • Ability to apply titration theory to real data
• Examiners look for:
  • Correct selection of concordant titres
  • Proper unit conversions
  • Correct mole calculations
  • Accurate final answer with appropriate significant figures

Core Examiner Rule

• Use mean of concordant titres for calculations
• Ignore rough titrations or non-concordant titres
• Use correct units: cm³ → dm³ conversion where required
• Show all working steps
• Final answer must match significant figures of apparatus

Step 1: Identifying Data

• Check titration table for:
  • Initial burette reading
  • Final burette reading
  • Titre = Final − Initial
• Select only concordant titres
• Calculate mean titre

Example Table

• Concordant titres: 23.55, 23.55, 23.55
• Mean titre = 23.55 cm³

Step 2: Converting cm³ to dm³

• Conversion: Volume in dm³ = Volume in cm³ ÷ 1000
• Example:
  23.55 cm³ ÷ 1000 = 0.02355 dm³

Step 3: Calculating Moles of Solution in Burette

• Formula: Moles = Concentration (mol/dm³) × Volume (dm³)
• Example: If 0.100 mol/dm³ HCl used:
  Moles HCl = 0.100 × 0.02355 = 0.002355 mol

Step 4: Using Stoichiometry

• Use balanced chemical equation to relate moles of titrant and analyte

Example reaction:
HCl + Na₂CO₃ → 2 NaCl + H₂O + CO₂

• Mole ratio: 2 HCl : 1 Na₂CO₃
• Moles of analyte:
  Moles Na₂CO₃ = Moles HCl ÷ 2 = 0.002355 ÷ 2 = 0.0011775 mol

Step 5: Calculating Concentration of Analyte

• Formula: C = Moles ÷ Volume (dm³)
• If solution volume = 25.0 cm³ = 0.0250 dm³:
  C = 0.0011775 ÷ 0.0250 = 0.0471 mol/dm³

Step 6: Calculating Mass or Percentage Purity

• Mass = Moles × Relative Formula Mass (RFM)
• Percentage purity = (Mass of pure substance ÷ Sample mass) × 100
• Example:
  Moles Na₂CO₃ = 0.0011775 mol
  RFM = 106 g/mol
  Mass of pure substance = 0.0011775 × 106 = 0.1248 g
  Sample mass = 0.200 g
  % Purity = 0.1248 ÷ 0.200 × 100 = 62.4%

Written and Compiled By Sir Hunain Zia, World Record Holder With 154 Total A Grades, 7 Distinctions and 11 World Records For Educate A Change O Level And IGCSE Chemistry Full Scale Course

Step 7: Significant Figures and Units

• ATP requires:
  • Consistent significant figures based on apparatus
  • Correct units for concentration, mass, volume
• Example:
  • Concentration in mol/dm³
  • Volume in cm³ or dm³
  • Mass in g

Common Examiner Traps

• Using rough titres instead of mean concordant titres
• Forgetting cm³ → dm³ conversion
• Using wrong mole ratio
• Ignoring significant figures
• Units missing or incorrect

ATP-Focused Calculation Checklist

1. Identify concordant titres
2. Calculate mean
3. Convert cm³ → dm³
4. Calculate moles of titrant
5. Apply mole ratio to find analyte moles
6. Calculate concentration, mass, or % purity
7. Check significant figures and units
8. Box final answer

Core Scientific Principle

• Titration calculations require:
  • Accuracy
  • Logic
  • Correct stoichiometry
• ATP rewards:
  • Methodical, stepwise working
  • Clear unit conversion
  • Correct handling of significant figures
• Mastery of titration calculations:
  • Guarantees high quantitative ATP marks