Chemistry · 8. Acids, bases and salts

Titration calculations (as required)

Lesson 3

Titration calculations (as required)

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Why This Matters

# Titration Calculations Summary ## Key Learning Outcomes Students learn to perform quantitative calculations using titration data to determine unknown concentrations of acids and bases. The process involves using the formula n = cv (moles = concentration × volume in dm³) and applying mole ratios from balanced equations to calculate concentrations, typically in mol/dm³. Students must accurately convert volumes from cm³ to dm³ and work through multi-step calculations involving reacting volumes and stoichiometry. ## Exam Relevance Titration calculations are high-value questions in IGCSE Chemistry papers, often appearing as structured 6-8 mark questions requiring systematic working and unit awareness. These calculations integrate multiple topics including moles, concentration, and chemical equations, making them essential for achieving higher grades in both Core and Extended syllabuses.

Key Words to Know

01
Titration — A laboratory method used to find the unknown concentration of a solution by reacting it with a solution of known concentration.
02
Concentration — A measure of how much solute (the 'stuff' dissolved) is present in a given volume of solvent (the liquid it's dissolved in).
03
Endpoint (or Equivalence Point) — The point in a titration where the reaction between the two solutions is complete, usually indicated by a colour change.
04
Standard Solution — A solution whose concentration is accurately known and is used to determine the concentration of another solution.
05
Mole — A unit used in chemistry to count very large numbers of atoms or molecules, like a 'dozen' but much bigger.
06
Mole Ratio — The ratio of the moles of reactants and products in a balanced chemical equation, showing how they combine.
07
Volume — The amount of space a substance occupies, often measured in cm³ (millilitres) or dm³ (litres).
08
Molarity (mol/dm³) — A common unit for concentration, meaning moles of solute per cubic decimetre of solution.
09
Balanced Chemical Equation — A chemical equation where the number of atoms of each element is the same on both sides, showing the correct mole ratios.

Core Concepts & Theory

Titration is a quantitative technique used to determine the concentration of an unknown solution by reacting it with a solution of known concentration. The solution of known concentration is called the standard solution, while the solution being analyzed is the analyte.

Key Terms:

  • Titre: The volume of solution added from the burette during a titration
  • End point: The point at which the indicator changes color, signaling that the reaction is complete
  • Concordant results: Titres that agree within 0.10 cm³ of each other
  • Mean titre: The average of concordant titres only

Essential Equipment:

  • Burette: Measures variable volumes accurately to ±0.05 cm³
  • Pipette: Measures fixed volumes (e.g., 25.0 cm³) accurately
  • Conical flask: Contains the analyte solution
  • Indicator: Shows the end point (phenolphthalein or methyl orange)

The Titration Formula:

Mole relationship: n = c × V (where n = moles, c = concentration in mol/dm³, V = volume in dm³)

Conversion: Volume in cm³ ÷ 1000 = Volume in dm³

Calculation Steps:

  1. Write the balanced equation
  2. Calculate moles of known substance using n = c × V
  3. Use mole ratio from equation to find moles of unknown
  4. Calculate concentration: c = n ÷ V

Mnemonic: BURP

  • Balanced equation first
  • Use n = cV for known
  • Ratio from equation
  • Product calculation (unknown concentration)

Detailed Explanation with Real-World Examples

Why Titrations Matter:

Titrations are used extensively in real-world applications. Water treatment plants use titrations to monitor acidity levels in drinking water. Pharmaceutical companies verify medication concentrations to ensure patient safety. Winemakers test acidity in wine to maintain quality. Even swimming pool maintenance relies on titrations to check chlorine and pH levels.

The Coffee Shop Analogy:

Think of titration like making the perfect cup of coffee. You know exactly how much coffee grounds you need (your standard solution), but you're testing different water amounts (the analyte) to find the ideal strength. The "end point" is when you taste it and say "perfect!" The indicator is like your taste buds – it tells you when you've reached the right balance.

Understanding the Process:

During an acid-base titration, imagine hydrogen ions (H⁺) from the acid seeking out hydroxide ions (OH⁻) from the base. They combine to form water in a neutralization reaction. The indicator molecule is sensitive to pH changes – it exists in two different colored forms. At the end point, the last drop of titrant causes a sudden pH change, triggering the indicator to switch color dramatically.

Real Laboratory Application:

A quality control chemist at a vinegar factory needs to verify that bottles contain 5% acetic acid (ethanoic acid). By titrating a measured sample against standardized sodium hydroxide, they can calculate the exact concentration. If it's too low, customers complain about weak flavor; too high, and it becomes unsafe for consumption. This single titration protects both product quality and consumer safety.

Worked Examples & Step-by-Step Solutions

Example 1: Classic Acid-Base Titration

Question: 25.0 cm³ of sulfuric acid (H₂SO₄) required 23.50 cm³ of 0.100 mol/dm³ sodium hydroxide (NaOH) for complete neutralization. Calculate the concentration of the sulfuric acid.

Step 1: Write balanced equation H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O

Step 2: Calculate moles of NaOH (the known)

  • Volume in dm³ = 23.50 ÷ 1000 = 0.02350 dm³
  • Moles = c × V = 0.100 × 0.02350 = 0.002350 mol

Step 3: Use mole ratio (1:2)

  • Moles of H₂SO₄ = 0.002350 ÷ 2 = 0.001175 mol

Step 4: Calculate concentration

  • Volume H₂SO₄ = 25.0 ÷ 1000 = 0.0250 dm³
  • Concentration = n ÷ V = 0.001175 ÷ 0.0250 = 0.0470 mol/dm³

Examiner note: Always show the mole ratio clearly – this is worth at least 1 mark

Example 2: Mean Titre Calculation

Results: Rough = 24.80 cm³, Titre 1 = 24.15 cm³, Titre 2 = 24.20 cm³, Titre 3 = 24.10 cm³

Solution:

  • Identify concordant: 24.15, 24.20, 24.10 (all within 0.10 cm³)
  • Mean = (24.15 + 24.20 + 24.10) ÷ 3 = 24.15 cm³
  • Never include rough titre in the mean

Examiner note: State which values you're using – examiners award marks for working

Common Exam Mistakes & How to Avoid Them

Mistake 1: Forgetting Volume Conversion

Why it happens: Students rush and use cm³ instead of dm³ in the formula n...

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Cambridge Exam Technique & Mark Scheme Tips

Understanding Command Words:

  • Calculate: Show ALL working steps. Even if your final answer is wrong, you gain ...
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Exam Tips

  • 1.Always write down the balanced chemical equation first to get the correct mole ratio (n₁ and n₂).
  • 2.Convert all volumes from cm³ to dm³ by dividing by 1000 *before* you start calculations.
  • 3.Clearly label all your known values (C, V, n for both solutions) to avoid mixing them up.
  • 4.Show all your working steps clearly; even if your final answer is wrong, you might get marks for correct method.
  • 5.Double-check your final answer with correct units (e.g., mol/dm³ for concentration).
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