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Complete Guide For GCSE/A-Level Chemistry (Edexcel, AQA, OCR, IB)
"Mastering titrations unlocks 10–15% of your GCSE/A-Level Chemistry exam—think buffer calculations, drug purity tests, and even how your stomach neutralises acid. One wrong step? You lose 4–6 marks. Today, we’ll turn this into a 100% predictable process."
If volume is in cm³, divide by 1000 first.
Dilution: c₁V₁ = c₂V₂
Used when diluting solutions (MEMORISE THIS).
% Purity = (mass of pure substance / mass of impure sample) × 100
Used in back titrations (MEMORISE THIS).
pH = -log[H⁺]
Goal: Find unknown concentration of acid/base.
Goal: Find % purity or concentration when direct titration is impossible (e.g., insoluble sample).
Question: 25.0 cm³ of NaOH neutralises 20.0 cm³ of 0.100 mol/dm³ HCl. Find [NaOH].
Steps:1. Balanced equation: HCl + NaOH → NaCl + H₂O (1:1 ratio).2. Moles HCl = 0.100 × (20.0/1000) = 0.00200 mol.3. Moles NaOH = 0.00200 mol (1:1 ratio).4. [NaOH] = 0.00200 / (25.0/1000) = 0.0800 mol/dm³.
What we did and why: - Used n = c × V to find moles of HCl. - 1:1 ratio → moles NaOH = moles HCl. - Rearranged c = n / V to find [NaOH].
Question: 2.50 g of impure CaCO₃ is reacted with 50.0 cm³ of 1.00 mol/dm³ HCl. Excess HCl requires 25.0 cm³ of 0.500 mol/dm³ NaOH for neutralisation. Find % purity of CaCO₃.
Steps:1. Equations: - CaCO₃ + 2HCl → CaCl₂ + H₂O + CO₂ - HCl + NaOH → NaCl + H₂O2. Moles NaOH = 0.500 × (25.0/1000) = 0.0125 mol.3. Moles excess HCl = 0.0125 mol (1:1 ratio).4. Initial moles HCl = 1.00 × (50.0/1000) = 0.0500 mol.5. Moles HCl reacted with CaCO₃ = 0.0500 – 0.0125 = 0.0375 mol.6. Moles CaCO₃ = 0.0375 / 2 = 0.01875 mol (1:2 ratio).7. Mass CaCO₃ = 0.01875 × 100.1 = 1.877 g.8. % purity = (1.877 / 2.50) × 100 = 75.1%.
What we did and why: - Used back titration to find moles of HCl that reacted with CaCO₃. - 1:2 ratio → moles CaCO₃ = ½ moles HCl. - Calculated mass and % purity.
Question: A student titrates 0.100 mol/dm³ CH₃COOH with 0.100 mol/dm³ NaOH. Sketch the pH curve and choose the best indicator.
Steps:1. pH curve features: - Starts at pH ~3 (weak acid). - Equivalence point at pH ~8.5 (weak acid + strong base). - Gradual rise, then sharp jump at equivalence point.2. Indicator choice: - Phenolphthalein (pH 8.3–10.0) overlaps equivalence point. - Methyl orange (pH 3.1–4.4) is wrong (too low).
What we did and why: - Weak acid + strong base → equivalence point > pH 7. - Phenolphthalein matches the pH jump.
MISTAKE: Using cm³ directly in n = c × V. WHY IT HAPPENS: Forgetting to convert to dm³. CORRECT APPROACH: Divide cm³ by 1000 first.
MISTAKE: Ignoring mole ratios. WHY IT HAPPENS: Assuming 1:1 ratio for all reactions. CORRECT APPROACH: Write balanced equation first.
MISTAKE: Choosing wrong indicator. WHY IT HAPPENS: Not checking equivalence point pH. CORRECT APPROACH: Match indicator range to equivalence point.
MISTAKE: Misreading burette. WHY IT HAPPENS: Reading from top instead of bottom of meniscus. CORRECT APPROACH: Always read at eye level, bottom of meniscus.
MISTAKE: Not rinsing pipette/burette. WHY IT HAPPENS: Contamination from previous solutions. CORRECT APPROACH: Rinse with solution to be used.
TRAP: Back titration with two different mole ratios. HOW TO SPOT IT: Two equations given (e.g., 1:1 and 1:2). HOW TO AVOID IT: Label each step clearly.
TRAP: Dilution before titration. HOW TO SPOT IT: "25 cm³ of solution was diluted to 250 cm³". HOW TO AVOID IT: Use c₁V₁ = c₂V₂ to find new concentration.
TRAP: pH curve with two equivalence points. HOW TO SPOT IT: Diprotic acid (e.g., H₂SO₄) or carbonate (CO₃²⁻). HOW TO AVOID IT: Identify all possible reactions.
"Here’s the night-before cheat sheet:1. Direct titration: Write equation → find moles of known → use ratio → find unknown concentration.2. Back titration: Two equations → find leftover moles → subtract → use ratio → find purity.3. Indicators: Strong acid + strong base = pH 7 (any indicator). Weak acid + strong base = pH > 7 (phenolphthalein). Strong acid + weak base = pH < 7 (methyl orange).4. pH curves: Sketch the jump—sharp for strong, gradual for weak.5. Units: Always dm³ for volume in n = c × V. Now go ace that exam!"
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