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Flashcards · 9620 · A Level Topic 10
Acids and bases (A Level) flashcards
28 cards on A Level Topic 10, subtopics 3.1.10.1 to 3.1.10.6. Pick a subtopic or study them all.
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3.1.10.1 Brønsted–Lowry acid–base equilibria in aqueous solution
- Brønsted–Lowry acid and base
- Acid: proton donor. Base: proton acceptor.
- Acid–base equilibrium
- One in which protons (H⁺) are transferred. Check both directions: the species that loses H⁺ in the reverse reaction is also an acid.
- In HCO₃⁻ + OH⁻ → CO₃²⁻ + H₂O, which species is the Brønsted–Lowry acid?
- HCO₃⁻: it donates a proton to OH⁻.
- Strong vs concentrated (acid)
- “Strong” describes the extent of dissociation (complete); “concentrated” describes how many moles are in each dm³.
3.1.10.2 Definition and determination of pH
- Definition of pH
- pH = −log₁₀[H⁺], so [H⁺] = 10^(−pH).
- pH of a strong monoprotic acid
- [H⁺] = concentration of the acid; pH = −log₁₀[H⁺]. Example: 0.0072 mol dm⁻³ HI has pH = −log₁₀(0.0072) = 2.14.
- Calculate [H⁺] in a solution of pH 10.46
- [H⁺] = 10^(−10.46) = 3.47 × 10⁻¹¹ mol dm⁻³
- Method: pH after mixing a strong acid with a strong base
- Find n(H⁺) and n(OH⁻), subtract the smaller from the larger, divide the excess by the combined volume. Excess H⁺ gives pH directly; excess OH⁻ needs Kw.
3.1.10.3 The ionic product of water, Kw
- Ionic product of water, Kw
- Kw = [H⁺][OH⁻] = 1.00 × 10⁻¹⁴ mol² dm⁻⁶ at 25 °C. It comes from Kc for H₂O ⇌ H⁺ + OH⁻, with [H₂O] effectively constant.
- pH of a strong base
- [OH⁻] = concentration × OH⁻ per formula unit; [H⁺] = Kw ÷ [OH⁻]; pH = −log₁₀[H⁺]. Example: 0.0028 mol dm⁻³ NaOH → [H⁺] = 3.57 × 10⁻¹² → pH 11.45.
- Neutral solution
- [H⁺] = [OH⁻]. This is pH 7 only at 25 °C.
- What happens to Kw and the pH of pure water as temperature rises?
- Water’s dissociation is endothermic, so Kw increases and the pH falls below 7. The water is still neutral because [H⁺] = [OH⁻].
3.1.10.4 Weak acids and bases; Ka for weak acids
- Weak acid or base
- Dissociates only slightly in water.
- Ka expression and pKa
- Ka = [H⁺][A⁻] ÷ [HA] (mol dm⁻³). pKa = −log₁₀ Ka. Larger Ka = stronger acid; larger pKa = weaker acid.
- pH of a weak acid from Ka
- Assume [H⁺] = [A⁻] and [HA] ≈ starting concentration, so [H⁺] = √(Ka × [HA]). Example: Ka = 2.0 × 10⁻⁶, [HA] = 0.500 → [H⁺] = 1.0 × 10⁻³, pH 3.00.
- Ka from the pH of a weak acid solution
- [H⁺] = 10^(−pH), then Ka = [H⁺]² ÷ [HA].
3.1.10.5 pH curves, titrations and indicators
- Indicator colour-change ranges: methyl orange and phenolphthalein
- Methyl orange: 3.1–4.4 (red to yellow). Phenolphthalein: 8.3–10.0 (colourless to pink).
- When is an indicator suitable for a titration?
- When its entire colour-change range lies within the vertical (steep) section of the pH curve.
- Indicator for weak acid–strong base and for strong acid–weak base titrations
- Weak acid–strong base (vertical about 7–11): phenolphthalein. Strong acid–weak base (vertical about 3–7): methyl orange. Weak–weak: no sharp change; use a pH meter.
- How can you find pKa from a weak acid–strong base pH curve?
- Read the pH at half the equivalence volume. There [HA] = [A⁻], so pH = pKa.
- End point vs equivalence point
- The end point is where the indicator changes; the equivalence point is where reacting amounts match.
- Required practical 7: key steps for a pH curve
- Calibrate the pH meter with buffers. Add base from a burette to the acid, stir, record pH and total volume, using smaller additions near the end point. Plot pH against volume.
3.1.10.6 Buffer action
- Buffer
- A solution whose pH stays nearly constant when it is diluted or when a small amount of acid or base is added.
- What does an acidic buffer contain? A basic buffer?
- Acidic: a weak acid and its salt (or a weak acid partly neutralised by NaOH). Basic: a weak base and its salt, e.g. methylamine with methylammonium chloride.
- How does an acidic buffer resist added acid and added base?
- Added H⁺ reacts with A⁻ to form HA (HA ⇌ H⁺ + A⁻ shifts left). Added OH⁻ removes H⁺, and HA dissociates to replace it (shifts right). Large amounts of HA and A⁻ keep [HA]/[A⁻] almost constant.
- [H⁺] in an acidic buffer
- [H⁺] = Ka × [HA] ÷ [A⁻] (or the ratio of moles). Do not use √(Ka[HA]) for a buffer.
- Buffer: 0.024 mol HA and 0.016 mol A⁻, Ka = 3.0 × 10⁻⁵ mol dm⁻³. pH?
- [H⁺] = 3.0 × 10⁻⁵ × 0.024 ÷ 0.016 = 4.5 × 10⁻⁵ mol dm⁻³, so pH = 4.35.
- Applications of buffers
- Calibrating pH meters, controlling the pH of blood, controlling acidity in foods and drinks, and any process (such as with enzymes) that needs a steady pH.