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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.