Edexcel International A Level Chemistry: Acid-base equilibria: titration curves and buffers (YCH11) – Revision Notes

These notes condense outcomes 14.15 to 14.23 of Topic 14: Acid-base Equilibria in the Pearson Edexcel International Advanced Subsidiary/Advanced Level in Chemistry specification (YCH11), Issue 1, September 2017, including Core Practical 11. This content sits in Unit 4: Rates, Equilibria and Further Organic Chemistry, which is IA2. If a point here is unclear, go back to the titration curves and buffers study guide, which explains each idea with worked examples.

Once you can recall all of this, try the practice questions. Other topics are on the course hub, you can tick off outcomes on the Edexcel Chemistry checklist, and a free diagnostic shows where to focus.

Key definitions

• Equivalence point: the volume at which acid and base have reacted exactly in the ratio of the equation.
• End point: the volume at which the indicator changes colour. A good indicator makes the end point match the equivalence point.
• Buffer solution (14.17): a solution whose pH changes very little when a small amount of acid or alkali is added to it.
• Acidic buffer: weak acid + its conjugate base (salt), e.g. propanoic acid + sodium propanoate.
• Basic buffer: weak base + its conjugate acid (salt), e.g. NH₃ + NH₄Cl.
• Half-neutralisation point: half the equivalence volume; here [HA] = [A⁻] and pH = pKa.

Titration curves at a glance (14.15)

Base added to 25 cm³ of acid of similar concentration:

Pair	Start pH	Equivalence pH	Steep section
Strong acid, strong base	~1	7	~3 to ~11
Weak acid, strong base	~3	>7	~7 to ~11
Strong acid, weak base	~1	<7	~3 to ~7
Weak acid, weak base	~3	~7	none sharp

Why the equivalence pH moves:

• Weak acid + strong base gives a salt whose anion (A⁻) accepts protons from water, so the solution is alkaline.
• Strong acid + weak base gives a salt whose cation (e.g. NH₄⁺) donates protons, so the solution is acidic.

Acid added to base: same shapes, reflected; the pH starts high and falls.

Sketch checklist: label both axes (pH; volume added / cm³), mark the equivalence volume, the start pH, the half-way pH (weak species only) and a final pH that levels off near the burette solution’s pH.

Diprotic acids and bases

• Diprotic weak acid H₂X with NaOH: two steep sections if pKa₁ and pKa₂ are well apart. Second equivalence volume = 2 × first.
• pH at half the first equivalence volume = pKa₁; pH midway between the two equivalence volumes = pKa₂.
• Sulfuric acid with NaOH: one steep section; twice the moles of NaOH needed compared with HCl.
• Sodium carbonate with HCl (diprotic base): CO₃²⁻ + H⁺ → HCO₃⁻, then HCO₃⁻ + H⁺ → H₂O + CO₂. Two steep sections, the second at twice the first volume.

Indicators (14.16)

Rule: the indicator’s whole colour-change range must lie inside the steep section of the curve.

Titration	Indicator
Strong acid, strong base	almost any (methyl orange, phenolphthalein, bromothymol blue)
Weak acid, strong base	phenolphthalein (range about 8.2 to 10.0)
Strong acid, weak base	methyl orange (range about 3.1 to 4.4)
Weak acid, weak base	none: use a pH meter
Na₂CO₃ with HCl	phenolphthalein first, then methyl orange

Colours: methyl orange red in acid, yellow in alkali; phenolphthalein colourless in acid, pink in alkali. In an exam, use the ranges you are given.

How a buffer works (14.18)

Acidic buffer equilibrium: HA ⇌ H⁺ + A⁻. The solution holds big reservoirs of HA and of A⁻.

Added	What removes it	Equation
H⁺	conjugate base A⁻	H⁺ + A⁻ → HA
OH⁻	weak acid HA	OH⁻ + HA → A⁻ + H₂O

For NH₃/NH₄⁺: H⁺ + NH₃ → NH₄⁺ and OH⁻ + NH₄⁺ → NH₃ + H₂O.

Key phrase for full answers: the ratio [HA]/[A⁻] changes only slightly because both are present in large amounts compared with the acid or alkali added, so [H⁺] changes only slightly.

Formulas

Use	Formula
Buffer pH (14.19)	[H⁺] = Ka × [HA]/[A⁻]
Same, log form	pH = pKa − log([HA]/[A⁻])
Buffer recipe (14.20)	[A⁻] = Ka × [HA]/[H⁺]
Half-neutralisation (14.21)	pH = pKa, so Ka = 10^-pH
Start of a weak acid curve	[H⁺] = √(Ka × [HA])
Excess strong base	[H⁺] = Kw/[OH⁻], Kw = 1.00 × 10⁻¹⁴ mol² dm⁻⁶ at 298 K

Assumptions for buffer pH: the weak acid’s own dissociation is negligible, and all A⁻ comes from the salt (or from neutralisation).

Method boxes

Buffer pH by partial neutralisation

1. Moles of weak acid at the start.
2. Moles of OH⁻ added.
3. HA left = step 1 − step 2; A⁻ formed = step 2.
4. [H⁺] = Ka × (HA left)/(A⁻ formed). Moles are fine: the volume cancels.
5. pH = −log[H⁺].

New buffer pH after adding a little strong acid: add the moles of H⁺ to HA and subtract them from A⁻, then repeat step 4. For strong alkali, do the reverse.

Making a buffer of a chosen pH

1. [H⁺] = 10^-pH.
2. [A⁻] = Ka × [HA]/[H⁺].
3. Moles of salt = [A⁻] × volume in dm³; mass = moles × Mr.
4. State the assumption that the volume does not change when the solid dissolves.

Ka from a titration curve

1. Find the equivalence volume at the middle of the steep section.
2. Halve it.
3. Read the pH at that volume: this is pKa. Ka = 10^-pKa.

Buffers in living things and food (14.22)

• Blood pH stays close to 7.4. Main buffer: H₂CO₃ ⇌ H⁺ + HCO₃⁻.
• Extra acid from cells (e.g. lactic acid): H⁺ + HCO₃⁻ → H₂CO₃, removed as CO₂ through the lungs.
• Cells: phosphate and protein buffers keep enzymes working in their narrow pH range.
• Foods: bacteria and fungi can change pH as they grow; a buffer resists that change and slows deterioration.

Core Practical 11: Ka for a weak acid (14.23)

• Calibrate the pH meter with buffers of known pH.
• Titrate a portion of the weak acid with NaOH (phenolphthalein) to find the end-point volume.
• Make a half-neutralised solution: a second identical portion of acid plus half that volume of NaOH (or add an equal portion of acid to the neutralised flask).
• pH of this solution = pKa. No concentrations are needed.

Must-know distinctions

• Equivalence point vs end point: one is chemistry, the other is the indicator.
• Equivalence point vs half-neutralisation point: pH = pKa only at the half-way volume.
• Buffer region vs steep section: flat part = buffering; steep part = buffer used up.
• Diluting a buffer vs adding acid: dilution leaves the [HA]/[A⁻] ratio and pH almost unchanged; adding acid shifts the ratio.

Quick self-test

1. Define a buffer solution.
2. Methyl orange changes over pH 3.1 to 4.4 and phenolphthalein over 8.2 to 10.0. Which suits hydrochloric acid titrated against aqueous ammonia?
3. A buffer is 0.30 mol dm⁻³ HA and 0.20 mol dm⁻³ NaA. Ka = 6.0 × 10⁻⁵ mol dm⁻³. Find the pH.
4. A weak acid’s curve has its equivalence point at 26.0 cm³; at 13.0 cm³ the pH is 4.40. Find Ka.
5. What ratio [A⁻]/[HA] gives a pH 0.30 above the pKa?
6. Why do titres for a weak acid against a weak base vary so much when an indicator is used?
7. In a sodium carbonate titration with HCl, the first end point is at 9.0 cm³. Where is the second?
8. Write the equation for how blood removes extra H⁺.
9. A buffer is diluted to twice its volume with water. What happens to its pH?
10. A solution contains 0.0200 mol HA (pKa = 4.35) and 0.0050 mol NaOH. Find the pH.
11. A diprotic acid’s curve has equivalence points at 11.0 and 22.0 cm³. At which volumes does the pH equal pKa₁ and pKa₂?

Answers

1. A solution whose pH changes very little when a small amount of acid or alkali is added.
2. Methyl orange: strong acid with weak base has its steep section below pH 7.
3. [H⁺] = 6.0 × 10⁻⁵ × 0.30/0.20 = 9.0 × 10⁻⁵; pH = 4.05.
4. pKa = 4.40; Ka = 10^-4.40 = 4.0 × 10⁻⁵ mol dm⁻³.
5. log([A⁻]/[HA]) = 0.30, so the ratio is 2.0.
6. The pH rises gradually near equivalence with no steep section, so the colour change is spread over several cm³; use a pH meter instead.
7. 18.0 cm³ (twice the first).
8. H⁺ + HCO₃⁻ → H₂CO₃.
9. Almost no change: both concentrations fall by the same factor, so the ratio is unchanged.
10. HA = 0.0150 mol, A⁻ = 0.0050 mol; pH = 4.35 − log 3.0 = 3.87.
11. pKa₁ at 5.5 cm³; pKa₂ at 16.5 cm³.

Where marks are usually lost

• Putting every equivalence point at pH 7.
• Choosing an indicator because its range touches the steep section, rather than lying wholly inside it.
• Using starting moles in the buffer ratio instead of moles left after neutralisation.
• Inverting the ratio, writing Ka × [A⁻]/[HA].
• Reading the pH at the equivalence volume and calling it pKa.
• Explaining buffer action without an equation, or without saying that HA and A⁻ are present in large amounts.
• Forgetting that the second end point for a diprotic species is at twice the first volume.
• Describing blood buffering without naming HCO₃⁻ and H₂CO₃.

Official syllabus

Pearson Edexcel International Advanced Subsidiary/Advanced Level in Chemistry (XCH11/YCH11) specification, Issue 1, September 2017, published by Pearson Education Limited. Topic 14: Acid-base Equilibria, outcomes 14.15 to 14.23, including Core Practical 11.