The shape of the lesson
By the end of the lesson, learners can
- How a buffer works: A buffer needs a large amount of both partners.
- Calculating buffer pH: Write K a for the weak acid, rearrange for [H + ] and use the concentrations of acid and salt directly.
- Titration curves and indicators: Run each acid and base pairing in the simulator.
- Added H+: A− + H+ → HA.
- Added OH−: HA + OH− → A− + H2O.
- pH: [H+] = Ka × [HA] ÷ [A−].
- Indicator: its range must lie in the steep section of the curve.
One picture every idea returns to
A buffer solution is a solution that resists changes in pH when small amounts of acid or alkali are added. An acidic buffer contains a weak acid and its conjugate base, for example CH3COOH with CH3COONa: added H+ is removed by CH3COO− + H+ → CH3COOH, and added OH− is removed by CH3COOH + OH− → CH3COO− + H2O. An alkaline buffer such as NH3 with NH4Cl works the same way using NH3 and NH4+. Because both reservoirs are large, their ratio hardly changes, so the pH, given by [H+] = Ka × [HA] ÷ [A−], hardly changes. Blood is buffered by H2CO3 / HCO3−.
Forty-five minutes, phase by phase
| Time | Phase | What happens in the room | Grouping |
|---|---|---|---|
| 0 to 5 min | Hook: predict | Open with the question in the title, Holding the pH steady.. A buffer keeps its pH almost constant when small amounts of acid or alkali are added. Put the figure on the board: it is the one picture every idea in the lesson returns to. | Whole class, then pairs |
| 5 to 13 min | Teach and check: How a buffer works | A buffer needs a large amount of both partners. Project tab 3 of the lab (Acid-base equilibria lab, buffers) and let a student drive it while the class predicts. | Whole class, then pairs |
| 13 to 21 min | Teach and check: Calculating buffer pH | Write K a for the weak acid, rearrange for [H + ] and use the concentrations of acid and salt directly. Project tab 3 of the lab (Acid-base equilibria lab, buffers) and let a student drive it while the class predicts. | Pairs |
| 21 to 30 min | Teach and check: Titration curves and indicators | Run each acid and base pairing in the simulator. Project tab 4 of the lab (Acid-base equilibria lab, titration curves) and let a student drive it while the class predicts. | Rally Coach |
| 30 to 36 min | Check | The four Stage 1 questions from the student lesson, answered on mini whiteboards or by show of fingers. Re-teach the one most get wrong. | Individual, then whole class |
| 36 to 43 min | Plenary | The exam-style questions from Stage 2, one at a time: thirty seconds alone, then a random call. The feedback lines in the key give the examiner's wording. | Individual, random call |
| 43 to 45 min | Exit and homework | Set the worksheet below (6 questions, 15 marks). Learners self-assess against the objectives before they leave. | Individual |
Protect the parts that carry the learning
The sequence carries about fifty minutes of material, so the lab is driven by one student while the class predicts rather than explored freely. Times are a guide.
Protected: the check at 30 minutes. It is the only individual read on whether the key idea landed, and it is not cut.
Cut first if the build overruns: the last check in Titration curves and indicators, never the check or the plenary.
In a 60 minute block: run worksheet questions 1 to 3 in class as a Rally Coach set and model both the hardest worked example and its check live.
Rally Coach
Pairs share one worksheet. One partner solves the first question aloud while the other coaches, checking each line and offering a prompt rather than the answer; they swap roles for the next question.
The full structure, with its variations and the other twelve, is on the cooperative-learning page.
What to head off, and how
| Trap learners fall into | Teaching move that pre-empts it |
|---|---|
| A buffer is not a solution whose pH never changes: it resists change from small additions, and fails once one reservoir is used up. | In buffer calculations use the salt concentration for [A − ], not [H + ]; the √(K a c) shortcut is for a weak acid on its own. When acid and alkali are mixed to make a buffer, work in moles after reaction before taking the ratio. |
Buffer solutions: worksheet
6 questions, 15 marks. Written in the style of the papers; no past paper question is reproduced.
Explain, with equations, how a mixture of ethanoic acid and sodium ethanoate keeps its pH almost constant when small amounts of acid or alkali are added.
4.10 g of sodium ethanoate (Mr 82.0) is dissolved in 500 cm3 of 0.200 mol dm−3 ethanoic acid. Calculate the pH of the buffer (Ka = 1.74 × 10−5 mol dm−3).
Calculate the ratio [CH3COO−] ÷ [CH3COOH] needed to make a buffer of pH 5.00 (Ka = 1.74 × 10−5 mol dm−3).
A buffer contains 0.200 mol dm−3 NH3 and 0.100 mol dm−3 NH4Cl. For NH4+, Ka = 5.62 × 10−10 mol dm−3. Calculate the pH.
Explain how the H2CO3 / HCO3− system controls the pH of blood.
Choose a suitable indicator for (a) ethanoic acid titrated with NaOH(aq) and (b) aqueous ammonia titrated with HCl(aq). Explain each choice.
Answer key and marking notes
- The mixture contains large amounts of CH3COOH and CH3COO− (from the fully ionised salt).
- Added H+: CH3COO− + H+ → CH3COOH. Added OH−: CH3COOH + OH− → CH3COO− + H2O.
- Both reservoirs are large, so the ratio [HA] ÷ [A−] and hence [H+] barely change.
- n(salt) = 4.10 ÷ 82.0 = 0.0500 mol, so [A−] = 0.0500 ÷ 0.500 = 0.100 mol dm−3.
- [H+] = 1.74 × 10−5 × 0.200 ÷ 0.100 = 3.48 × 10−5 mol dm−3.
- pH = 4.46.
- [H+] = 10−5.00 = 1.00 × 10−5 mol dm−3.
- [A−] ÷ [HA] = Ka ÷ [H+] = 1.74 × 10−5 ÷ 1.00 × 10−5 = 1.74.
- NH4+ is the acid and NH3 its conjugate base: [H+] = Ka × [NH4+] ÷ [NH3].
- [H+] = 5.62 × 10−10 × 0.100 ÷ 0.200 = 2.81 × 10−10 mol dm−3.
- pH = 9.55.
- H2CO3 ⇌ H+ + HCO3−; extra H+ is removed by HCO3− + H+ → H2CO3.
- Extra OH− is removed by H2CO3 + OH− → HCO3− + H2O, keeping blood near pH 7.4.
- (a) Phenolphthalein: the equivalence point is about pH 9 and the steep section lies in its range, 8.3 to 10.0.
- (b) Methyl orange: the equivalence point is about pH 5 and the steep section covers its range, 3.1 to 4.4.
8 multiple-choice questions on this lesson
From the topic question bank. Assign them in the app for a marked score in your gradebook, or run the topic as a class game.
Which statement best defines a buffer solution?
- Aa solution that resists changes in pH when small amounts of acid or alkali are added
- Ba solution whose pH stays exactly the same whatever amount of acid or alkali is added
- Ca neutral solution with a pH of 7.00
- Da solution that neutralises any acid added to it, giving pH 7
Which mixture, with both substances in similar large amounts, acts as a buffer solution?
- Aaqueous ammonia and ammonium chloride
- Bhydrochloric acid and sodium chloride
- Csodium hydroxide and sodium chloride
- Dsodium chloride and water
Blood is buffered by carbonic acid and hydrogencarbonate ions. Which equation shows how a small amount of added acid is removed?
- AHCO3− + H+ → H2CO3
- BH2CO3 + OH− → HCO3− + H2O
- CH2CO3 → H+ + HCO3−
- DCO2 + H2O → H2CO3
A buffer contains 0.100 mol dm−3 ethanoic acid and 0.100 mol dm−3 sodium ethanoate. Ka(ethanoic acid) = 1.74 × 10−5 mol dm−3. What is the pH?
- A4.76
- B2.88
- C7.00
- D9.24
The graph shows the titration curve for ethanoic acid with sodium hydroxide, with the colour-change ranges of two indicators shaded. Which indicator is suitable and why?
- Athe indicator with range B (phenolphthalein), because it changes colour within the steep part of the curve
- Bthe indicator with range A (methyl orange), because it changes colour at the start of the titration
- Ceither indicator, because all acid-alkali titrations end at pH 7
- Dneither indicator, because a weak acid cannot be titrated
Calculate the pH of a buffer containing 0.200 mol dm−3 ethanoic acid and 0.100 mol dm−3 sodium ethanoate. Ka = 1.74 × 10−5 mol dm−3.
- A4.46
- B5.06
- C4.76
- D2.73
50.0 cm3 of 0.200 mol dm−3 ethanoic acid is mixed with 25.0 cm3 of 0.100 mol dm−3 sodium hydroxide. What is the pH of the resulting solution? (Ka = 1.74 × 10−5 mol dm−3)
- A4.28
- B4.46
- C4.76
- D5.24
1.00 dm3 of buffer contains 0.0100 mol ethanoic acid and 0.0100 mol sodium ethanoate (pH 4.76). 0.00100 mol of HCl is added. What is the new pH? (Ka = 1.74 × 10−5 mol dm−3)
- A4.67
- B3.00
- C4.76
- D4.85
Quiz key
A buffer minimises the change in pH when small amounts of acid or alkali are added; its pH changes slightly, not at all. No buffer is unlimited: once one reservoir (the weak acid or its conjugate base) is used up, the pH changes sharply, so 'whatever amount' is wrong. Buffers can work at any pH, for example about 4.8 for ethanoic acid and ethanoate or about 9 for ammonia and ammonium, so they are not defined by pH 7. A buffer holds the pH near its own value; it does not bring the solution to pH 7.
NH3 is a weak base and NH4+ is its conjugate acid: added H+ is removed by NH3 + H+ → NH4+, and added OH− by NH4+ + OH− → NH3 + H2O, giving an alkaline buffer (about pH 9). HCl is a strong acid and Cl− is too weak a base to take up added H+, so HCl and NaCl cannot buffer. NaOH is a strong base with no partner that can remove added OH−. Sodium chloride in water contains no weak acid or weak base at all.
The conjugate base HCO3− takes up the added H+, forming H2CO3, so [H+] and the pH (about 7.4) barely change. H2CO3 + OH− is how the buffer removes added alkali, not acid. H2CO3 → H+ + HCO3− releases H+, which would lower the pH rather than resist the change. Dissolving CO2 to form H2CO3 is how carbonic acid forms in blood, but it does not remove added H+.
With [HA] = [A−], [H+] = Ka = 1.74 × 10−5 mol dm−3, so pH = pKa = 4.76. 2.88 is the pH of 0.100 mol dm−3 ethanoic acid alone; the ethanoate ions suppress its ionisation and raise the pH. 7.00 assumes the acid and its salt neutralise each other, but they do not react. 9.24 is 14 − 4.76, which mixes up pKa with pOH.
The equivalence point is at 25.0 cm3 and about pH 8.7, and the steep section runs from roughly pH 7 to 11. Range B (8.3 to 10.0, phenolphthalein) lies inside it, so the colour change happens at the equivalence volume. Range A (3.1 to 4.4) lies in the gentle buffer region near the start, so methyl orange would change colour gradually and far too early. Only strong acid and strong base titrations have an equivalence point near pH 7; a weak acid and strong base give an alkaline equivalence point because the ethanoate ion is basic. Weak acids titrate perfectly well against strong bases; it is weak acid against weak base that has no sharp end point. Note also that at 12.5 cm3, half-neutralisation, pH = pKa = 4.76.
[H+] = 1.74 × 10−5 × 0.200 ÷ 0.100 = 3.48 × 10−5 mol dm−3, so pH = 4.46. More acid than salt makes the buffer more acidic than pKa. 5.06 inverts the ratio, using [A−] ÷ [HA]. 4.76 is the pH only when the two concentrations are equal. 2.73 ignores the salt and treats 0.200 mol dm−3 ethanoic acid on its own with √(Ka c).
n(acid) = 0.0100 mol and n(NaOH) = 0.00250 mol. The reaction CH3COOH + OH− → CH3COO− + H2O leaves 0.00750 mol acid and forms 0.00250 mol ethanoate in the same volume, a 3 : 1 ratio. [H+] = 1.74 × 10−5 × 3 = 5.22 × 10−5 mol dm−3, pH 4.28. 4.46 takes the 0.200 : 0.100 ratio of the two starting concentrations and forgets that the NaOH uses up acid. 4.76 assumes half-neutralisation, but only a quarter of the acid has reacted. 5.24 inverts the 3 : 1 ratio.
The ethanoate removes the acid: CH3COO− + H+ → CH3COOH. Now n(CH3COOH) = 0.0110 mol and n(CH3COO−) = 0.00900 mol, so [H+] = 1.74 × 10−5 × 0.0110 ÷ 0.00900 = 2.13 × 10−5 mol dm−3, pH 4.67: a fall of only 0.09. 3.00 is what the same HCl would give in 1.00 dm3 of pure water, ignoring the buffer. 4.76 claims the pH does not change at all; a buffer reduces the change but does not remove it. 4.85 shifts the ratio the wrong way, as if alkali had been added.
What to project
The student lesson embeds these tabs. Open them full screen on the projector; every slider has a drag handle on the stage.
Slides for this lesson
A projectable deck for every pack is on the way. Until then, present the student lesson full screen: its sections, figure and lab tabs follow the same order as the plan above.