The shape of the lesson
By the end of the lesson, learners can
- Water, Kw and strong acids and bases: Choose a strong acid, then a strong base, in the simulator and change the concentration.
- Weak acids and Ka: Switch the simulator to a weak acid and compare its pH with a strong acid of the same concentration.
- pH = −log[H+]; Kw = [H+][OH−] = 1.00 × 10−14.
- Strong base: [OH−] first, then [H+] = Kw ÷ [OH−].
- Weak acid: Ka = [H+]2 ÷ [HA].
- pKa = −log Ka: smaller pKa, stronger acid.
One picture every idea returns to
Water ionises slightly: H2O ⇌ H+ + OH−, and the ionic product of water Kw = [H+][OH−] = 1.00 × 10−14 mol2 dm−6 at 298 K. pH = −log[H+]; for a strong acid [H+] equals the acid concentration, and for a strong base [H+] = Kw ÷ [OH−], so 0.0100 mol dm−3 NaOH has pH 12.0. A weak acid HA is only partly ionised, described by the acid dissociation constant Ka = [H+][A−] ÷ [HA], and pKa = −log Ka (smaller pKa, stronger acid). Assuming [H+] = [A−] and that very little HA ionises, [H+] = √(Ka × c): 0.100 mol dm−3 ethanoic acid (Ka 1.74 × 10−5 mol dm−3) has pH 2.88.
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, Putting a number on acid strength.. Water itself ionises a tiny amount, and that fixes the link between [H + ] and [OH − ]. Put the figure on the board: it is the one picture every idea in the lesson returns to. | Whole class, then pairs |
| 5 to 17 min | Teach and check: Water, Kw and strong acids and bases | Choose a strong acid, then a strong base, in the simulator and change the concentration. Project tab 1 of the lab (Acid-base equilibria lab, Kw and strong acids) and let a student drive it while the class predicts. | Whole class, then pairs |
| 17 to 30 min | Teach and check: Weak acids and Ka | Switch the simulator to a weak acid and compare its pH with a strong acid of the same concentration. Project tab 2 of the lab (Acid-base equilibria lab, weak acids and Ka) and let a student drive it while the class predicts. | Round Table |
| 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 Weak acids and Ka, 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.
Round Table
One sheet travels round the team. Each student adds one line of working or one idea and passes it on, so the solution is built four hands at a time and every contribution is visible.
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 |
|---|---|
| Do not set [H + ] equal to the concentration of a weak acid; that is only true for strong acids. | For a strong base, find [OH − ] first, then use K w; quoting −log[OH − ] as the pH is the classic slip (that is pOH). Remember that Ba(OH) 2 releases two OH − per formula unit, and that pure water is neutral at every temperature even when its pH is not 7. |
Ka, pKa and the pH of weak acids: worksheet
6 questions, 15 marks. Written in the style of the papers; no past paper question is reproduced.
Calculate the pH of 0.0500 mol dm−3 barium hydroxide, Ba(OH)2, at 298 K.
Methanoic acid has Ka = 1.78 × 10−4 mol dm−3. Calculate the pH of a 0.0500 mol dm−3 solution.
A 0.200 mol dm−3 solution of a weak acid HA has pH 2.72. Calculate Ka and pKa.
State the two assumptions made when the pH of a weak acid is calculated using [H+] = √(Ka × c).
Calculate the pH of 0.00200 mol dm−3 potassium hydroxide at 298 K.
At 323 K, Kw = 5.48 × 10−14 mol2 dm−6. Calculate the pH of pure water at 323 K and explain whether it is acidic, neutral or alkaline.
Answer key and marking notes
- [OH−] = 2 × 0.0500 = 0.100 mol dm−3.
- [H+] = 1.00 × 10−14 ÷ 0.100 = 1.00 × 10−13 mol dm−3, so pH = 13.0.
- [H+]2 = Ka × c = 1.78 × 10−4 × 0.0500 = 8.90 × 10−6.
- [H+] = 2.98 × 10−3 mol dm−3.
- pH = −log(2.98 × 10−3) = 2.53.
- [H+] = 10−2.72 = 1.91 × 10−3 mol dm−3.
- Ka = (1.91 × 10−3)2 ÷ 0.200 = 1.81 × 10−5 mol dm−3.
- pKa = −log(1.81 × 10−5) = 4.74.
- [H+] = [A−]: all the H+ comes from the acid, and the ionisation of water is ignored.
- The acid is so slightly ionised that [HA] at equilibrium is equal to its initial concentration c.
- [H+] = 1.00 × 10−14 ÷ 0.00200 = 5.00 × 10−12 mol dm−3.
- pH = 11.3.
- In pure water [H+] = [OH−], so [H+] = √(5.48 × 10−14) = 2.34 × 10−7 mol dm−3.
- pH = 6.63.
- It is still neutral, because [H+] = [OH−]; Kw is larger because the ionisation of water is endothermic.
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 expression and value describe the ionic product of water, Kw, at 298 K?
- AKw = [H+][OH−] = 1.00 × 10−14 mol2 dm−6
- BKw = [H+][OH−] ÷ [H2O] = 1.00 × 10−14 mol dm−3
- CKw = [H+] = 1.00 × 10−7 mol dm−3
- DKw = [H+] + [OH−] = 14
What is the pH of 0.0100 mol dm−3 aqueous sodium hydroxide at 298 K?
- A12.0
- B2.00
- C10.0
- D7.01
What is the pH of 0.0200 mol dm−3 nitric acid, HNO3?
- A1.70
- B2.00
- C0.0200
- D3.40
Four acids have these pKa values: methanoic acid 3.75, ethanoic acid 4.76, propanoic acid 4.87, ammonium ion 9.25. Which is the strongest acid?
- Amethanoic acid, because it has the smallest pKa and so the largest Ka
- Bammonium ion, because it has the largest pKa
- Cpropanoic acid, because it has the most carbon atoms
- Dthey are equally strong, because all four are weak acids
Ethanoic acid has Ka = 1.74 × 10−5 mol dm−3. What is the pH of a 0.100 mol dm−3 solution?
- A2.88
- B1.00
- C4.76
- D5.76
A 0.200 mol dm−3 solution of a weak monobasic acid HA has pH 2.72 at 298 K. What is Ka?
- A1.81 × 10−5 mol dm−3
- B9.53 × 10−3 mol dm−3
- C3.63 × 10−6 mol dm−3
- D1.91 × 10−3 mol dm−3
The ionisation of water is endothermic. At 323 K, Kw = 5.48 × 10−14 mol2 dm−6. Which statement about pure water at 323 K is correct?
- Aits pH is 6.63 and it is neutral
- Bits pH is 6.63 and it is slightly acidic
- Cits pH is 7.00, because pure water always has pH 7
- Dits pH is 7.37 and it is slightly alkaline
Methanoic acid has Ka = 1.78 × 10−4 mol dm−3. What is the pH of a 0.0500 mol dm−3 solution?
- A2.53
- B1.30
- C3.75
- D5.05
Quiz key
Water ionises very slightly. [H2O] is so large that it is effectively constant and is absorbed into the constant, giving Kw = [H+][OH−] = 1.00 × 10−14 mol2 dm−6 at 298 K; the units are those of two concentrations multiplied. Dividing by [H2O] gives the ordinary Kc, not Kw, and the units would be wrong. 1.00 × 10−7 mol dm−3 is [H+] in pure water at 298 K, which is the square root of Kw. The ion concentrations multiply, they do not add; pH + pOH = 14 is the logarithmic form of the product.
NaOH is fully dissociated, so [OH−] = 0.0100 mol dm−3. [H+] = 1.00 × 10−14 ÷ 0.0100 = 1.00 × 10−12 mol dm−3, so pH = 12.0. 2.00 is −log[OH−], the pOH, quoted as if it were the pH. 10.0 comes from dividing 1.00 × 10−14 by 1.00 × 10−4, misreading the concentration. 7.01 treats the alkali as if it barely changed the pH of water, but a strong base at this concentration is clearly alkaline.
A strong monobasic acid is fully ionised, so [H+] = 0.0200 mol dm−3 and pH = −log 0.0200 = 1.70. 2.00 would be the pH of 0.0100 mol dm−3 acid. 0.0200 is the concentration itself, not its negative logarithm. 3.40 doubles the logarithm, as if [H+] were (0.0200)2; there is no square or square root for a strong acid.
pKa = −log Ka, so the smallest pKa belongs to the largest Ka: methanoic acid (Ka = 1.78 × 10−4 mol dm−3) ionises the most and is the strongest. The ammonium ion has the largest pKa, so it is the weakest acid of the four. A longer alkyl chain is electron-donating, which makes propanoic acid slightly weaker than ethanoic acid, not stronger. 'Weak' only means partly ionised; weak acids still differ in strength, which is exactly what Ka measures.
[H+]2 = 1.74 × 10−5 × 0.100 = 1.74 × 10−6, so [H+] = 1.32 × 10−3 mol dm−3 and pH = 2.88. 1.00 treats ethanoic acid as fully ionised, which is only true for a strong acid. 4.76 is the pKa; the pH only equals pKa in a buffer with equal amounts of acid and salt. 5.76 is −log(Ka × c), forgetting to take the square root.
[H+] = 10−2.72 = 1.91 × 10−3 mol dm−3. With [H+] = [A−] and [HA] ≈ 0.200, Ka = (1.91 × 10−3)2 ÷ 0.200 = 1.81 × 10−5 mol dm−3 (pKa 4.74). 9.53 × 10−3 divides [H+] by c without squaring: that is the fraction ionised. 3.63 × 10−6 squares [H+] but forgets to divide by [HA]. 1.91 × 10−3 is just [H+].
[H+] = √(5.48 × 10−14) = 2.34 × 10−7 mol dm−3, so pH = 6.63. Neutral means [H+] = [OH−], which is still true, so the water is neutral even though its pH is below 7. Calling it acidic confuses pH 7 with neutrality, which only coincide at 298 K. pH 7.00 ignores the rise in Kw with temperature (Le Chatelier: heating shifts the endothermic ionisation to the right). 7.37 is 14 minus 6.63, which wrongly assumes pH + pOH = 14 still holds at 323 K (it is 13.26 there); [H+] has increased, so the pH must fall, not rise.
[H+]2 = 1.78 × 10−4 × 0.0500 = 8.90 × 10−6, so [H+] = 2.98 × 10−3 mol dm−3 and pH = 2.53. 1.30 is −log 0.0500, treating methanoic acid as a strong acid. 3.75 is the pKa, not the pH of the solution. 5.05 is −log(8.90 × 10−6), forgetting to take the square root.
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.