Acid-base Equilibria
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12.1 Brønsted–Lowry acids and bases
A Brønsted–Lowry acid is a proton (H+) donor. A Brønsted–Lowry base is a proton acceptor.
Acids and bases react by transferring a proton. In HCl + H2O → H3O+ + Cl−, HCl is the acid and water is the base. In NH3 + H2O ⇌ NH4+ + OH−, water is the acid. Water can act as either.
In aqueous solution H+ exists as H3O+, but H+(aq) is used for short.
Acid + base → salt + water. Acid + carbonate → salt + water + CO2.
Check yourself: In NH3 + HCl → NH4Cl, which species is the Brønsted–Lowry base?
NH3 (it accepts a proton)
12.2 pH and the pH of strong acids
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12.3 Strong and weak acids
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12.4 Ka, pKa and the pH of a weak acid
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12.5 The ionic product of water and strong bases
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12.6 Interpreting pH data
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12.7 pH titration curves
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12.8 Choosing an indicator
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12.9 Buffer solutions
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12.10 Buffer calculations
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12.11 Using a weak acid–strong base titration curve
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12.12 Core Practical 9: finding Ka for a weak acid
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12.13 Enthalpy changes of neutralisation
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12.14 Controlling the pH of blood
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Extra worked questions
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All 8 worked solutions are free for members. The first one is open to everyone.
pH of a strong acid
(a) Calculate the pH of 0.020 mol dm−3 hydrochloric acid. (b) Calculate [H+] in a solution of pH 3.40.
Try it first, then show the worked solution
| (a) | Fully dissociated: [H+] = 0.020; pH = −log 0.020 = 1.70 | 1 |
|---|---|---|
| (b) | [H+] = 10−3.40 = 3.98 × 10−4 mol dm−3 | 1 |
Answer: (a) 1.70 (b) 3.98 × 10−4 mol dm−3
Examiner tip: Give pH to two decimal places unless told otherwise.
pH of sodium hydroxide
Calculate the pH of 0.050 mol dm−3 sodium hydroxide at 298 K. Kw = 1.00 × 10−14 mol2 dm−6.
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pH of ethanoic acid
Calculate the pH of 0.10 mol dm−3 ethanoic acid. Ka = 1.74 × 10−5 mol dm−3.
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Ka from a pH
A 0.25 mol dm−3 solution of a weak acid HA has pH 2.70. Calculate Ka and pKa.
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Mixing acid and alkali
25.0 cm3 of 0.100 mol dm−3 HCl is mixed with 30.0 cm3 of 0.100 mol dm−3 NaOH. Calculate the pH of the mixture. Kw = 1.00 × 10−14.
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Neutralising a strong and a weak acid
50.0 cm3 of 1.00 mol dm−3 HCl is mixed with 50.0 cm3 of 1.00 mol dm−3 NaOH and the temperature rises by 6.8 °C. With ethanoic acid instead, the rise is 6.6 °C. Calculate both enthalpy changes of neutralisation (c = 4.18 J g−1 K−1, 1 cm3 of solution = 1 g) and explain the difference.
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A buffer by partial neutralisation
20.0 cm3 of 0.200 mol dm−3 NaOH is added to 50.0 cm3 of 0.200 mol dm−3 ethanoic acid (Ka = 1.74 × 10−5). Calculate the pH.
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A buffer under attack
A buffer contains 0.200 mol ethanoic acid and 0.100 mol sodium ethanoate in 1.00 dm3. Ka = 1.74 × 10−5 mol dm−3. (a) Calculate its pH. (b) Calculate the pH after adding 0.010 mol HCl. (c) Explain why the pH barely changes.
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Video walkthroughs
Narrated walkthroughs of exam-style questions from this topic of the book, showing where each mark comes from. Try the question first, then watch. Captions are available in the player.
pH of a weak acid
Ka and pKa from a measured pH
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Buffer pH before and after adding acid
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Ka by the half-neutralisation method
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Enthalpy of neutralisation of a weak acid
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Making a basic buffer of a chosen pH
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The blood buffer
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Recall cards
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