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Questions on Brønsted–Lowry acids and bases, pH and strong acids

Brønsted–Lowry acids as proton donors and bases as proton acceptors, conjugate acid–base pairs, strong and weak acids, strength against concentration, pH = −log[H⁺] and its inverse, the pH of strong monoprotic acids and of sulfuric acid, basicity (monobasic, dibasic, tribasic), ionic equations for acids with metals, carbonates and oxides, the effect of dilution, and the pH after mixing a strong acid with a strong base, with either one in excess.

18 original questions · 66 marks · the Brønsted–Lowry acids and bases, pH and strong acids notes · Acids and bases · Flashcards for this unit

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• 6 marks · Discuss · BN-Q2

  Discuss a student's claim that 'watering hydrochloric acid down turns it into a weak acid'. Use 0.100 mol dm⁻³ hydrochloric acid diluted tenfold, and 0.100 mol dm⁻³ ethanoic acid of pH 2.88, in your answer.

  Mark scheme

  Indicative content

  1. a strong acid is one that dissociates completely in aqueous solution, and a weak acid is one that dissociates only partially
  2. strength is a property of the acid, the extent of its dissociation, whereas concentration is the amount of acid in each cubic decimetre of solution
  3. diluting 0.100 mol dm⁻³ hydrochloric acid tenfold gives 0.0100 mol dm⁻³ acid, still completely dissociated, so [H⁺] = 0.0100 mol dm⁻³ and the pH is 2.00
  4. dilution lowers the concentration and raises the pH, but the acid stays strong because no HCl molecules re-form at any concentration
  5. 0.100 mol dm⁻³ ethanoic acid has pH 2.88, higher than the 1.00 of hydrochloric acid at the same concentration, because it is only partly dissociated
  6. the claim is wrong because it confuses 'dilute' with 'weak': the watered-down acid is a dilute solution of a strong acid

  Levels

  • Level 3 (5–6 marks) strength and concentration are defined and kept apart throughout, the diluted hydrochloric acid is shown to stay fully dissociated with its pH worked to 2.00, the ethanoic acid comparison is used to show what weak actually looks like, and the claim is judged wrong for confusing dilute with weak
  • Level 2 (3–4 marks) the strong/weak distinction is stated correctly and the claim is rejected, with one of the two comparisons worked through; the other pH value goes unused, or strength and concentration slide together in places
  • Level 1 (1–2 marks) the answer agrees with the claim, or takes the higher pH of the diluted acid as proof that it has turned weak; strength and concentration are used as if they meant the same thing

• 5 marks · Explain · BN-Q9

  Explain why diluting a solution of a strong acid tenfold raises its pH by one unit, and why hydrochloric acid of concentration 1.0 × 10⁻⁸ mol dm⁻³ does not have a pH of 8.00.

  Mark scheme

  1. B1 the acid is fully dissociated, so a tenfold dilution divides [H⁺] by ten
  2. B1 pH is a logarithm to base 10, so a factor of ten in [H⁺] changes the pH by exactly one
  3. B1 water itself dissociates and supplies H⁺ at about 1.0 × 10⁻⁷ mol dm⁻³
  4. B1 at this dilution the acid supplies less H⁺ than the water does, so the water's H⁺ cannot be ignored
  5. B1 the total [H⁺] is a little above 1.0 × 10⁻⁷ mol dm⁻³, so the pH is just below 7 and the solution is still acidic

• 4 marks · State · BN-Q1

  State what is meant by a Brønsted–Lowry acid and by a Brønsted–Lowry base, and identify the two conjugate acid–base pairs in NH₃(aq) + H₂O(l) ⇌ NH₄⁺(aq) + OH⁻(aq), saying which member of each pair is the acid.

  Mark scheme

  1. B1 a Brønsted–Lowry acid is a proton donor; allow 'H⁺ ion donor', reject 'hydrogen donor'
  2. B1 a Brønsted–Lowry base is a proton acceptor, a species that accepts an H⁺ ion
  3. B1 NH₄⁺ and NH₃ are a conjugate pair, with NH₄⁺ the acid and NH₃ the base
  4. B1 H₂O and OH⁻ are a conjugate pair, with H₂O the acid and OH⁻ the base

• 4 marks · Calculate · BN-Q4

  Calculate the pH of 0.0350 mol dm⁻³ nitric acid, HNO₃, a strong monoprotic acid, and the hydrogen ion concentration in a solution of pH 3.65.

  Mark scheme

  1. M1 nitric acid is fully dissociated, so [H⁺] = 0.0350 mol dm⁻³ and pH = −log(0.0350)
  2. A1 pH = 1.46, to two decimal places
  3. M1 [H⁺] = 10^(−pH) = 10^(−3.65)
  4. A1 [H⁺] = 2.24 × 10⁻⁴ mol dm⁻³, to three significant figures

• 4 marks · Calculate · BN-Q5

  Calculate the pH of 7.50 × 10⁻³ mol dm⁻³ sulfuric acid, first assuming that both protons of each molecule are fully dissociated, and then assuming that only the first dissociation takes place.

  Mark scheme

  1. M1 fully dissociated, each H₂SO₄ releases two H⁺, so [H⁺] = 2 × 7.50 × 10⁻³ = 0.0150 mol dm⁻³
  2. A1 pH = −log(0.0150) = 1.82
  3. M1 first dissociation only, [H⁺] = 7.50 × 10⁻³ mol dm⁻³
  4. A1 pH = −log(7.50 × 10⁻³) = 2.12, to two decimal places

• 4 marks · Calculate · BN-Q6

  Calculate the pH at 298 K of the solution formed when the two solutions in the table are mixed. Kw = 1.00 × 10⁻¹⁴ mol² dm⁻⁶.

  Solution	Volume / cm³	Concentration / mol dm⁻³
  Hydrochloric acid	25.0	0.120
  Sodium hydroxide	35.0	0.100

  Mark scheme

  1. M1 moles of H⁺ = 0.120 × 0.0250 = 3.00 × 10⁻³ mol and moles of OH⁻ = 0.100 × 0.0350 = 3.50 × 10⁻³ mol
  2. M1 OH⁻ is in excess by 5.0 × 10⁻⁴ mol, so [OH⁻] = 5.0 × 10⁻⁴ ÷ 0.0600 = 8.33 × 10⁻³ mol dm⁻³, using the total volume
  3. M1 [H⁺] = Kw ÷ [OH⁻] = 1.00 × 10⁻¹⁴ ÷ 8.33 × 10⁻³ = 1.20 × 10⁻¹² mol dm⁻³; allow error carried forward
  4. A1 pH = 11.92, to two decimal places

• 4 marks · Calculate · BN-Q8

  Calculate the pH of the solution formed when the two solutions in the table are mixed.

  Solution	Volume / cm³	Concentration / mol dm⁻³
  Hydrochloric acid	25.0	0.150
  Sodium hydroxide	15.0	0.200

  Mark scheme

  1. M1 moles of H⁺ = 0.150 × 0.0250 = 3.75 × 10⁻³ mol and moles of OH⁻ = 0.200 × 0.0150 = 3.00 × 10⁻³ mol
  2. M1 H⁺ is in excess by 3.75 × 10⁻³ − 3.00 × 10⁻³ = 7.5 × 10⁻⁴ mol
  3. M1 [H⁺] = 7.5 × 10⁻⁴ ÷ 0.0400 = 0.01875 mol dm⁻³, using the total volume of 40.0 cm³; dividing by 25.0 cm³ alone is rejected
  4. A1 pH = −log(0.01875) = 1.73, to two decimal places

• 4 marks · Calculate · BN-Q16

  Calculate the volume of 0.150 mol dm⁻³ sodium hydroxide needed to neutralise completely 20.0 cm³ of 0.0500 mol dm⁻³ phosphoric(V) acid, H₃PO₄, and state what is meant by describing H₃PO₄ as tribasic.

  Mark scheme

  1. B1 a tribasic acid is one of which each molecule can donate three protons (H⁺ ions)
  2. M1 moles of H₃PO₄ = 0.0500 × 20.0 ÷ 1000 = 1.00 × 10⁻³ mol
  3. M1 H₃PO₄ is tribasic, so 3 × 1.00 × 10⁻³ = 3.00 × 10⁻³ mol of NaOH is needed
  4. A1 volume = 3.00 × 10⁻³ ÷ 0.150 = 0.0200 dm³ = 20.0 cm³, to three significant figures

• 4 marks · Give · BN-Q18

  Give the ionic equations, with state symbols, for the reactions of dilute nitric acid with solid calcium carbonate and with magnesium ribbon.

  Mark scheme

  1. B1 CaCO₃ + 2H⁺ → Ca²⁺ + CO₂ + H₂O, balanced, with the nitrate ion left out as a spectator
  2. B1 state symbols CaCO₃(s), H⁺(aq), Ca²⁺(aq), CO₂(g) and H₂O(l)
  3. B1 Mg + 2H⁺ → Mg²⁺ + H₂, balanced, with no nitrate ion
  4. B1 state symbols Mg(s), H⁺(aq), Mg²⁺(aq) and H₂(g); a missing state symbol on one species loses this mark only

• 3 marks · Give · BN-Q3

  Give the formula of the conjugate base of each of these acids: HSO₄⁻, H₂O and NH₄⁺.

  Mark scheme

  1. B1 the conjugate base of HSO₄⁻ is SO₄²⁻
  2. B1 the conjugate base of H₂O is OH⁻
  3. B1 the conjugate base of NH₄⁺ is NH₃

• 3 marks · Calculate · BN-Q7

  Calculate the pH of the solution made when 10.0 cm³ of 0.200 mol dm⁻³ hydrochloric acid is diluted to 250 cm³ with water.

  Mark scheme

  1. M1 moles of HCl = 0.200 × 10.0 ÷ 1000 = 2.00 × 10⁻³ mol
  2. M1 new [H⁺] = 2.00 × 10⁻³ ÷ 0.250 = 8.00 × 10⁻³ mol dm⁻³
  3. A1 pH = −log(8.00 × 10⁻³) = 2.10, to two decimal places

• 3 marks · Calculate · BN-Q10

  Calculate the volume of water that must be added to 50.0 cm³ of hydrochloric acid of pH 1.00 to give a solution of pH 2.00.

  Mark scheme

  1. M1 a rise of one pH unit means [H⁺] must fall by a factor of ten, from 0.100 to 0.0100 mol dm⁻³
  2. M1 the amount of HCl is unchanged, so the final volume is ten times the original, 500 cm³
  3. A1 volume of water added = 500 − 50.0 = 450 cm³, to three significant figures

• 3 marks · Explain · BN-Q11

  Explain why 0.100 mol dm⁻³ hydrochloric acid has a pH of 1.00 but 0.100 mol dm⁻³ ethanoic acid has a pH of 2.88.

  Mark scheme

  1. B1 hydrochloric acid is a strong acid that dissociates completely, so [H⁺] = 0.100 mol dm⁻³
  2. B1 ethanoic acid is a weak acid that dissociates only partially, in an equilibrium lying well to the left
  3. B1 so [H⁺] in the ethanoic acid is much lower than 0.100 mol dm⁻³ and its pH is higher

• 3 marks · Suggest · BN-Q12

  Suggest the role of nitric acid in the reaction HNO₃ + H₂SO₄ → H₂NO₃⁺ + HSO₄⁻, which takes place in a mixture of the two pure acids, and identify the two conjugate pairs.

  Mark scheme

  1. B1 nitric acid acts as a Brønsted–Lowry base, accepting a proton from sulfuric acid
  2. B1 H₂SO₄ and HSO₄⁻ are a conjugate pair, with H₂SO₄ the acid
  3. B1 H₂NO₃⁺ and HNO₃ are a conjugate pair, with H₂NO₃⁺ the acid

• 3 marks · Suggest · BN-Q13

  Suggest why the measured pH of 0.0500 mol dm⁻³ sulfuric acid is about 1.24, rather than 1.30 or 1.00.

  Mark scheme

  1. B1 the first dissociation, H₂SO₄ → H⁺ + HSO₄⁻, is complete, which alone would give [H⁺] = 0.0500 mol dm⁻³ and pH 1.30
  2. B1 the second dissociation, HSO₄⁻ ⇌ H⁺ + SO₄²⁻, is partial, so some extra H⁺ is released and the pH is lower than 1.30
  3. B1 the second proton is not fully released, so [H⁺] is less than 0.100 mol dm⁻³ and the pH is higher than 1.00

• 3 marks · Compare · BN-Q14

  Compare 2.0 mol dm⁻³ ethanoic acid with 0.010 mol dm⁻³ hydrochloric acid in terms of strength and concentration, and say which has the lower pH. The ethanoic acid has a pH of 2.23.

  Mark scheme

  1. B1 ethanoic acid is weak, partially dissociated, while hydrochloric acid is strong, fully dissociated
  2. B1 the ethanoic acid is the more concentrated solution, by a factor of 200
  3. B1 the hydrochloric acid has pH 2.00, lower than 2.23, so it has the higher [H⁺] even though it is far less concentrated

• 3 marks · Evaluate · BN-Q15

  Evaluate a student's claim that the pH scale runs from 0 to 14 and that no solution can have a pH below 0. Use 2.00 mol dm⁻³ hydrochloric acid in your answer.

  Mark scheme

  1. B1 2.00 mol dm⁻³ HCl is fully dissociated, so [H⁺] = 2.00 mol dm⁻³
  2. B1 pH = −log(2.00) = −0.30, a negative value
  3. B1 the claim is wrong: any strong acid with [H⁺] above 1 mol dm⁻³ has a negative pH, and 0 to 14 covers only the usual range

• 3 marks · Explain · BN-Q17

  Explain why 25.0 cm³ of 0.100 mol dm⁻³ hydrochloric acid reacts faster with excess magnesium than 25.0 cm³ of 0.100 mol dm⁻³ ethanoic acid, yet both give the same total amount of hydrogen.

  Mark scheme

  1. B1 the reacting species is H⁺: Mg + 2H⁺ → Mg²⁺ + H₂, and [H⁺] is much higher in the fully dissociated hydrochloric acid, so the rate is greater
  2. B1 as H⁺ is used up in the ethanoic acid, the equilibrium CH₃COOH ⇌ CH₃COO⁻ + H⁺ shifts to the right and more acid dissociates
  3. B1 both solutions contain 2.50 × 10⁻³ mol of acid, all of which eventually reacts, giving 1.25 × 10⁻³ mol of H₂ in each

Practise Brønsted–Lowry acids and bases, pH and strong acids one question at a time