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Questions on buffer solutions and buffer calculations

Buffer solutions defined, acidic buffers from a weak acid and its salt or by partial neutralisation, basic buffers from ammonia and ammonium chloride, how a buffer resists added acid and added alkali, the pH of an acidic buffer from Ka and the ratio of acid to salt, the pH after a small amount of acid or base is added, the amount of salt needed for a chosen pH, the Henderson–Hasselbalch equation, the carbonic acid–hydrogencarbonate buffer in blood, and, for Cambridge, the solubility product Ksp with the common ion effect.

19 original questions · 69 marks · the buffer solutions and buffer calculations notes · Acids and bases · Flashcards for this unit

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• 6 marks · Explain · BS-Q2

  Explain, with equations, how a solution of propanoic acid and sodium propanoate resists a change in pH when a small amount of acid or of alkali is added, and why it stops working when a large amount is added.

  Mark scheme

  Indicative content

  1. the buffer contains large amounts of both CH₃CH₂COOH and CH₃CH₂COO⁻, the propanoate coming from the fully dissociated salt
  2. added H⁺ reacts with propanoate ions, CH₃CH₂COO⁻(aq) + H⁺(aq) → CH₃CH₂COOH(aq), so most of it is removed
  3. added OH⁻ reacts with propanoic acid molecules, CH₃CH₂COOH(aq) + OH⁻(aq) → CH₃CH₂COO⁻(aq) + H₂O(l)
  4. both reservoirs are large compared with the amount added, so the ratio [HA]/[A⁻] changes very little
  5. [H⁺] = Ka × [HA]/[A⁻], so with the ratio almost unchanged [H⁺] and the pH change only slightly
  6. once the added acid or alkali exceeds the propanoate or the propanoic acid present, that reservoir is used up and the pH changes sharply

  Levels

  • Level 3 (5–6 marks) both reservoirs are identified with the salt as the source of propanoate, both equations are correct and matched to added acid and added alkali, the small change in the [HA]/[A⁻] ratio is linked to [H⁺] through Ka, and the limit is explained as a reservoir being used up
  • Level 2 (3–4 marks) both equations are given and the removal of added H⁺ and OH⁻ is explained; the answer does not get from the [HA]/[A⁻] ratio to [H⁺] through Ka, or does not say why a large addition overwhelms the buffer
  • Level 1 (1–2 marks) the buffer is said to contain a weak acid and its salt, usually with the claim that its pH cannot change at all; one equation at most, and nothing on large additions

• 5 marks · Calculate · BS-Q12

  Calculate the mass of sodium ethanoate, CH₃COONa, that must be dissolved in the ethanoic acid described in the table to give a buffer of pH 5.00. Assume no change in volume.

  Quantity	Value
  Volume of ethanoic acid	250 cm³
  Concentration of ethanoic acid	0.100 mol dm⁻³
  Ka of ethanoic acid	1.74 × 10⁻⁵ mol dm⁻³
  Mr of CH₃COONa	82.0

  Mark scheme

  1. M1 [H⁺] = 10^(−5.00) = 1.00 × 10⁻⁵ mol dm⁻³
  2. M1 [CH₃COO⁻] = Ka × [CH₃COOH] ÷ [H⁺] = 1.74 × 10⁻⁵ × 0.100 ÷ 1.00 × 10⁻⁵ = 0.174 mol dm⁻³
  3. M1 moles of CH₃COONa = 0.174 × 0.250 = 0.0435 mol
  4. M1 mass = moles × Mr = 0.0435 × 82.0
  5. A1 mass = 3.57 g, to three significant figures

• 5 marks · Calculate · BS-Q16 CAIE

  Calculate the solubility of lead(II) iodide, PbI₂, in water at 298 K, in mol dm⁻³, given that Ksp = 9.80 × 10⁻⁹ mol³ dm⁻⁹, and give the expression for its Ksp with its units.

  Mark scheme

  1. B1 Ksp = [Pb²⁺][I⁻]², for PbI₂(s) ⇌ Pb²⁺(aq) + 2I⁻(aq)
  2. B1 the units are mol³ dm⁻⁹
  3. M1 if the solubility is s, [Pb²⁺] = s and [I⁻] = 2s
  4. M1 Ksp = s × (2s)² = 4s³ = 9.80 × 10⁻⁹, so s³ = 2.45 × 10⁻⁹
  5. A1 s = 1.35 × 10⁻³ mol dm⁻³, to three significant figures

• 4 marks · Calculate · BS-Q5

  Calculate the pH change when hydrogen chloride is added to the propanoate buffer described in the table, and compare it with the change when the same amount is added to 500 cm³ of pure water. Assume no change in volume.

  Quantity	Value
  Volume of buffer	500 cm³
  Concentration of propanoic acid	0.200 mol dm⁻³
  Concentration of sodium propanoate	0.200 mol dm⁻³
  Ka of propanoic acid	1.35 × 10⁻⁵ mol dm⁻³
  Hydrogen chloride added	0.00500 mol

  Mark scheme

  1. M1 the buffer holds 0.100 mol of each, so its pH = pKa = 4.87 before the addition
  2. M1 after the addition, propanoic acid = 0.105 mol and propanoate = 0.095 mol
  3. M1 [H⁺] = 1.35 × 10⁻⁵ × 0.105 ÷ 0.095 = 1.49 × 10⁻⁵ mol dm⁻³, so the pH falls to 4.83
  4. A1 a fall of 0.04, against a fall from 7.00 to 2.00 in water, where [H⁺] = 0.0100 mol dm⁻³

• 4 marks · Calculate · BS-Q6

  Calculate the volume of sodium hydroxide that must be added to the ethanoic acid in the table to make a buffer of pH 5.00. Ka = 1.74 × 10⁻⁵ mol dm⁻³.

  Solution	Volume / cm³	Concentration / mol dm⁻³
  Ethanoic acid	50.0	0.200
  Sodium hydroxide	to be found	0.100

  Mark scheme

  1. M1 [A⁻]/[HA] = Ka ÷ [H⁺] = 1.74 × 10⁻⁵ ÷ 1.00 × 10⁻⁵ = 1.74
  2. M1 with x mol of OH⁻ added, x ÷ (0.0100 − x) = 1.74, since the acid starts at 0.0100 mol
  3. M1 x = 0.0174 ÷ 2.74 = 6.35 × 10⁻³ mol of NaOH
  4. A1 volume = 6.35 × 10⁻³ ÷ 0.100 = 0.0635 dm³ = 63.5 cm³

• 4 marks · Explain · BS-Q7

  Explain, with equations, how a solution of ammonia and ammonium chloride acts as a buffer.

  Mark scheme

  1. B1 the solution contains large amounts of NH₃ and NH₄⁺, the NH₄⁺ coming from the fully dissociated salt
  2. B1 added acid is removed by ammonia: NH₃ + H⁺ → NH₄⁺
  3. B1 added alkali is removed by ammonium ions: NH₄⁺ + OH⁻ → NH₃ + H₂O
  4. B1 both reservoirs are large, so the ratio of NH₃ to NH₄⁺, and the pH, change very little

• 4 marks · Calculate · BS-Q10

  Calculate the pH of the buffer formed when the sodium hydroxide in the table is added to the ethanoic acid. Ka = 1.74 × 10⁻⁵ mol dm⁻³.

  Solution	Volume / cm³	Concentration / mol dm⁻³
  Ethanoic acid	40.0	0.250
  Sodium hydroxide	20.0	0.200

  Mark scheme

  1. M1 moles of CH₃COOH at the start = 0.250 × 0.0400 = 0.0100 mol and moles of OH⁻ = 0.200 × 0.0200 = 4.00 × 10⁻³ mol
  2. M1 after reaction, CH₃COOH = 6.00 × 10⁻³ mol and CH₃COO⁻ = 4.00 × 10⁻³ mol
  3. M1 [H⁺] = 1.74 × 10⁻⁵ × 6.00 × 10⁻³ ÷ 4.00 × 10⁻³ = 2.61 × 10⁻⁵ mol dm⁻³
  4. A1 pH = 4.58, to two decimal places

• 4 marks · Calculate · BS-Q11

  Calculate the pH after sodium hydroxide is added to the ethanoate buffer described in the table, assuming no change in volume.

  Quantity	Value
  Volume of buffer	500 cm³
  Concentration of ethanoic acid	0.100 mol dm⁻³
  Concentration of sodium ethanoate	0.100 mol dm⁻³
  Ka of ethanoic acid	1.74 × 10⁻⁵ mol dm⁻³
  Sodium hydroxide added	5.00 × 10⁻³ mol

  Mark scheme

  1. M1 moles at the start: CH₃COOH = CH₃COO⁻ = 0.100 × 0.500 = 0.0500 mol
  2. M1 OH⁻ converts 5.00 × 10⁻³ mol of acid to ethanoate: CH₃COOH = 0.0450 mol, CH₃COO⁻ = 0.0550 mol
  3. M1 [H⁺] = 1.74 × 10⁻⁵ × 0.0450 ÷ 0.0550 = 1.42 × 10⁻⁵ mol dm⁻³
  4. A1 pH = 4.85, to two decimal places, up from 4.76

• 3 marks · State · BS-Q1

  State what is meant by a buffer solution, and give two different ways of making an acidic buffer from ethanoic acid.

  Mark scheme

  1. B1 a buffer is a solution that resists changes in pH when small amounts of acid or base are added; reject 'keeps the pH constant'
  2. B1 mix ethanoic acid with sodium ethanoate, a salt of the weak acid
  3. B1 add sodium hydroxide to an excess of ethanoic acid, so that only part of the acid is neutralised

• 3 marks · Identify · BS-Q3

  Identify which of these mixtures is a buffer: (A) hydrochloric acid with sodium chloride; (B) ammonia with ammonium chloride. Give reasons for your choice.

  Mark scheme

  1. B1 mixture B, ammonia with ammonium chloride, is the buffer
  2. B1 it contains a weak base and its conjugate acid in large amounts, to remove added H⁺ and added OH⁻
  3. B1 in mixture A, Cl⁻ is too weak a base to remove added H⁺, and there are no HCl molecules to react with added OH⁻

• 3 marks · Calculate · BS-Q4

  Calculate the ratio [CH₃COO⁻]/[CH₃COOH] needed to make an ethanoic acid buffer of pH 4.50, and state the pH of the buffer when the ratio is 1. Ka = 1.74 × 10⁻⁵ mol dm⁻³.

  Mark scheme

  1. B1 when [HA] = [A⁻] the pH equals the pKa, 4.76
  2. M1 [H⁺] = 10^(−4.50) = 3.16 × 10⁻⁵ mol dm⁻³, and the ratio = Ka ÷ [H⁺]
  3. A1 [CH₃COO⁻]/[CH₃COOH] = 1.74 × 10⁻⁵ ÷ 3.16 × 10⁻⁵ = 0.550, to three significant figures

• 3 marks · Explain · BS-Q8

  Explain why diluting a buffer with water changes its pH very little, and why the diluted buffer is less useful.

  Mark scheme

  1. B1 dilution lowers [HA] and [A⁻] by the same factor
  2. B1 the ratio [HA]/[A⁻] is unchanged, so [H⁺] = Ka × [HA]/[A⁻] and the pH are almost unchanged
  3. B1 there is less HA and A⁻ in each cubic decimetre, so a smaller addition of acid or alkali uses up a reservoir

• 3 marks · Calculate · BS-Q9

  Calculate the pH of a buffer that is 0.200 mol dm⁻³ in propanoic acid and 0.100 mol dm⁻³ in sodium propanoate. Ka = 1.35 × 10⁻⁵ mol dm⁻³.

  Mark scheme

  1. M1 [H⁺] = Ka × [HA] / [A⁻], with [A⁻] taken from the salt
  2. M1 [H⁺] = 1.35 × 10⁻⁵ × 0.200 ÷ 0.100 = 2.70 × 10⁻⁵ mol dm⁻³
  3. A1 pH = 4.57, to two decimal places

• 3 marks · Suggest · BS-Q13

  Suggest how the carbonic acid–hydrogencarbonate buffer keeps the pH of blood close to 7.4 when muscles release lactic acid during hard exercise, and what the body does with the product.

  Mark scheme

  1. B1 H⁺ from the lactic acid reacts with hydrogencarbonate ions: HCO₃⁻ + H⁺ → H₂CO₃
  2. B1 the large reservoir of HCO₃⁻ means [H⁺] rises only slightly
  3. B1 the H₂CO₃ breaks down to CO₂ and H₂O, and the CO₂ is removed by breathing it out, faster during exercise

• 3 marks · Suggest · BS-Q14

  Suggest why a buffer of 0.00100 mol dm⁻³ ethanoic acid and 0.00100 mol dm⁻³ sodium ethanoate would be of little use for holding the pH of a reaction that releases 0.0100 mol dm⁻³ of H⁺, and how the buffer should be changed.

  Mark scheme

  1. B1 the amount of ethanoate available to take up H⁺ is far smaller than the amount of H⁺ released
  2. B1 the ethanoate is used up, after which the extra H⁺ is not removed and the pH falls sharply, as it would in water
  3. B1 use the same ratio at much higher concentrations, well above 0.0100 mol dm⁻³ of each, so the pH is unchanged but the reservoirs are large

• 3 marks · Evaluate · BS-Q15

  Evaluate the statement: 'A buffer keeps the pH of a solution constant, whatever is added to it.'

  Mark scheme

  1. B1 a buffer resists a change in pH, but the pH still changes a little with each addition
  2. B1 it works only for small additions, compared with the amounts of weak acid and conjugate base present
  3. B1 once the added acid or base exceeds the reservoir of A⁻ or HA, the buffer is used up and the pH changes sharply, so the statement is incorrect

• 3 marks · Deduce · BS-Q17 CAIE

  Deduce whether a precipitate of silver chloride forms when equal volumes of 2.0 × 10⁻⁴ mol dm⁻³ silver nitrate and 2.0 × 10⁻⁴ mol dm⁻³ sodium chloride are mixed at 298 K. Ksp(AgCl) = 1.80 × 10⁻¹⁰ mol² dm⁻⁶.

  Mark scheme

  1. M1 mixing equal volumes halves each concentration, so [Ag⁺] = [Cl⁻] = 1.0 × 10⁻⁴ mol dm⁻³
  2. M1 the ionic product [Ag⁺][Cl⁻] = 1.0 × 10⁻⁸ mol² dm⁻⁶
  3. B1 1.0 × 10⁻⁸ is greater than Ksp, so silver chloride precipitates until the product falls to 1.80 × 10⁻¹⁰

• 3 marks · Explain · BS-Q18 CAIE

  Explain why silver chloride is much less soluble in 0.100 mol dm⁻³ sodium chloride than in pure water.

  Mark scheme

  1. B1 a saturated solution has AgCl(s) ⇌ Ag⁺(aq) + Cl⁻(aq) at equilibrium, with Ksp = [Ag⁺][Cl⁻] constant at a given temperature
  2. B1 sodium chloride supplies a high concentration of Cl⁻, the common ion, which shifts the equilibrium to the left
  3. B1 with [Cl⁻] fixed near 0.100 mol dm⁻³, [Ag⁺] can reach only Ksp ÷ 0.100, so far less silver chloride dissolves

• 3 marks · Suggest · BS-Q19 CAIE

  Suggest why a suspension of barium sulfate, Ksp = 1.1 × 10⁻¹⁰ mol² dm⁻⁶, can be swallowed before an X-ray of the gut although barium ions are toxic, and why adding a little sodium sulfate to the suspension would make it safer still.

  Mark scheme

  1. B1 in saturated barium sulfate [Ba²⁺] = √(1.1 × 10⁻¹⁰) = 1.0 × 10⁻⁵ mol dm⁻³, so very little toxic Ba²⁺ is in solution
  2. B1 sodium sulfate raises [SO₄²⁻], the common ion, shifting BaSO₄(s) ⇌ Ba²⁺(aq) + SO₄²⁻(aq) to the left
  3. B1 since [Ba²⁺][SO₄²⁻] must stay equal to Ksp, [Ba²⁺] falls further

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