Buffer Practice Problems

Buffers are a favourite topic for multi-step exam questions because they combine several ideas at once: weak acid equilibria, stoichiometry, logarithms and Le Chatelier’s principle. The good news is that almost every buffer question can be answered with one equation and one habit.

The equation: pH = pKa + log([A⁻] ÷ [HA]) (see the Henderson–Hasselbalch equation).

The habit: when anything is added or mixed, work in moles first, do any neutralisation, and only then put the ratio into the equation.

Use pKa(ethanoic acid) = 4.76 and pKa(NH₄⁺) = 9.25 unless told otherwise. Try every question before looking at the answers.

Section A: the basics

1. A buffer contains 0.30 mol/dm³ ethanoic acid and 0.20 mol/dm³ sodium ethanoate. Calculate its pH.

2. What ratio of [ethanoate] to [ethanoic acid] gives a buffer at pH 5.20?

3. Which would make a better buffer at pH 5.0: ethanoic acid/ethanoate (pKa 4.76) or methanoic acid/methanoate (pKa 3.75)? Explain.

4. Explain why a mixture of hydrochloric acid and sodium chloride is not a buffer.

Section B: making buffers

5. 50.0 cm³ of 0.200 mol/dm³ ethanoic acid is mixed with 50.0 cm³ of 0.100 mol/dm³ NaOH. Calculate the pH.

6. 50.0 cm³ of 0.200 mol/dm³ ammonia is mixed with 25.0 cm³ of 0.200 mol/dm³ HCl. Calculate the pH.

7. What mass of ammonium chloride (Mr 53.5) must be dissolved in 250 cm³ of 0.100 mol/dm³ ammonia to make a buffer at pH 9.00? Assume no volume change.

8. A buffer at pH 4.50 is to be made from 0.0500 mol benzoic acid (pKa 4.20) and sodium benzoate. How many moles of sodium benzoate are needed?

9. Which phosphate pair would you use to make a buffer at pH 7.40, and in what ratio? (pKa₁ 2.15, pKa₂ 7.20, pKa₃ 12.35)

Section C: adding acid, base and water

A buffer contains 0.100 mol ethanoic acid and 0.100 mol sodium ethanoate in 1.00 dm³.

10. Calculate the pH after adding 0.020 mol NaOH.

11. Calculate the pH after adding 0.020 mol HCl (to a fresh sample).

12. Calculate the pH after the original buffer is diluted to 10.0 dm³ with water.

13. A different buffer contains 0.050 mol ethanoic acid and 0.050 mol ethanoate in 1.00 dm³. 0.060 mol of HCl is added. Calculate the pH.

Section D: capacity and biology

14. Buffer X contains 0.50 mol/dm³ of both HA and A⁻. Buffer Y contains 0.050 mol/dm³ of both. They have the same pH. Which has the greater buffer capacity, and why?

15. A blood sample has [HCO₃⁻] = 24 mmol/dm³ and dissolved CO₂ (treated as H₂CO₃) = 1.2 mmol/dm³. Using pKa = 6.1, calculate the blood pH.

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Answers

1. pH = 4.76 + log(0.20 ÷ 0.30) = 4.76 + log(0.667) = 4.76 − 0.18 = 4.58

2. 5.20 = 4.76 + log(ratio) → log(ratio) = 0.44 → ratio = 10⁰·⁴⁴ = 2.75

3. Ethanoic acid/ethanoate. Its pKa (4.76) is within 0.24 of the target, so it needs a ratio close to 1 : 1 and has good capacity. Methanoic acid’s pKa (3.75) is 1.25 units away, outside the useful pKa ± 1 range. See buffer capacity.

4. HCl is a strong acid, so Cl⁻ is an extremely weak base. There’s nothing in the mixture to react with added H⁺, so the pH changes as it would in water.

5.

• n(CH₃COOH) = 0.200 × 0.0500 = 0.0100 mol
• n(NaOH) = 0.100 × 0.0500 = 0.00500 mol
• After reaction: CH₃COOH = 0.00500 mol; CH₃COO⁻ = 0.00500 mol
• pH = 4.76 + log(1) = 4.76

6.

• n(NH₃) = 0.200 × 0.0500 = 0.0100 mol
• n(HCl) = 0.200 × 0.0250 = 0.00500 mol
• After reaction: NH₃ = 0.00500 mol; NH₄⁺ = 0.00500 mol
• pH = 9.25 + log(1) = 9.25

7.

• n(NH₃) = 0.100 × 0.250 = 0.0250 mol
• 9.00 = 9.25 + log([NH₃] ÷ [NH₄⁺]) → ratio = 10⁻⁰·²⁵ = 0.562
• n(NH₄⁺) = 0.0250 ÷ 0.562 = 0.0445 mol
• Mass = 0.0445 × 53.5 = 2.38 g

8. 4.50 = 4.20 + log(n(A⁻) ÷ 0.0500) → ratio = 10⁰·³⁰ = 2.00 → n(A⁻) = 0.100 mol

9. Use H₂PO₄⁻/HPO₄²⁻ (pKa₂ = 7.20, closest to 7.40). Ratio [HPO₄²⁻] ÷ [H₂PO₄⁻] = 10^(7.40 − 7.20) = 10⁰·²⁰ = 1.58.

10. OH⁻ converts HA to A⁻: HA = 0.080 mol, A⁻ = 0.120 mol. pH = 4.76 + log(1.50) = 4.76 + 0.18 = 4.94

11. H⁺ converts A⁻ to HA: HA = 0.120 mol, A⁻ = 0.080 mol. pH = 4.76 + log(0.667) = 4.58

12. The ratio is unchanged (both amounts still 0.100 mol), so pH = 4.76. Dilution lowers capacity but not pH.

13. Only 0.050 mol A⁻ is available to react with 0.060 mol H⁺. All the ethanoate is used up and 0.010 mol H⁺ is left over. The buffer is destroyed. The excess strong acid controls the pH: [H⁺] ≈ 0.010 mol/dm³, pH ≈ 2.0. (The 0.100 mol of ethanoic acid now present adds very little H⁺, because its ionisation is suppressed by the strong acid.)

14. Buffer X. It contains ten times more acid and conjugate base, so it can absorb ten times more added H⁺ or OH⁻ before the ratio changes significantly. Same ratio means same pH; more material means more capacity.

15. pH = 6.1 + log(24 ÷ 1.2) = 6.1 + log(20) = 6.1 + 1.30 = 7.40. See the blood buffer system.

A strategy for any buffer question

Almost every buffer problem can be cracked by asking three questions in order. What are the two buffer components? Identify the weak acid and its conjugate base, or the weak base and its conjugate acid, and look up the right pKa (for a basic buffer, use the pKa of the conjugate acid). Has anything reacted? If a strong acid or base has been added, or if the buffer was made by partial neutralisation, set up a simple before-and-after table in moles and let the strong reagent react completely. What’s the final ratio? Put the moles of base form over the moles of acid form into Henderson–Hasselbalch. If either amount is zero, the buffer has failed, and you need a different method.

How to check your answers quickly

• Direction check: adding acid should lower pH; adding base should raise it; dilution shouldn’t change it.
• Size check: a small addition (a few percent of the buffer components) should change pH by only a few hundredths or tenths.
• Range check: a working buffer’s pH should be within about 1 unit of its pKa.
• Sign check: if [A⁻] > [HA], pH > pKa; if [A⁻] < [HA], pH < pKa.

Where marks are lost

1. Flipping the ratio (acid over base instead of base over acid).
2. Using concentrations before accounting for mixing volumes. Moles avoid this.
3. Forgetting to neutralise first when strong acid or base is added.
4. Using Henderson–Hasselbalch after the buffer is exhausted.
5. Using pKb instead of pKa for a basic buffer. For NH₃/NH₄⁺, use pKa(NH₄⁺) = 9.25.

Key takeaways

• pH = pKa + log([A⁻] ÷ [HA]); use moles, since the volume cancels.
• Partial neutralisation of a weak acid by a strong base makes a buffer.
• Added acid converts A⁻ to HA; added base converts HA to A⁻; water changes neither.
• Once a component runs out, the buffer fails and the excess strong acid or base controls pH.
• Review worked examples in buffer calculations.