These are original Marlbridge practice questions, not past exam questions. The mark points are a suggested Marlbridge rubric to help you check your work; they are not an official scoring guideline. Data at 25 °C: pKa of acetic acid = 4.76; pKa of NH₄⁺ = 9.25. HA is a generic weak acid and NaA its soluble sodium salt. HCl and NaOH are strong and react completely with the buffer components.
Question 1 (multiple choice · foundation)
Buffer X is 0.10 M CH₃COOH and 0.10 M CH₃COONa. Buffer Y is 1.0 M CH₃COOH and 1.0 M CH₃COONa. Equal volumes of each are compared. Which statement is correct?
- (A) X and Y have the same pH, and Y has the greater buffer capacity.
- (B) Y has a lower pH than X because it contains more acid.
- (C) X and Y have the same pH and the same buffer capacity.
- (D) X has the greater buffer capacity because a larger fraction of its acid is ionised.
Answer and explanation
Answer: (A). Both have [A⁻]/[HA] = 1, so both have pH = pKa = 4.76. Y holds ten times as many moles of each component, so it can neutralise ten times as much added acid or base.
- (B) ignores the base: Y also has ten times more acetate, so the ratio, and the pH, are unchanged.
- (C) confuses pH with capacity. The same ratio gives the same pH, but capacity depends on amounts.
- (D) is irrelevant: capacity comes from the moles of HA and A⁻ available to react, not from the tiny fraction of HA that ionises.
Question 2 (multiple choice · core)
A buffer contains 0.040 mol HA and 0.010 mol A⁻. Which addition will cause the largest change in pH?
- (A) 0.005 mol HCl
- (B) 0.020 mol HCl
- (C) 0.005 mol NaOH
- (D) 0.020 mol NaOH
Answer and explanation
Answer: (B). Added HCl is used up by A⁻, and there is only 0.010 mol of A⁻. Adding 0.020 mol HCl uses all of it and leaves 0.010 mol of excess H₃O⁺, so the pH falls sharply.
- (A) is less than the 0.010 mol of A⁻, so it is neutralised. The pH falls, but much less.
- (C) and (D) are both less than the 0.040 mol of HA, so the buffer neutralises them. (D) shifts the ratio noticeably, but HA is not used up and no excess OH⁻ remains, so the change is much smaller than in (B). This HA-rich buffer has a large capacity for added base.
Question 3 (multiple choice · core)
A buffer is 200. mL of solution that is 0.25 M in HA and 0.15 M in NaA. What is the maximum amount of NaOH it can neutralise before the HA is used up?
- (A) 0.030 mol
- (B) 0.050 mol
- (C) 0.080 mol
- (D) 0.25 mol
Answer and explanation
Answer: (B). Added base reacts with HA, 1 : 1. n(HA) = 0.200 L × 0.25 mol L⁻¹ = 0.050 mol.
- (A) is the amount of A⁻ (0.200 × 0.15). That limits added acid, not base.
- (C) adds HA and A⁻ together. A⁻ cannot neutralise added base.
- (D) is the concentration of HA, not the amount. You must multiply by the volume.
Question 4 (multiple choice · core)
Four 1.0 L buffers are made from the same weak acid HA and its salt NaA. Which has the greatest capacity for added acid?
- (A) 0.50 M HA and 0.10 M NaA
- (B) 0.20 M HA and 0.20 M NaA
- (C) 0.10 M HA and 0.30 M NaA
- (D) 0.050 M HA and 0.050 M NaA
Answer and explanation
Answer: (C). Added acid is neutralised by A⁻. Buffer (C) has the most A⁻: 0.30 mol.
- (A) has the largest total amount (0.60 mol), but only 0.10 mol of A⁻. It is strong against added base, weak against added acid.
- (B) has the same total as (C) (0.40 mol), but only 0.20 mol of A⁻.
- (D) has the least of everything.
Question 5 (calculation · core)
A buffer is made of 250. mL of solution that is 0.20 M NH₃ and 0.20 M NH₄Cl.
(a) Calculate the pH. (b) Calculate the maximum amount of HCl, and the maximum amount of NaOH, the buffer can neutralise. (c) 0.060 mol of HCl is added. Describe and explain what happens to the pH. (d) Describe how to make 250. mL of a buffer with the same pH but twice the capacity.
Worked solution
(a) [NH₃]/[NH₄⁺] = 1, so pH = pKa(NH₄⁺) = 9.25.
(b) n(NH₃) = n(NH₄⁺) = 0.250 L × 0.20 mol L⁻¹ = 0.050 mol. Added HCl reacts with the base, NH₃ + H₃O⁺ → NH₄⁺ + H₂O: at most 0.050 mol HCl. Added NaOH reacts with the acid, NH₄⁺ + OH⁻ → NH₃ + H₂O: at most 0.050 mol NaOH.
(c) 0.060 mol exceeds the 0.050 mol of NH₃. All the NH₃ is used up and 0.010 mol of H₃O⁺ is left over with nothing to react with it. The buffer’s capacity has been exceeded, so the pH falls sharply, far below 9.25 (into the acidic range).
(d) Use 0.40 M NH₃ and 0.40 M NH₄Cl (0.10 mol of each in 250. mL). The ratio is still 1, so the pH stays 9.25, and there are twice as many moles of each component.
Suggested mark points (4): 1 for pH 9.25 with the reason (ratio 1); 1 for 0.050 mol in both directions, matched to the correct component; 1 for (c) linking the sharp fall to the excess H₃O⁺ after NH₃ runs out; 1 for (d) doubling both concentrations at the same ratio.
Question 6 (constructed response · core)
A student measures the acid capacity of a commercial buffer. She titrates 20.00 mL samples with HCl labelled 0.200 M and records the volume needed to lower the pH by exactly 1.00 unit. The mean volume is 12.40 mL.
(a) Calculate the amount of HCl absorbed by one sample, and the capacity per litre of buffer. (b) Before filling the burette she rinsed it with distilled water but not with the HCl. State whether her calculated capacity is too high, too low or unaffected, and explain. (c) Her pH meter reads 0.15 units too low throughout. State the effect on her calculated capacity, and explain. (d) A second bottle of the same buffer had been diluted with an equal volume of water by mistake. Predict how its measured starting pH and its measured capacity per 20.00 mL sample compare with the first bottle’s.
Worked solution and suggested Marlbridge rubric
(a) n(HCl) = 0.01240 L × 0.200 mol L⁻¹ = 0.00248 mol per sample. Capacity per litre = 0.00248 mol ÷ 0.02000 L = 0.124 mol L⁻¹.
(b) Too high. Water left in the burette dilutes the HCl, so each millilitre delivers fewer moles than 0.200 M. She needs a larger volume to reach the same pH change, and multiplying that larger volume by 0.200 M overstates the moles absorbed.
(c) Unaffected. A constant offset shifts every reading by the same amount, so the change of 1.00 unit is still measured correctly and the endpoint volume is the same.
(d) Starting pH: about the same, because dilution does not change [A⁻]/[HA]. Capacity per sample: about half, because each 20.00 mL now contains half as many moles of the buffer components, so about 6.2 mL of HCl would be needed.
| Point | What earns it |
|---|---|
| 1 | 0.00248 mol HCl per sample |
| 1 | 0.124 mol L⁻¹ capacity |
| 1 | (b) too high, with the diluted-titrant reasoning |
| 1 | (c) unaffected, because a constant offset cancels in a pH change |
| 1 | (d) pH unchanged because the ratio is unchanged |
| 1 | (d) capacity halved because each sample holds half the moles |
Question 7 (argument · stretch)
A student claims: “Buffer A (0.010 M HA, 0.010 M NaA) and Buffer B (0.10 M HA, 0.10 M NaA) are equally good buffers, because they have the same pH.”
(a) Evaluate the claim. (b) The student now needs 0.500 L of an acetate buffer with pH 5.06 that can neutralise at least 0.030 mol of NaOH while still holding its pH. Find the minimum concentrations of CH₃COOH and CH₃COONa needed. (c) Will the buffer in (b) be better at resisting added acid or added base? Explain.
Worked solution and suggested Marlbridge rubric
(a) The claim is partly right and partly wrong. The pH values are equal because the ratio [A⁻]/[HA] = 1 in both. But for equal volumes, B contains ten times as many moles of HA and A⁻, so it can neutralise ten times as much added acid or base. B is the better buffer: same pH, greater capacity.
(b) Ratio: log([A⁻]/[HA]) = 5.06 − 4.76 = 0.30, so [A⁻]/[HA] = 10^0.30 = 2.0. Added NaOH is used up by CH₃COOH, so n(CH₃COOH) must be at least 0.030 mol: [CH₃COOH] ≥ 0.030 mol ÷ 0.500 L = 0.060 M. Then [CH₃COONa] = 2.0 × 0.060 M = 0.12 M. (In practice, use somewhat more than these minimums, because the pH drifts before the acid is completely used.)
(c) Better against added acid. It contains twice as much conjugate base (0.060 mol) as weak acid (0.030 mol), and added acid is absorbed by the conjugate base.
| Point | What earns it |
|---|---|
| 1 | Agrees the pH is the same, giving the equal ratio as the reason |
| 1 | Rejects “equally good”: B has more moles, so greater capacity |
| 1 | Ratio of 2.0 from pH − pKa = 0.30 |
| 1 | 0.060 M CH₃COOH from the NaOH requirement, linked to HA as the component that reacts with base |
| 1 | 0.12 M CH₃COONa |
| 1 | (c) resists acid better, because there is more A⁻ than HA |
How did you do?
- Q1 or Q7(a) wrong: re-read “Same ratio, more concentrated” in the study guide.
- Q2, Q3 or Q4 wrong: practise the short method: moles first, then match acid to A⁻ and base to HA.
- Q6 incomplete: for each error, ask “does it change the ratio, the amounts, or neither?” (Worked example 3).
Then tick off the topic checklist.