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: 25 °C; acetic acid K_a = 1.8 × 10⁻⁵ (pK_a 4.74). HF, HCOOH (methanoic acid), CH₃CH₂COOH (propanoic acid) and CH₃NH₂ (methylamine) are weak; HCl, HNO₃ and NaOH are strong. Laboratory readings in Question 6 are invented for practice.
Question 1 (multiple choice · foundation)
Which mixture, dissolved in water, is a buffer?
- (A) HNO₃ and NaNO₃
- (B) HCOOH and HCOONa
- (C) NaOH and NaCl
- (D) HCl and NH₄Cl
Answer and explanation
Answer: (B). Methanoic acid is a weak acid and HCOO⁻ (from HCOONa) is its conjugate base. Both members of the pair are present in large amounts.
- (A) HNO₃ is a strong acid. NO₃⁻ is too weak a base to remove added H₃O⁺.
- (C) contains a strong base and a neutral salt; there is no weak conjugate pair.
- (D) contains a strong acid and the weak acid NH₄⁺, but no NH₃. There is no base member to remove added acid.
Question 2 (multiple choice · core)
A buffer contains methylamine, CH₃NH₂, and methylammonium chloride, CH₃NH₃Cl. Which equation shows the main reaction when a small amount of NaOH is added?
- (A) CH₃NH₂(aq) + H₃O⁺(aq) → CH₃NH₃⁺(aq) + H₂O(l)
- (B) CH₃NH₃⁺(aq) + OH⁻(aq) → CH₃NH₂(aq) + H₂O(l)
- (C) CH₃NH₂(aq) + OH⁻(aq) → CH₃NH⁻(aq) + H₂O(l)
- (D) Cl⁻(aq) + H₂O(l) → HCl(aq) + OH⁻(aq)
Answer and explanation
Answer: (B). Added OH⁻ is removed by the acid member of the buffer, CH₃NH₃⁺.
- (A) is the reaction for added acid, not added base.
- (C) has a base reacting with a base. Methylamine has no tendency to give up a proton to OH⁻ in this course’s model.
- (D) Cl⁻ is a spectator ion; it is the conjugate base of a strong acid and does not react with water.
Question 3 (multiple choice · core)
0.20 mol of HF and 0.10 mol of NaOH are dissolved together in 1.0 L of water. Which best describes the solution after reaction?
- (A) A buffer containing 0.10 mol HF and 0.10 mol F⁻
- (B) A solution of NaF only, which is not a buffer
- (C) A buffer containing 0.20 mol HF and 0.10 mol F⁻
- (D) A solution containing 0.10 mol F⁻ and 0.10 mol OH⁻, which is not a buffer
Answer and explanation
Answer: (A). The strong base reacts completely: HF + OH⁻ → F⁻ + H₂O. 0.10 mol of OH⁻ converts 0.10 mol of HF into F⁻, leaving 0.20 − 0.10 = 0.10 mol HF. Both members of the pair remain, so it is a buffer.
- (B) would need equal moles of HF and NaOH.
- (C) forgets that the F⁻ was made by using up HF.
- (D) has the OH⁻ surviving, but OH⁻ is the limiting reactant and is used up completely.
Question 4 (multiple choice · core)
A buffer contains 0.050 mol of a weak acid HA and 0.050 mol of its conjugate base A⁻. 0.010 mol of NaOH is added. What are the amounts after reaction?
- (A) 0.040 mol HA and 0.060 mol A⁻
- (B) 0.060 mol HA and 0.040 mol A⁻
- (C) 0.050 mol HA, 0.050 mol A⁻ and 0.010 mol OH⁻
- (D) 0.040 mol HA and 0.050 mol A⁻
Answer and explanation
Answer: (A). HA + OH⁻ → A⁻ + H₂O. Each mole of OH⁻ removes one mole of HA and makes one mole of A⁻.
- (B) is the result for adding 0.010 mol of strong acid.
- (C) assumes no reaction, which is exactly what a buffer prevents.
- (D) removes HA but forgets that A⁻ is formed.
Question 5 (calculation · core)
A buffer contains 0.150 mol of propanoic acid, CH₃CH₂COOH, and 0.100 mol of sodium propanoate, CH₃CH₂COONa.
(a) 0.030 mol of HCl is added. Write the net ionic equation and give the amounts of CH₃CH₂COOH and CH₃CH₂COO⁻ afterwards. (b) 0.030 mol of NaOH is added to a fresh sample of the same buffer. Write the net ionic equation and give the amounts afterwards. (c) What is the greatest amount of HCl that this buffer can react with before one member is used up? Describe what would happen to the pH if more HCl were added beyond that point.
Worked solution
(a) CH₃CH₂COO⁻(aq) + H₃O⁺(aq) → CH₃CH₂COOH(aq) + H₂O(l) CH₃CH₂COOH: 0.150 + 0.030 = 0.180 mol. CH₃CH₂COO⁻: 0.100 − 0.030 = 0.070 mol.
(b) CH₃CH₂COOH(aq) + OH⁻(aq) → CH₃CH₂COO⁻(aq) + H₂O(l) CH₃CH₂COOH: 0.150 − 0.030 = 0.120 mol. CH₃CH₂COO⁻: 0.100 + 0.030 = 0.130 mol.
(c) Added acid is removed by propanoate, of which there is 0.100 mol. After that, no base member is left. Any further HCl stays as free H₃O⁺, because nothing is left to remove it, so the pH falls sharply.
Suggested mark points (5): 1 for each correct equation in (a) and (b) (2 total); 1 for both sets of amounts correct; 1 for 0.100 mol in (c); 1 for explaining that extra H₃O⁺ is no longer removed, so the pH drops sharply.
Common error: using the 0.150 mol of acid as the limit for added HCl. Added acid reacts with the base member.
Question 6 (constructed response · core)
A student sets up two beakers. Beaker 1 holds 50.0 mL of pure water. Beaker 2 holds 50.0 mL of a solution that is 0.10 M in acetic acid and 0.10 M in sodium acetate. The student adds 1.00 mL of 0.10 M NaOH to each and records the pH with a meter.
| Beaker | pH before | pH after |
|---|---|---|
| 1 (water) | 7.00 | 11.29 |
| 2 (acetic acid + sodium acetate) | 4.74 | 4.76 |
(a) Calculate the amount of OH⁻ added to each beaker, and the amounts of CH₃COOH and CH₃COO⁻ in beaker 2 before and after the addition. (b) Using particle-level reasoning, explain why the pH of beaker 1 rose by about 4.3 units while the pH of beaker 2 rose by only about 0.02 units. (c) Predict the effect of adding 1.00 mL of 0.10 M HCl to a fresh beaker 2 instead, and write the equation for the reaction.
Worked solution and suggested Marlbridge rubric
(a) OH⁻ added = 0.10 mol L⁻¹ × 0.00100 L = 1.0 × 10⁻⁴ mol in each beaker. Beaker 2 before: CH₃COOH = CH₃COO⁻ = 0.10 mol L⁻¹ × 0.0500 L = 5.0 × 10⁻³ mol each. After: CH₃COOH + OH⁻ → CH₃COO⁻ + H₂O, so CH₃COOH = 4.9 × 10⁻³ mol and CH₃COO⁻ = 5.1 × 10⁻³ mol.
(b) In water nothing reacts with the added OH⁻, so all 1.0 × 10⁻⁴ mol stays free in about 51 mL. [OH⁻] jumps from 10⁻⁷ M to about 2 × 10⁻³ M, so the pH rises sharply. In the buffer, the acetic acid molecules react with the OH⁻ ions and remove them, forming acetate ions and water. The strong base is replaced by acetate, a weak base. Only 2% of the acetic acid is used, so both members remain in large and nearly equal amounts. The ratio [CH₃COO⁻]/[CH₃COOH] changes only from 1.00 to about 1.04, so the pH changes very little.
(c) The pH falls only slightly (to about 4.73; calculating this value is not required in this course). Acetate removes the added acid: CH₃COO⁻(aq) + H₃O⁺(aq) → CH₃COOH(aq) + H₂O(l).
| Point | What earns it |
|---|---|
| 1 | 1.0 × 10⁻⁴ mol OH⁻ and 5.0 × 10⁻³ mol of each buffer member |
| 1 | Correct amounts after the addition (4.9 × 10⁻³ and 5.1 × 10⁻³ mol) |
| 1 | Water: no species reacts with OH⁻, so it remains free and [OH⁻] rises greatly |
| 1 | Buffer: CH₃COOH reacts with (removes) the added OH⁻, replacing it with a weak base |
| 1 | Both members remain in large amounts, so their ratio (and pH) changes only slightly |
| 1 | (c) small decrease in pH with the correct equation showing acetate reacting with H₃O⁺ |
A numerical pH in (c) is not required. Accept “falls slightly” with the correct equation.
Question 7 (explanation · stretch)
Two students make claims.
- Student X: “A solution containing 0.10 M HCl and 0.10 M NaCl is a buffer, because it contains an acid and its conjugate base.”
- Student Y: “A 0.10 M solution of acetic acid on its own is a good buffer, because the acetic acid can react with any base that is added.”
Evaluate each claim.
Model answer and suggested Marlbridge rubric
Student X is wrong. HCl and Cl⁻ are a conjugate pair, but HCl is a strong acid, so Cl⁻ is a negligibly weak base. Cl⁻ does not react with added H₃O⁺, so added acid stays free and the pH falls. (Added base would simply neutralize some HCl, and the solution’s pH would still be set by a strong acid with no buffering pair.) A buffer needs a weak acid and its conjugate base, both in large amounts.
Student Y is only half right. Acetic acid can remove added OH⁻ (CH₃COOH + OH⁻ → CH₃COO⁻ + H₂O). But a solution of acetic acid alone contains very little acetate, because the acid is only slightly ionized. There is almost nothing to remove added H₃O⁺, so the solution does not buffer against acid. It also starts well away from its pK_a, where small additions change the ratio a lot. Adding sodium acetate (or partly neutralizing the acid with NaOH) would make it a buffer.
| Point | What earns it |
|---|---|
| 1 | X: Cl⁻ is the conjugate base of a strong acid and is too weak to react with added H₃O⁺ |
| 1 | X: so the mixture is not a buffer / a buffer needs a weak acid–conjugate base pair |
| 1 | Y: acetic acid does remove added OH⁻ (with equation or clear description) |
| 1 | Y: almost no CH₃COO⁻ is present, so added H₃O⁺ is not removed |
| 1 | Y: a valid fix, such as adding sodium acetate or partly neutralizing with NaOH |
How did you do?
- Q1, Q3 or Q7 wrong: revisit “How buffers are made” and Worked example 2 in the study guide.
- Q2, Q4 or Q5 wrong: practise the two reactions and the bar diagram (Figure 1); remember which member removes which.
- Q6 incomplete: your explanation needs the particle step (which species reacts with the added ion) as well as the numbers.
Then tick off the topic checklist.