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. Experimental data in Question 6 are invented for practice. Strong acids you may assume: HCl, HBr, HI, HNO₃, HClO₄ and H₂SO₄ (first proton).
Question 1 (multiple choice · foundation)
What is the conjugate acid of the hydrogen phosphate ion, HPO₄²⁻?
- (A) PO₄³⁻
- (B) H₂PO₄⁻
- (C) H₃PO₄
- (D) H₂PO₄²⁻
Answer and explanation
Answer: (B). A conjugate acid has one more H⁺, so add one H and raise the charge by 1: HPO₄²⁻ + H⁺ → H₂PO₄⁻.
- (A) is the conjugate base (one H⁺ removed).
- (C) adds two protons; it is the conjugate acid of H₂PO₄⁻, not of HPO₄²⁻.
- (D) adds the H but leaves the charge unchanged. Adding a positive proton must raise the charge.
Question 2 (multiple choice · core)
CH₃NH₃⁺(aq) + CO₃²⁻(aq) ⇌ CH₃NH₂(aq) + HCO₃⁻(aq)
Which reactant acts as the Brønsted-Lowry base, and what is its conjugate acid?
- (A) CO₃²⁻; its conjugate acid is HCO₃⁻
- (B) CH₃NH₃⁺; its conjugate acid is CH₃NH₂
- (C) CO₃²⁻; its conjugate acid is CH₃NH₃⁺
- (D) CO₃²⁻; its conjugate acid is H₂CO₃
Answer and explanation
Answer: (A). CO₃²⁻ gains an H⁺ to become HCO₃⁻, so it is the base and HCO₃⁻ is its conjugate acid. CH₃NH₃⁺ donates the proton, so it is the acid and CH₃NH₂ is its conjugate base.
- (B) has the roles reversed: CH₃NH₃⁺ loses a proton, so it is the acid, and CH₃NH₂ is its conjugate base.
- (C) pairs CO₃²⁻ with the other acid in the equation. A conjugate pair must differ by only one H⁺.
- (D) adds two protons to CO₃²⁻. H₂CO₃ does not appear in this reaction.
Question 3 (multiple choice · core)
Two acids, HA and HB, are dissolved in separate beakers at the same concentration. A particle model of a small part of each solution (water not shown) contains:
| Solution | HA or HB molecules | H₃O⁺ ions | A⁻ or B⁻ ions |
|---|---|---|---|
| HA | 0 | 10 | 10 |
| HB | 9 | 1 | 1 |
Which statement is correct?
- (A) B⁻ is a stronger base than A⁻.
- (B) HB is the stronger acid, because more HB molecules are present.
- (C) The two solutions have the same H₃O⁺ concentration, because the acid concentrations are equal.
- (D) A⁻ is the stronger base, because there are more A⁻ ions than B⁻ ions.
Answer and explanation
Answer: (A). HA has ionised completely (10 of 10), so it is a strong acid. HB has ionised only 1 of 10 (10%), so it is weak. The weaker acid has the stronger conjugate base, so B⁻ is the stronger base: it keeps most of the protons attached.
- (B) Intact molecules show that HB does not give up its proton easily, which makes it the weaker acid.
- (C) The model shows 10 H₃O⁺ against 1 H₃O⁺. Equal acid concentrations do not give equal H₃O⁺ concentrations when the strengths differ.
- (D) The number of A⁻ ions shows how fully HA ionised, not how strongly A⁻ attracts protons. A⁻ stays unprotonated because it is a very weak base.
Question 4 (constructed response · core)
(a) Hydrocyanic acid, HCN, is a weak acid. Write the equation for its reaction with water. Label the acid, the base, the conjugate acid and the conjugate base. (b) The hypochlorite ion, ClO⁻, is a weak base. Write the equation for its reaction with water and label the four species in the same way. (c) Using your answers, explain what feature of the water molecule lets it play both roles.
Worked solution and suggested Marlbridge rubric
(a) HCN(aq) + H₂O(l) ⇌ H₃O⁺(aq) + CN⁻(aq) HCN = acid; H₂O = base; H₃O⁺ = conjugate acid (of H₂O); CN⁻ = conjugate base (of HCN).
(b) ClO⁻(aq) + H₂O(l) ⇌ HClO(aq) + OH⁻(aq) ClO⁻ = base; H₂O = acid; HClO = conjugate acid (of ClO⁻); OH⁻ = conjugate base (of H₂O).
(c) In (a) water accepts a proton, using one of the two lone pairs on its oxygen atom to bond to H⁺. In (b) water donates a proton: an H atom on a polar O–H bond leaves as H⁺. A molecule with both a lone pair to accept H⁺ and an H it can give away can act as a base or an acid, depending on its partner.
| Point | What earns it |
|---|---|
| 1 | Correct equation for (a) |
| 1 | All four labels correct in (a) |
| 1 | Correct equation for (b), with OH⁻ as a product |
| 1 | All four labels correct in (b) |
| 1 | (c) links lone pairs on O to accepting H⁺ and polar O–H bonds / available H to donating H⁺ |
If (a) is written as HCN(aq) ⇌ H⁺(aq) + CN⁻(aq), water’s role is hidden: award the equation point, but the labels point only if the answer also names H₂O as the base and H₃O⁺ as its conjugate acid.
Question 5 (constructed response · core)
The hydrogen oxalate ion, HC₂O₄⁻, is amphiprotic.
(a) Write the formula of its conjugate acid and of its conjugate base. (b) Write an equation for HC₂O₄⁻ acting as an acid with hydroxide ions. (c) Write an equation for HC₂O₄⁻ acting as a base with hydronium ions. (d) Explain why the oxalate ion, C₂O₄²⁻, can act as a Brønsted-Lowry base but not as a Brønsted-Lowry acid.
Worked solution and suggested Marlbridge rubric
(a) Conjugate acid: H₂C₂O₄. Conjugate base: C₂O₄²⁻.
(b) HC₂O₄⁻(aq) + OH⁻(aq) → C₂O₄²⁻(aq) + H₂O(l)
(c) HC₂O₄⁻(aq) + H₃O⁺(aq) → H₂C₂O₄(aq) + H₂O(l)
(d) An acid must donate a proton, and C₂O₄²⁻ has no hydrogen atoms to donate. It does have lone pairs on its oxygen atoms, so it can accept a proton (becoming HC₂O₄⁻) and act as a base.
| Point | What earns it |
|---|---|
| 1 | Both formulas in (a) with correct charges |
| 1 | Balanced equation (b), atoms and charge |
| 1 | Balanced equation (c), atoms and charge |
| 1 | (d) no H to donate, but lone pairs to accept H⁺ |
Question 6 (calculation · core)
A student prepares two solutions, each with an acid concentration of 0.050 M: perchloric acid, HClO₄, and hydrofluoric acid, HF. The student measures the H₃O⁺ concentration of the HF solution as 5.6 × 10⁻³ M (invented data).
(a) State the H₃O⁺ concentration in the HClO₄ solution, and explain your answer. (b) Calculate the percentage of HF molecules that have ionised. (c) A particle drawing of the HF solution is to show 25 HF units added. How many F⁻ ions and how many HF molecules should it show? (d) Which is the stronger base, ClO₄⁻ or F⁻? Justify your answer using (a) and (b).
Worked solution and suggested Marlbridge rubric
(a) HClO₄ is a strong acid, so it ionises essentially completely: each HClO₄ gives one H₃O⁺. [H₃O⁺] = 0.050 M.
(b) Each ionised HF gives one H₃O⁺, so the concentration of HF ionised is 5.6 × 10⁻³ M. Percentage = (5.6 × 10⁻³ ÷ 0.050) × 100 = 11% (11.2% before rounding).
(c) 11.2% of 25 = 2.8, so about 3 F⁻ ions (with 3 H₃O⁺) and 22 HF molecules.
(d) F⁻ is the stronger base. HClO₄ ionises completely, so its conjugate base ClO₄⁻ has a negligible tendency to accept a proton. HF ionises only about 11%, which shows that F⁻ pulls protons back strongly enough to keep most of the HF intact. The weaker acid (HF) has the stronger conjugate base (F⁻).
| Point | What earns it |
|---|---|
| 1 | 0.050 M with “strong acid / ionises completely” |
| 1 | 11% (accept 11.2%) with working |
| 1 | 3 F⁻ and 22 HF (accept 2.8 rounded to 3, with HF = 25 − 3) |
| 1 | F⁻ stronger, linked to HF being weaker (partial ionisation) than HClO₄ |
Common error: in (c), drawing 3 F⁻ but still 25 HF molecules. Each F⁻ comes from one HF, so the HF count must fall by the same number.
Question 7 (explanation · stretch)
A student writes: “Cl⁻ is the conjugate base of HCl, and F⁻ is the conjugate base of HF. So a solution of NaCl should be basic, just like a solution of NaF.” Evaluate the claim. Include an equation in your answer.
Model answer and suggested Marlbridge rubric
Model answer. The claim is only half right. Both ions are conjugate bases, but their strengths are very different. HCl is a strong acid: it gives its proton to water completely, so Cl⁻ has a negligible tendency to take a proton back. Cl⁻ therefore does not react with water to any useful extent, and a NaCl solution is not made basic by it. HF is a weak acid, so its conjugate base F⁻ is a weak base that accepts some protons from water:
F⁻(aq) + H₂O(l) ⇌ HF(aq) + OH⁻(aq)
The OH⁻ formed makes the NaF solution slightly basic. (Na⁺ is a spectator in both solutions.) The student has ignored the rule that the stronger the acid, the weaker its conjugate base.
| Point | What earns it |
|---|---|
| 1 | States that Cl⁻ is a negligibly weak base because HCl is a strong acid |
| 1 | States that F⁻ is a (weak) base because HF is a weak acid |
| 1 | Correct equation for F⁻ accepting a proton from water, giving OH⁻ |
| 1 | Concludes the claim is wrong for NaCl but right that NaF is basic, using the strength rule |
How did you do?
- Q1 or Q5(a) wrong: practise writing conjugates in the study guide: one H⁺ and a charge of 1 each time.
- Q2 or Q4 wrong: label all four species, following Figure 1 of the study guide.
- Q3 or Q6 wrong: revisit Worked example 3 and count particles before and after ionisation.
- Q7 incomplete: your answer needs the link between acid strength and conjugate base strength, not just the word “conjugate”.
Then tick off the topic checklist.