These are original Marlbridge practice questions, not past exam questions. The acids and bases named HZ, HW, B and H₂R are fictional. The mark points are a suggested Marlbridge rubric to help you check your work; they are not an official scoring guideline. Data for every question: Kw = 1.0 × 10⁻¹⁴ at 25 °C. All titration reactions go to completion and volumes add.
Question 1 (multiple choice · foundation)
A solution of acetic acid is titrated with sodium hydroxide. Why is the pH at the equivalence point greater than 7?
- (A) Excess hydroxide ions remain in the flask at the equivalence point.
- (B) Acetate ions, the conjugate base of acetic acid, accept protons from water and produce OH⁻ ions.
- (C) Sodium ions react with water to form sodium hydroxide.
- (D) Some acetic acid molecules have not yet reacted at the equivalence point.
Answer and explanation
Answer: (B). At equivalence the acid and the added base have both been used up, leaving acetate ions. Acetate is a weak base: CH₃COO⁻ + H₂O ⇌ CH₃COOH + OH⁻, so the solution is basic.
- (A) By definition there is no excess OH⁻ at equivalence; that describes the region after it.
- (C) Na⁺ is the cation of a strong base and does not react with water to any useful extent.
- (D) Unreacted acid would make the solution more acidic, not basic, and at equivalence the acid has been used up.
Question 2 (multiple choice · core)
A 20.00 mL sample of the weak acid HZ is titrated with 0.100 M NaOH. The equivalence point is at 32.00 mL. Some readings are:
| Volume of NaOH (mL) | 0.00 | 8.00 | 16.00 | 24.00 | 32.00 | 40.00 |
|---|---|---|---|---|---|---|
| pH | 2.33 | 3.39 | 3.86 | 4.34 | 8.33 | 12.12 |
What is Ka for HZ?
- (A) 4.7 × 10⁻³
- (B) 4.1 × 10⁻⁴
- (C) 1.4 × 10⁻⁴
- (D) 4.7 × 10⁻⁹
Answer and explanation
Answer: (C). Half of 32.00 mL is 16.00 mL. There [HZ] = [Z⁻], so pH = pKa = 3.86 and Ka = 10^(−3.86) = 1.4 × 10⁻⁴.
- (A) is 10^(−2.33), from the starting pH. That is [H₃O⁺] of the acid alone, not Ka.
- (B) is 10^(−3.39), from 8.00 mL, which is a quarter of the way to equivalence, not halfway.
- (D) is 10^(−8.33), from the equivalence point, where almost no HZ is left.
Question 3 (multiple choice · core)
A 20.00 mL sample of a weak base is titrated with 0.0750 M HCl. The pH curve has its steepest point at 26.40 mL. What was the concentration of the base?
- (A) 0.0495 M
- (B) 0.0568 M
- (C) 0.0990 M
- (D) 0.198 M
Answer and explanation
Answer: (C). n(HCl) = 26.40 mL × 0.0750 M = 1.980 mmol. The base accepts one proton, so n(base) = 1.980 mmol, and 1.980 mmol ÷ 20.00 mL = 0.0990 M.
- (A) uses the half-equivalence volume, 13.20 mL, as if it were the equivalence volume.
- (B) inverts the volume ratio: 0.0750 × 20.00 ÷ 26.40.
- (D) doubles the moles, as if each base particle needed two protons.
Question 4 (multiple choice · core)
A diprotic acid H₂Y is titrated with NaOH. The first equivalence point is at 15.0 mL and the second at 30.0 mL. Which species are present in the largest amounts after 22.5 mL of NaOH has been added?
- (A) H₂Y and HY⁻, in roughly equal amounts
- (B) HY⁻ and Y²⁻, in roughly equal amounts
- (C) Y²⁻ only
- (D) H₂Y and Y²⁻, in roughly equal amounts
Answer and explanation
Answer: (B). By 15.0 mL all H₂Y has become HY⁻. 22.5 mL is halfway between the two equivalence points, so half of the HY⁻ has become Y²⁻. That is also why pH ≈ pKa₂ at this point.
- (A) describes 7.5 mL, halfway to the first equivalence point.
- (C) describes the second equivalence point at 30.0 mL.
- (D) H₂Y and Y²⁻ are never both major species: H₂Y is used up before much Y²⁻ forms.
Question 5 (calculation · core)
A 40.00 mL sample of 0.0500 M weak base B (Kb = 2.0 × 10⁻⁵) is titrated with 0.100 M HCl.
(a) Calculate the volume of HCl needed to reach the equivalence point. (b) Calculate the pH before any HCl is added. (c) Calculate the pH after 15.00 mL of HCl has been added. (d) Calculate the pH at the equivalence point. (e) Calculate the pH after 25.00 mL of HCl has been added.
Worked solution and suggested Marlbridge rubric
(a) n(B) = 40.00 mL × 0.0500 M = 2.000 mmol. V = 2.000 mmol ÷ 0.100 M = 20.00 mL.
(b) [OH⁻] ≈ √(Kb × [B]) = √(2.0 × 10⁻⁵ × 0.0500) = 1.0 × 10⁻³ M, so pOH = 3.00 and pH = 11.00.
(c) 1.500 mmol H₃O⁺ reacts with 1.500 mmol B: 0.500 mmol B and 1.500 mmol BH⁺ remain (a buffer). pKa(BH⁺) = 14.00 − pKb = 14.00 − 4.70 = 9.30. pH = 9.30 + log(0.500 / 1.500) = 9.30 − 0.48 = 8.82.
(d) 2.000 mmol BH⁺ in 60.00 mL: [BH⁺] = 0.03333 M. Ka(BH⁺) = Kw ÷ Kb = 5.0 × 10⁻¹⁰. [H₃O⁺] ≈ √(5.0 × 10⁻¹⁰ × 0.03333) = 4.08 × 10⁻⁶ M, pH = 5.39.
(e) 2.500 − 2.000 = 0.500 mmol excess H₃O⁺ in 65.00 mL: [H₃O⁺] = 7.69 × 10⁻³ M, pH = 2.11.
| Point | What earns it |
|---|---|
| 1 | 20.00 mL from the 1:1 mole ratio |
| 1 | pH 11.00 from the Kb equilibrium |
| 1 | Correct amounts of B and BH⁺ after reaction at 15.00 mL |
| 1 | pH 8.82, using pKa of BH⁺ (or Kb with pOH) |
| 1 | pH 5.39 using Ka = Kw/Kb and the total volume of 60.00 mL |
| 1 | pH 2.11 from excess H₃O⁺ in 65.00 mL |
Common error in (d): using 40.00 mL instead of 60.00 mL gives pH 5.30. In (c), working with pKb gives pOH = 4.70 + log(1.500 / 0.500) = 5.18 and the same pH, 8.82; accept either route.
Question 6 (constructed response · core)
Two flasks each hold 25.00 mL of a 0.100 M acid. Flask 1 holds HCl. Flask 2 holds a weak acid HW (Ka = 1.0 × 10⁻⁵). Each is titrated with 0.100 M NaOH.
A student claims: “Flask 2 will need less NaOH to reach the equivalence point, because HW is only partly ionised.”
(a) Evaluate the claim. (b) Calculate the pH in each flask after 12.50 mL of NaOH has been added. (c) The equivalence-point pH is 7.00 for flask 1 and 8.85 for flask 2. Explain the difference in terms of the species present.
Worked solution and suggested Marlbridge rubric
(a) The claim is incorrect. Each flask holds 2.500 mmol of acid, so each needs 2.500 mmol of OH⁻, which is 25.00 mL. As OH⁻ removes HW, the equilibrium HW ⇌ H⁺ + W⁻ shifts right (and OH⁻ reacts with HW directly), so all of the HW is used. The degree of ionisation before titration does not change the amount of base needed.
(b) Flask 1: 2.500 − 1.250 = 1.250 mmol H₃O⁺ in 37.50 mL, [H₃O⁺] = 0.03333 M, pH = 1.48. Flask 2: 1.250 mmol HW and 1.250 mmol W⁻ remain. This is the half-equivalence point, so pH = pKa = 5.00.
(c) In flask 1 the only ions left are Na⁺ and Cl⁻. Neither reacts with water, so the pH is 7.00. In flask 2 the solution holds W⁻, the conjugate base of a weak acid. W⁻ accepts protons from water, W⁻ + H₂O ⇌ HW + OH⁻, so the solution is basic.
| Point | What earns it |
|---|---|
| 1 | States the claim is wrong, with both flasks needing 25.00 mL (equal moles of acid) |
| 1 | Explains that the weak acid equilibrium shifts as OH⁻ reacts, so all HW is neutralised |
| 1 | pH 1.48 for flask 1, using excess H₃O⁺ and the total volume |
| 1 | pH 5.00 for flask 2, identified as pH = pKa at half-equivalence |
| 1 | Explains the basic equivalence point by W⁻ reacting with water to make OH⁻, and the neutral one by spectator ions |
Do not award the first point for “the claim is wrong” without the equal-moles reason.
Question 7 (data analysis · stretch)
A 10.00 mL sample of an acid H₂R is titrated with 0.200 M NaOH. The curve shows two steep rises, centred at 12.50 mL and 25.00 mL. Selected readings:
| Volume of NaOH (mL) | 0.00 | 6.25 | 11.50 | 12.50 | 13.50 | 18.75 | 24.00 | 25.00 | 26.00 | 30.00 |
|---|---|---|---|---|---|---|---|---|---|---|
| pH | 2.45 | 4.30 | 5.35 | 6.30 | 7.25 | 8.30 | 9.36 | 10.58 | 11.75 | 12.40 |
(a) How many acidic protons does H₂R give up in this titration? Justify your answer from the data. (b) Calculate the concentration of H₂R. (c) Estimate pKa₁ and pKa₂. (d) Identify the major acid-containing species at 12.50 mL and at 18.75 mL. (e) Explain why the pH at 25.00 mL is above 7.
Worked solution and suggested Marlbridge rubric
(a) Two. There are two steep rises (around 12.50 mL and 25.00 mL), and the second comes after exactly the same extra volume as the first. Each rise marks the removal of one proton from every acid molecule.
(b) Use the first equivalence point: 12.50 mL × 0.200 M = 2.50 mmol OH⁻ removes one proton from each H₂R, so n(H₂R) = 2.50 mmol and [H₂R] = 2.50 mmol ÷ 10.00 mL = 0.250 M. (Using 25.00 mL with a 1:1 ratio would give 0.500 M, twice the true value.)
(c) Halfway to the first equivalence point (6.25 mL), pH ≈ pKa₁ = 4.30. Halfway between the two equivalence points (18.75 mL), pH ≈ pKa₂ = 8.30.
(d) At 12.50 mL: HR⁻ (nearly all of it). At 18.75 mL: HR⁻ and R²⁻ in roughly equal amounts.
(e) At 25.00 mL both protons have been removed, so the main species is R²⁻, a weak base. It accepts protons from water, R²⁻ + H₂O ⇌ HR⁻ + OH⁻, making the solution basic.
| Point | What earns it |
|---|---|
| 1 | Two protons, justified by two jumps with equal volume spacing |
| 1 | 0.250 M from the first equivalence volume (or half the second) |
| 1 | pKa₁ ≈ 4.30 and pKa₂ ≈ 8.30, read at the correct volumes |
| 1 | HR⁻ at 12.50 mL; HR⁻ and R²⁻ roughly equal at 18.75 mL |
| 1 | R²⁻ reacts with water to produce OH⁻ |
You are not asked to calculate the concentration of every species along this curve; that is outside the course. Reasoning about which species are major is what matters.
How did you do?
- Q1, Q6(c) or Q7(e) wrong: re-read “What sets the pH at the equivalence point” in the study guide.
- Q2 or Q3 wrong: check you halve the volume for the half-equivalence point and use the full equivalence volume for moles.
- Q4 or Q7 wrong: work through Figure 2 and Worked example 3 in the guide.
- Q5 incomplete: compare your moles tables with Worked examples 1 and 2.
Then tick off the topic checklist.