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 for every question: O 16.00, S 32.06, Cl 35.45, Ba 137.33 g mol⁻¹. All titration reactions go to completion.
Question 1 (multiple choice · foundation)
Which statement best describes the equivalence point of a titration?
- (A) The point at which the indicator changes colour.
- (B) The point at which the volume of titrant added equals the volume of analyte solution.
- (C) The point at which the amount of titrant added has exactly used up the analyte, in the mole ratio of the balanced equation.
- (D) The point at which the solution in the flask always has a pH of 7.
Answer and explanation
Answer: (C). The equivalence point is defined by amounts: all of the analyte has reacted and no extra titrant has been added.
- (A) describes the endpoint, the observed change. A good method makes it close to the equivalence point, but they are not the same thing.
- (B) confuses volumes with moles. The volumes are equal only by coincidence.
- (D) is true only for a strong acid with a strong base, and many titrations are not acid–base at all.
Question 2 (multiple choice · core)
A 20.00 mL sample of sodium hydroxide solution is titrated with 0.150 M nitric acid, HNO₃. The equivalence point is reached after 16.40 mL of acid. What is the concentration of the sodium hydroxide?
- (A) 0.0615 M
- (B) 0.123 M
- (C) 0.183 M
- (D) 0.246 M
Answer and explanation
Answer: (B). HNO₃ + NaOH → NaNO₃ + H₂O is 1:1. n(HNO₃) = 0.150 × 0.01640 = 0.00246 mol = n(NaOH). c = 0.00246 ÷ 0.02000 = 0.123 M.
- (A) halves the answer, as if the ratio were 1:2.
- (C) puts the volumes the wrong way round (0.150 × 20.00 ÷ 16.40). The base needed less acid volume than its own volume, so it must be less concentrated than the acid.
- (D) doubles the answer, as if the ratio were 2:1.
Question 3 (multiple choice · core)
What volume of 0.200 M HCl is needed to reach the equivalence point with 15.0 mL of 0.100 M barium hydroxide, Ba(OH)₂?
- (A) 3.75 mL
- (B) 7.50 mL
- (C) 15.0 mL
- (D) 30.0 mL
Answer and explanation
Answer: (C). 2HCl + Ba(OH)₂ → BaCl₂ + 2H₂O. n(Ba(OH)₂) = 0.100 × 0.0150 = 0.00150 mol, so n(HCl) = 2 × 0.00150 = 0.00300 mol. V = 0.00300 ÷ 0.200 = 0.0150 L = 15.0 mL.
- (A) uses the ratio upside down (½ instead of 2).
- (B) uses M₁V₁ = M₂V₂, which assumes 1:1 and ignores that each Ba(OH)₂ releases two OH⁻.
- (D) multiplies by 2 twice.
Question 4 (calculation · core)
The chloride ion concentration in a fictional sample of well water is found by titration with silver nitrate. The silver ions precipitate the chloride: Ag⁺(aq) + Cl⁻(aq) → AgCl(s). A 10.00 mL sample needs 12.36 mL of 0.0500 M AgNO₃ to reach the equivalence point.
(a) Calculate the chloride concentration in mol L⁻¹. (b) Express it in milligrams of chloride per litre. (c) A second sample has a chloride concentration of 0.0450 M. Predict the volume of the same AgNO₃ solution needed for a 10.00 mL portion.
Worked solution
(a) n(Ag⁺) = 0.0500 × 0.01236 = 0.000618 mol = n(Cl⁻) (1:1). c = 0.000618 ÷ 0.01000 = 0.0618 M.
(b) 0.0618 mol L⁻¹ × 35.45 g mol⁻¹ = 2.191 g L⁻¹ = 2190 mg L⁻¹ (3 significant figures).
(c) n(Cl⁻) = 0.0450 × 0.01000 = 0.000450 mol, so V = 0.000450 ÷ 0.0500 = 0.00900 L = 9.00 mL.
Suggested mark points (3): 1 for moles of Ag⁺ with volume in litres and the 1:1 ratio; 1 for 2190 mg L⁻¹; 1 for 9.00 mL.
This is a precipitation titration: the same amount reasoning applies whatever type of reaction is used.
Question 5 (constructed response · core)
A student titrates a sample of acid with standard sodium hydroxide from a burette. For each mistake below, state whether the calculated concentration of the acid will be too high, too low or unchanged. Justify each answer.
(a) The burette was rinsed with distilled water but not with the NaOH solution before being filled. (b) The pipetted acid sample was run into a conical flask that still contained a few drops of distilled water. (c) There was an air bubble in the burette tip at the start, which was pushed out during the titration.
Worked solution and suggested Marlbridge rubric
(a) Too high. Water left in the burette dilutes the NaOH, so each millilitre carries fewer moles of OH⁻. A larger volume is needed to neutralise the acid, but the calculation uses the original (higher) concentration, so it overestimates the moles of NaOH and therefore of acid.
(b) Unchanged. The pipette delivered the same moles of acid. Extra water in the flask changes the concentration in the flask, not the moles, so the same volume of NaOH is needed.
(c) Too high. The bubble’s volume is counted in the burette reading, but no NaOH was delivered into the flask in its place. The recorded volume is larger than the volume that reacted, so the calculated moles of NaOH, and of acid, are too large.
| Point | What earns it |
|---|---|
| 1 | (a) too high, linked to diluted titrant needing a larger recorded volume |
| 1 | (b) unchanged, linked to moles of analyte not changing |
| 1 | (c) too high, linked to the recorded volume exceeding the volume actually delivered |
No mark for a direction without a reason based on moles or recorded volume.
Question 6 (graphing and analysis · core)
A conductivity probe is placed in 25.00 mL of a barium hydroxide solution of unknown concentration. A 0.100 M sulfuric acid solution is added from a burette:
Ba(OH)₂(aq) + H₂SO₄(aq) → BaSO₄(s) + 2H₂O(l)
The probe gives these readings (fictional data, arbitrary units):
| Volume of H₂SO₄ added (mL) | 0.0 | 4.0 | 8.0 | 12.0 | 18.0 | 22.0 | 26.0 | 30.0 |
|---|---|---|---|---|---|---|---|---|
| Conductivity reading | 4.50 | 3.30 | 2.10 | 0.90 | 1.05 | 2.45 | 3.85 | 5.25 |
(a) Plot the data with suitable axes, labels and units. Draw a straight line through each set of points and find the volume at which they meet. (b) Explain, in terms of ions, why the conductivity falls and then rises. (c) Calculate the concentration of the barium hydroxide solution. (d) Calculate the mass of barium sulfate formed at the equivalence point.
Worked solution and suggested Marlbridge rubric
(a) The first four points lie on a straight line falling by 0.30 units per mL; the last four lie on a straight line rising by 0.35 units per mL. Extending both lines, they meet at 15.0 mL, where the conductivity is close to zero.
(b) Before equivalence, each H₂SO₄ added removes ions: Ba²⁺ and SO₄²⁻ leave the solution as solid BaSO₄, and H⁺ and OH⁻ combine to form water. Fewer ions means lower conductivity. At equivalence almost no ions remain. After equivalence, the extra acid adds H⁺ and SO₄²⁻ ions that have nothing to react with, so the conductivity rises again.
(c) n(H₂SO₄) = 0.100 × 0.0150 = 0.00150 mol = n(Ba(OH)₂) (1:1). c = 0.00150 ÷ 0.02500 = 0.0600 M.
(d) n(BaSO₄) = 0.00150 mol; M(BaSO₄) = 137.33 + 32.06 + 4(16.00) = 233.39 g mol⁻¹; mass = 0.350 g.
| Point | What earns it |
|---|---|
| 1 | Axes labelled with quantity and unit (volume in mL; conductivity with its unit or “arbitrary units”), with even scales |
| 1 | Two straight lines drawn and the equivalence volume read as 15.0 mL (accept 14.8 to 15.2 mL) |
| 1 | Falling conductivity explained by ions being removed as BaSO₄(s) and H₂O |
| 1 | Rising conductivity explained by excess H⁺ and SO₄²⁻ ions |
| 1 | 0.0600 M, using the 1:1 ratio and the sample volume (carry forward from the graph reading) |
| 1 | 0.350 g of BaSO₄ |
Question 7 (planning and explanation · stretch)
A student must titrate 25.00 mL of hydrochloric acid that is roughly 0.08 M, using a 50.00 mL burette. Three sodium hydroxide solutions are available: 0.500 M, 0.100 M and 0.0100 M.
(a) For each solution, predict the volume needed to reach the equivalence point. (b) Choose the best solution and justify your choice. Assume each burette reading can be off by about 0.05 mL. (c) Describe the particles present in the flask at the equivalence point, apart from water, and explain why H⁺ is no longer present in a significant amount.
Model answer and suggested Marlbridge rubric
(a) n(HCl) ≈ 0.02500 × 0.08 = 0.0020 mol, and the reaction is 1:1. Volumes: 0.500 M → 4.0 mL; 0.100 M → 20 mL; 0.0100 M → 200 mL.
(b) Choose 0.100 M. The volume (about 20 mL) fits in one burette filling and is large enough that a 0.05 mL reading error is only about 0.25% of the titre. With 0.500 M, the same reading error is about 1.25% of only 4 mL. With 0.0100 M the burette would have to be refilled at least three times, adding more reading errors and taking much longer.
(c) Na⁺ and Cl⁻ ions (a sodium chloride solution). The OH⁻ added has reacted with the H⁺ in the 1:1 ratio, H⁺ + OH⁻ → H₂O, so at equivalence the moles of OH⁻ added equal the moles of H⁺ that were present and essentially none of either is left. Na⁺ and Cl⁻ are spectator ions and stay in solution throughout.
| Point | What earns it |
|---|---|
| 1 | All three volumes correct, from the 1:1 ratio |
| 1 | 0.100 M chosen, with a reason based on fitting the burette and keeping the relative reading error small |
| 1 | Na⁺ and Cl⁻ named as the main solute particles at equivalence |
| 1 | Explanation that H⁺ has all reacted with an equal number of moles of OH⁻ |
How did you do?
- Q1 wrong: re-read “Equivalence point and endpoint” in the study guide.
- Q2, Q3 or Q4 wrong: work through the four-step method and Worked example 1 again; check the mole ratio.
- Q5 incomplete: ask “does this change the volume of titrant I record?” for each mistake.
- Q6 or Q7 incomplete: compare your graph with Figure 1 and the particle table in the study guide.
Then tick off the topic checklist.