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. All solutions are at 25 °C: K_w = 1.0 × 10⁻¹⁴ and pH + pOH = 14.00. Molar masses: H 1.008, O 16.00, K 39.10, Ca 40.08 g mol⁻¹. Assume volumes add when solutions are mixed.
Question 1 (multiple choice · foundation)
What is the pH of 0.0040 M hydroiodic acid, HI?
- (A) 2.10
- (B) 2.40
- (C) 5.52
- (D) 11.60
Answer and explanation
Answer: (B). HI is a strong acid, so [H₃O⁺] = 0.0040 M and pH = −log(0.0040) = 2.40 (2 significant figures → 2 decimal places).
- (A) doubles the concentration (−log 0.0080), as if HI released two protons. It has only one.
- (C) uses the natural log (ln) instead of log to base 10.
- (D) is the pOH. An acid solution must have pH below 7.
Question 2 (multiple choice · core)
What is the pH of 0.015 M strontium hydroxide, Sr(OH)₂?
- (A) 1.52
- (B) 1.82
- (C) 12.18
- (D) 12.48
Answer and explanation
Answer: (D). Sr(OH)₂ → Sr²⁺ + 2 OH⁻, so [OH⁻] = 2 × 0.015 = 0.030 M. pOH = −log(0.030) = 1.52, and pH = 14.00 − 1.52 = 12.48.
- (A) is the pOH, reported as if it were the pH.
- (B) is the pOH without the factor of 2.
- (C) forgets the factor of 2: [OH⁻] = 0.015 M gives pH 12.18.
Question 3 (multiple choice · core)
Which row correctly gives the concentrations of the species in 0.10 M nitric acid, HNO₃?
- (A) [H₃O⁺] = 0.10 M, [NO₃⁻] = 0.10 M, [HNO₃] = 0.10 M, [OH⁻] = 1.0 × 10⁻¹³ M
- (B) [H₃O⁺] = 0.10 M, [NO₃⁻] = 0.10 M, [HNO₃] ≈ 0, [OH⁻] = 1.0 × 10⁻⁷ M
- (C) [H₃O⁺] = 0.10 M, [NO₃⁻] = 0.10 M, [HNO₃] ≈ 0, [OH⁻] = 1.0 × 10⁻¹³ M
- (D) [H₃O⁺] = 0.10 M, [NO₃⁻] = 0.10 M, [HNO₃] ≈ 0, [OH⁻] = 0
Answer and explanation
Answer: (C). HNO₃ ionises completely, so almost no HNO₃ molecules remain, and [H₃O⁺] = [NO₃⁻] = 0.10 M. [OH⁻] = K_w ÷ 0.10 = 1.0 × 10⁻¹³ M.
- (A) keeps all the HNO₃ as molecules as well as listing the ions, which counts the acid twice. In a strong acid solution, [HNO₃] ≈ 0.
- (B) uses the [OH⁻] of pure water. The added acid shifts the water equilibrium left, lowering [OH⁻].
- (D) says there is no OH⁻. K_w means a small amount is always present.
Question 4 (multiple choice · core)
10.0 mL of 0.10 M HCl (pH 1.00) is diluted with water to a total volume of 1.00 L. What is the pH of the diluted solution?
- (A) 1.00
- (B) 2.00
- (C) 3.00
- (D) 11.00
Answer and explanation
Answer: (C). The dilution factor is 1000 mL ÷ 10.0 mL = 100. So c = 0.10 ÷ 100 = 1.0 × 10⁻³ M, and pH = −log(1.0 × 10⁻³) = 3.00. A hundredfold dilution raises the pH of a strong acid by 2 units.
- (A) assumes that because HCl is strong, its pH cannot change. Dilution lowers [H₃O⁺].
- (B) treats the dilution as tenfold (10.0 mL to 100 mL).
- (D) is the pOH of the diluted solution.
Question 5 (calculation · core)
40.0 mL of 0.050 M HCl is mixed with 60.0 mL of 0.020 M HBr. No reaction occurs between them. Calculate the pH of the mixture, and the concentrations of Cl⁻, Br⁻ and OH⁻.
Worked solution
- Both acids are strong, so each mole of acid gives one mole of H₃O⁺.
- Moles of H₃O⁺ from HCl: 0.0400 L × 0.050 mol L⁻¹ = 2.0 × 10⁻³ mol. From HBr: 0.0600 L × 0.020 mol L⁻¹ = 1.2 × 10⁻³ mol.
- Total H₃O⁺ = 3.2 × 10⁻³ mol in a total volume of 0.1000 L, so [H₃O⁺] = 0.032 M.
- pH = −log(0.032) = 1.49.
- [Cl⁻] = 2.0 × 10⁻³ ÷ 0.1000 = 0.020 M; [Br⁻] = 1.2 × 10⁻³ ÷ 0.1000 = 0.012 M.
- [OH⁻] = 1.0 × 10⁻¹⁴ ÷ 0.032 = 3.1 × 10⁻¹³ M.
Check: [H₃O⁺] = [Cl⁻] + [Br⁻] = 0.032 M, so the charges balance.
Suggested mark points (3): 1 for total moles of H₃O⁺ from both acids; 1 for [H₃O⁺] using the total volume and pH 1.49; 1 for all three ion concentrations.
Common errors: adding the two concentrations (0.070 M, pH 1.15) ignores the different volumes and the dilution on mixing; averaging them (0.035 M, pH 1.46) ignores the different volumes.
Question 6 (constructed response · core)
A student has two solutions at 25 °C: 0.0100 M calcium hydroxide, Ca(OH)₂, and 0.0100 M potassium hydroxide, KOH. She claims: “Both are strong bases at the same concentration, so they have the same pH.”
(a) Calculate the pH of each solution. (b) Evaluate the student’s claim. (c) List the concentration of every ion in the Ca(OH)₂ solution. (d) Calculate the mass of KOH needed to make 250.0 mL of a KOH solution with the same pH as the Ca(OH)₂ solution.
Worked solution and suggested Marlbridge rubric
(a) Ca(OH)₂ → Ca²⁺ + 2 OH⁻, so [OH⁻] = 0.0200 M; pOH = −log(0.0200) = 1.699; pH = 12.301. KOH → K⁺ + OH⁻, so [OH⁻] = 0.0100 M; pOH = 2.000; pH = 12.000.
(b) The claim is incorrect. Both bases dissociate completely, but each Ca(OH)₂ formula unit releases two OH⁻ ions, while each KOH releases one. The Ca(OH)₂ solution has twice the [OH⁻], so its pH is higher by log 2 ≈ 0.30.
(c) [Ca²⁺] = 0.0100 M; [OH⁻] = 0.0200 M; [H₃O⁺] = 1.0 × 10⁻¹⁴ ÷ 0.0200 = 5.0 × 10⁻¹³ M.
(d) Same pH means [OH⁻] = 0.0200 M, so [KOH] = 0.0200 M. n = 0.0200 mol L⁻¹ × 0.2500 L = 5.00 × 10⁻³ mol. M(KOH) = 39.10 + 16.00 + 1.008 = 56.11 g mol⁻¹, so m = 5.00 × 10⁻³ × 56.11 = 0.281 g.
| Point | What earns it |
|---|---|
| 1 | pH 12.301 for Ca(OH)₂, using [OH⁻] = 2c |
| 1 | pH 12.000 for KOH |
| 1 | Claim rejected with the reason that Ca(OH)₂ releases two OH⁻ per formula unit |
| 1 | All three ion concentrations in (c), including the minor ion H₃O⁺ |
| 1 | 0.281 g KOH, from [KOH] = 0.0200 M |
Accept a pH from (a) carried forward into (d). Do not award the point in (b) for “Ca(OH)₂ is stronger”: both are strong; the difference is the number of OH⁻ per formula unit.
Question 7 (explanation · stretch)
A student calculates the pH of 1.0 × 10⁻⁸ M HCl at 25 °C as −log(1.0 × 10⁻⁸) = 8.00.
(a) Explain why this answer cannot be correct. (b) Explain what the student’s method leaves out, and why it does not matter for 0.010 M HCl. (c) State whether the true pH is below 7, exactly 7 or above 7, and justify your answer in terms of [H₃O⁺] and [OH⁻].
Model answer and suggested Marlbridge rubric
(a) pH 8.00 would mean the solution is basic ([OH⁻] > [H₃O⁺]). Adding an acid to neutral water can only increase [H₃O⁺], so it cannot make the solution basic.
(b) The method ignores the H₃O⁺ that comes from the autoionization of water. In pure water that is 1.0 × 10⁻⁷ M, which is ten times more than the 1.0 × 10⁻⁸ M from the acid, so it cannot be left out. In 0.010 M HCl, the acid supplies 0.010 M H₃O⁺, about 100 000 times more than water could, and the acid also pushes the water equilibrium to the left. Water’s contribution is then negligible.
(c) The pH is just below 7. The total [H₃O⁺] is slightly more than 1.0 × 10⁻⁷ M (the acid adds a little to water’s own supply), and because K_w is fixed, [OH⁻] is slightly less than 1.0 × 10⁻⁷ M. With [H₃O⁺] > [OH⁻], the solution is very slightly acidic. (A full calculation, beyond what this course asks, gives pH ≈ 6.98.)
| Point | What earns it |
|---|---|
| 1 | pH 8 would mean a basic solution, which adding acid cannot produce |
| 1 | Identifies the missing H₃O⁺ from water’s autoionization, comparable to or larger than the acid’s contribution |
| 1 | Explains why it is negligible at 0.010 M (acid’s H₃O⁺ far larger) |
| 1 | pH just below 7, justified by [H₃O⁺] slightly above, and [OH⁻] slightly below, 1.0 × 10⁻⁷ M |
How did you do?
- Q1 or Q2 wrong: check the mole ratios in “What makes an acid or base strong” in the study guide, and Worked example 2.
- Q3 wrong: re-read Worked example 1 and its species table.
- Q4 or Q5 wrong: find moles first, then divide by the final volume (Worked example 3).
- Q6 or Q7 incomplete: your reasoning needs the why (two OH⁻ per formula unit; water’s own ions), not only the numbers.
Then tick off the topic checklist.