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: K_w = 1.0 × 10⁻¹⁴ at 25 °C, so pK_w = 14.00. Where a question gives a different K_w, use that value. All samples are aqueous.
Question 1 (multiple choice · foundation)
At 25 °C, a solution has [H₃O⁺] = 2.5 × 10⁻⁵ M. What is [OH⁻]?
- (A) 4.0 × 10⁻¹⁰ M
- (B) 2.5 × 10⁻⁹ M
- (C) 1.0 × 10⁻⁷ M
- (D) 4.0 × 10⁴ M
Answer and explanation
Answer: (A). [OH⁻] = K_w ÷ [H₃O⁺] = 1.0 × 10⁻¹⁴ ÷ 2.5 × 10⁻⁵ = 4.0 × 10⁻¹⁰ M. Check: (2.5 × 10⁻⁵)(4.0 × 10⁻¹⁰) = 1.0 × 10⁻¹⁴.
- (B) subtracts the exponent from 14 but keeps the 2.5. You cannot split the number from its power of ten like that.
- (C) is [OH⁻] in pure water. Adding acid changes [OH⁻].
- (D) calculates 1 ÷ [H₃O⁺], leaving out K_w. A concentration of 40 000 M is impossible.
Question 2 (multiple choice · foundation)
The pH of a solution changes from 3.0 to 5.0 at constant temperature. What happens to [H₃O⁺]?
- (A) It increases by a factor of 100.
- (B) It decreases by a factor of 2.
- (C) It decreases by a factor of 100.
- (D) It decreases by 2.0 mol L⁻¹.
Answer and explanation
Answer: (C). [H₃O⁺] goes from 10⁻³·⁰ = 1 × 10⁻³ M to 10⁻⁵·⁰ = 1 × 10⁻⁵ M. A rise of 2 pH units is a fall of 10² = 100 times.
- (A) gets the size right but the direction wrong: higher pH means lower [H₃O⁺].
- (B) treats pH as a linear scale.
- (D) treats the pH difference as a concentration difference. The whole starting concentration is only 0.001 M.
Question 3 (multiple choice · core)
At a certain temperature above 25 °C, K_w = 4.0 × 10⁻¹⁴. Which statement about pure water at this temperature is correct?
- (A) pH = 6.70, and the water is neutral.
- (B) pH = 6.70, and the water is acidic.
- (C) pH = 7.00, and the water is neutral.
- (D) pH = 13.40, and the water is basic.
Answer and explanation
Answer: (A). In pure water [H₃O⁺] = [OH⁻] = √(4.0 × 10⁻¹⁴) = 2.0 × 10⁻⁷ M, so pH = −log(2.0 × 10⁻⁷) = 6.70. The two ion concentrations are equal, so the water is neutral.
- (B) uses the 25 °C rule “pH below 7 is acidic”. Acidic needs [H₃O⁺] > [OH⁻], which is not true here.
- (C) assumes K_w is 1.0 × 10⁻¹⁴ at every temperature.
- (D) gives pK_w (13.40) instead of half of it, and calls it basic.
Question 4 (multiple choice · core)
Which statement is true for every neutral aqueous solution, at any temperature?
- (A) pH = 7.00
- (B) [H₃O⁺] = [OH⁻]
- (C) [H₃O⁺] = [OH⁻] = 1.0 × 10⁻⁷ M
- (D) pH + pOH = 14.00
Answer and explanation
Answer: (B). Neutral is defined by equal concentrations of hydronium and hydroxide ions.
- (A) and (C) are true only at 25 °C, where K_w = 1.0 × 10⁻¹⁴.
- (D) is true for every solution (neutral or not) at 25 °C, but not at other temperatures, where pH + pOH = pK_w, a different number.
Question 5 (calculation · core)
A cleaning solution has pOH = 10.25 at 25 °C. Calculate its pH, [H₃O⁺] and [OH⁻], and state whether it is acidic or basic.
Worked solution
- pH = 14.00 − 10.25 = 3.75.
- [H₃O⁺] = 10⁻³·⁷⁵ = 1.8 × 10⁻⁴ M (2 decimal places in pH → 2 significant figures).
- [OH⁻] = 10⁻¹⁰·²⁵ = 5.6 × 10⁻¹¹ M.
- Check: (1.78 × 10⁻⁴)(5.62 × 10⁻¹¹) = 1.0 × 10⁻¹⁴.
- [H₃O⁺] > [OH⁻] (pH below 7 at 25 °C), so the solution is acidic.
Suggested mark points (3): 1 for pH 3.75; 1 for [H₃O⁺] with correct significant figures; 1 for [OH⁻] with “acidic”. Finding [OH⁻] from K_w ÷ [H₃O⁺] instead of from pOH is equally valid.
Common error: a high pOH can look “basic” at a glance. A high pOH means very little OH⁻, so the solution is acidic.
Question 6 (constructed response · core)
A student heats a sample of pure water to a temperature at which K_w = 2.9 × 10⁻¹⁴. She measures its pH and writes: “The pH is below 7, so heating has made the water acidic.”
(a) Calculate [H₃O⁺] and the pH of the pure water at this temperature. (b) Evaluate the student’s claim. (c) K_w at 25 °C is 1.0 × 10⁻¹⁴. Use this and Le Châtelier’s principle to state whether the autoionization of water is endothermic or exothermic, with a reason.
Worked solution and suggested Marlbridge rubric
(a) In pure water [H₃O⁺] = [OH⁻] = x, so x² = 2.9 × 10⁻¹⁴ and x = 1.7 × 10⁻⁷ M. pH = −log(1.70 × 10⁻⁷) = 6.77.
(b) The claim is incorrect. The pH is below 7, but [H₃O⁺] = [OH⁻] = 1.7 × 10⁻⁷ M. Equal concentrations mean the water is neutral. pH 7 is the neutral point only when K_w = 1.0 × 10⁻¹⁴ (25 °C). At this temperature the neutral point is pH 6.77.
(c) K_w is larger at the higher temperature, so heating shifts 2 H₂O ⇌ H₃O⁺ + OH⁻ to the right. By Le Châtelier’s principle, an equilibrium shifts in the endothermic direction when heated, so the forward reaction (autoionization) is endothermic.
| Point | What earns it |
|---|---|
| 1 | [H₃O⁺] = 1.7 × 10⁻⁷ M from √K_w |
| 1 | pH = 6.77 |
| 1 | States the claim is wrong and gives [H₃O⁺] = [OH⁻] as the reason |
| 1 | Endothermic, linked to K_w increasing with temperature (a shift to the right on heating) |
Do not award the point in (b) for “it is neutral” without the equal-concentration reason. Accept a pH from (a) carried forward.
Question 7 (explanation · stretch)
A drop of concentrated acid is added to a beaker of pure water at 25 °C. Once the solution is mixed, [H₃O⁺] = 1.0 × 10⁻⁴ M.
(a) Calculate [OH⁻] in the new solution. (b) Just after the acid mixes in, but before the water equilibrium adjusts, [OH⁻] is still 1.0 × 10⁻⁷ M. Calculate the reaction quotient Q for the water equilibrium at that moment, compare it with K_w, and use the comparison to explain why [OH⁻] falls. (c) Explain why the concentration of water does not appear in K_w.
Model answer and suggested Marlbridge rubric
(a) [OH⁻] = 1.0 × 10⁻¹⁴ ÷ 1.0 × 10⁻⁴ = 1.0 × 10⁻¹⁰ M.
(b) Q = [H₃O⁺][OH⁻] = (1.0 × 10⁻⁴)(1.0 × 10⁻⁷) = 1.0 × 10⁻¹¹. This is 1000 times larger than K_w. When Q > K, the reverse reaction runs faster than the forward one: H₃O⁺ and OH⁻ combine to form water. This uses up OH⁻ (and a tiny, negligible amount of the much larger H₃O⁺) until Q falls back to K_w, at [OH⁻] = 1.0 × 10⁻¹⁰ M. In Le Châtelier terms, adding a product shifts the equilibrium to the left.
(c) Water is the solvent. It is a pure liquid present in huge excess (about 55.5 M), and almost none of it reacts, so its concentration is effectively constant. Pure liquids and solids are left out of equilibrium-constant expressions.
| Point | What earns it |
|---|---|
| 1 | [OH⁻] = 1.0 × 10⁻¹⁰ M |
| 1 | Q = 1.0 × 10⁻¹¹ and the statement Q > K_w |
| 1 | Links Q > K_w to the reverse reaction (H₃O⁺ + OH⁻ → water), so [OH⁻] decreases |
| 1 | Water omitted because it is the pure liquid solvent with effectively constant concentration |
For the third point, a Le Châtelier argument (“adding H₃O⁺ shifts the equilibrium to the left, which uses up OH⁻”) is also accepted.
How did you do?
- Q1 or Q5 wrong: practise the “one number gives you all four” table in the study guide.
- Q2 wrong: re-read “pH and pOH: a log scale for tiny numbers” and Worked example 3.
- Q3, Q4 or Q6 wrong: re-read “Neutral means equal, not pH 7” and Worked example 2.
- Q7 incomplete: your reasoning needs Q compared with K_w (or the Le Châtelier shift), not only the final number.
Then tick off the topic checklist.