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 data are at 25 °C: K_w = 1.0 × 10⁻¹⁴ and pK_w = 14.00. Acids and bases named with capital letters (HR, HM, HE, HX, Y, B) are fictional and monoprotic. Assume volumes add when solutions are mixed.
Question 1 (multiple choice · foundation)
20.0 mL of 0.100 M HNO₃ is mixed with 30.0 mL of 0.0500 M KOH. What is the pH of the mixture?
- (A) 1.30
- (B) 1.60
- (C) 2.00
- (D) 7.00
Answer and explanation
Answer: (C). n(H₃O⁺) = 0.0200 × 0.100 = 2.00 × 10⁻³ mol; n(OH⁻) = 0.0300 × 0.0500 = 1.50 × 10⁻³ mol. Excess H₃O⁺ = 5.0 × 10⁻⁴ mol in 50.0 mL, so [H₃O⁺] = 0.0100 M and pH = 2.00.
- (A) subtracts the molarities (0.100 − 0.0500 = 0.050 M) and ignores the different volumes.
- (B) divides the excess by the acid volume only (20.0 mL), not the total.
- (D) assumes any acid-base mixture is neutral; the acid is in excess.
Question 2 (multiple choice · core)
Equal amounts in moles of a weak acid HA and NaOH are mixed. Which statement describes the resulting solution?
- (A) The major species are Na⁺, A⁻ and H₂O, and the pH is above 7 because A⁻ reacts with water to form OH⁻.
- (B) The major species are Na⁺, A⁻ and H₂O, and the pH is 7 because equal moles of acid and base reacted.
- (C) The major species are HA, OH⁻ and Na⁺, and the pH is above 7 because the reaction stops at equilibrium.
- (D) The major species are HA, A⁻ and Na⁺, and the solution is a buffer with pH equal to pK_a.
Answer and explanation
Answer: (A). HA + OH⁻ → A⁻ + H₂O goes to completion, so at equal moles only A⁻ (with Na⁺) is left. A⁻ is the conjugate base of a weak acid, so A⁻ + H₂O ⇌ HA + OH⁻ makes the solution slightly basic.
- (B) is only true for a strong acid with a strong base.
- (C) treats the reaction with a strong base as incomplete. It is essentially complete.
- (D) describes the mixture when only half the acid has been neutralised, not all of it.
Question 3 (multiple choice · core)
Which mixture, made up to 100. mL with water, is a buffer?
- (A) 0.010 mol HCl and 0.010 mol NaOH
- (B) 0.010 mol HR and 0.010 mol NaOH, where HR is a weak acid
- (C) 0.010 mol HR and 0.004 mol NaOH, where HR is a weak acid
- (D) 0.010 mol HCl and 0.004 mol NaOH
Answer and explanation
Answer: (C). NaOH is limiting. After reaction, 0.006 mol HR and 0.004 mol R⁻ remain: a weak acid with its conjugate base in comparable amounts.
- (A) leaves only water, Na⁺ and Cl⁻.
- (B) converts all HR to R⁻; there is no weak acid left to react with added base.
- (D) leaves excess strong acid; Cl⁻ is far too weak a base to buffer.
Question 4 (multiple choice · core)
The weak acid HX (pK_a = 3.00) is mixed with the weak base Y. The conjugate acid of Y, HY⁺, has pK_a = 8.00. What is K for HX + Y ⇌ X⁻ + HY⁺, and which side is favoured?
- (A) 1.0 × 10⁵, products favoured
- (B) 1.0 × 10⁻⁵, reactants favoured
- (C) 1.0 × 10⁻¹¹, reactants favoured
- (D) 1.0 × 10⁻³, reactants favoured
Answer and explanation
Answer: (A). K = K_a(HX) ÷ K_a(HY⁺) = 10^(−3.00) ÷ 10^(−8.00) = 1.0 × 10⁵. HX is a much stronger acid than HY⁺, so the proton moves to Y.
- (B) divides the wrong way round.
- (C) multiplies the two K_a values.
- (D) is just K_a of HX; it ignores the base.
Question 5 (calculation · core)
30.0 mL of 0.200 M HM (pK_a = 5.10) is mixed with 25.0 mL of 0.150 M NaOH. (a) Identify the major species after reaction. (b) Calculate the pH.
Worked solution
| HM | OH⁻ | M⁻ | |
|---|---|---|---|
| Before (mol) | 6.00 × 10⁻³ | 3.75 × 10⁻³ | 0 |
| After (mol) | 2.25 × 10⁻³ | 0 | 3.75 × 10⁻³ |
(a) HM, M⁻, Na⁺ and H₂O. HM and M⁻ together make a buffer.
(b) pH = 5.10 + log(3.75 × 10⁻³ ÷ 2.25 × 10⁻³) = 5.10 + 0.22 = 5.32. Using concentrations gives the same ratio: [M⁻] = 0.0682 M and [HM] = 0.0409 M in 55.0 mL.
Suggested mark points (3): 1 for the moles before and after; 1 for naming HM and M⁻ as the major species (a buffer); 1 for pH 5.32.
Common error: writing log(HM ÷ M⁻) gives 4.88. Check: there is more M⁻ than HM, so the pH must be above pK_a.
Question 6 (constructed response · core)
The weak base B has K_b = 1.6 × 10⁻⁹. A student mixes 50.0 mL of 0.120 M B with 50.0 mL of 0.120 M HCl.
(a) Calculate the pH of the mixture. (b) The student writes: “Equal moles of acid and base were mixed, so the solution is neutral.” Evaluate this claim. (c) Calculate the pH if 60.0 mL of the HCl had been used instead.
Worked solution and suggested Marlbridge rubric
(a) n(B) = n(H₃O⁺) = 0.0500 × 0.120 = 6.00 × 10⁻³ mol. B + H₃O⁺ → HB⁺ + H₂O goes to completion, leaving only HB⁺. [HB⁺] = 6.00 × 10⁻³ ÷ 0.100 L = 0.0600 M. K_a(HB⁺) = 1.0 × 10⁻¹⁴ ÷ 1.6 × 10⁻⁹ = 6.25 × 10⁻⁶. [H₃O⁺] ≈ √(6.25 × 10⁻⁶ × 0.0600) = 6.1 × 10⁻⁴ M (about 1.0% of 0.0600, so the shortcut is fine). pH = 3.21.
(b) The claim is wrong. Equal moles means the acid and base have reacted completely, but the product HB⁺ is the conjugate acid of a weak base. It gives H₃O⁺ to water, so the solution is acidic (pH 3.21). Only a strong acid with a strong base gives pH 7.
(c) n(H₃O⁺) = 0.0600 × 0.120 = 7.20 × 10⁻³ mol. Excess = 7.20 × 10⁻³ − 6.00 × 10⁻³ = 1.20 × 10⁻³ mol in 110.0 mL. [H₃O⁺] = 1.20 × 10⁻³ ÷ 0.1100 = 0.0109 M, so pH = 1.96. The HB⁺ adds a negligible amount of H₃O⁺ next to the excess strong acid.
| Point | What earns it |
|---|---|
| 1 | Recognises equal moles, so only HB⁺ remains, at 0.0600 M in the total volume |
| 1 | K_a = K_w ÷ K_b = 6.25 × 10⁻⁶ |
| 1 | pH 3.21 |
| 1 | Claim wrong, because HB⁺ is a weak acid that makes the solution acidic |
| 1 | pH 1.96 from excess H₃O⁺ divided by the total volume 110.0 mL |
Accept pH 3.21 or 3.22 in (a): the shortcut and the quadratic differ only in the last digit.
Question 7 (design and explanation · stretch)
A student needs a buffer with pH 4.60. She has 100.0 mL of 0.100 M HE (pK_a = 4.60), 0.100 M NaOH and 0.100 M HCl.
(a) What volume of NaOH should she add? Explain. (b) Her partner adds 100.0 mL of NaOH instead. Calculate the pH and explain why the result is not a buffer. (c) Another student adds only 40.0 mL of NaOH. Is the mixture still a buffer? Calculate its pH. (d) Explain why adding the HCl to the HE solution cannot make a buffer.
Model answer and suggested Marlbridge rubric
(a) pH = pK_a when [E⁻] = [HE]. n(HE) = 1.00 × 10⁻² mol, so she should convert half: 5.00 × 10⁻³ mol OH⁻ = 50.0 mL of NaOH. This leaves 5.00 × 10⁻³ mol each of HE and E⁻.
(b) 100.0 mL gives equal moles, so only E⁻ remains: [E⁻] = 1.00 × 10⁻² ÷ 0.2000 L = 0.0500 M. K_b = 10⁻¹⁴ ÷ 10^(−4.60) = 3.98 × 10⁻¹⁰; [OH⁻] ≈ 4.46 × 10⁻⁶ M; pOH 5.35; pH = 8.65. No HE is left to react with added base, so the solution cannot buffer.
(c) Yes. 4.00 × 10⁻³ mol E⁻ and 6.00 × 10⁻³ mol HE remain, both in comparable amounts. pH = 4.60 + log(4.00 / 6.00) = 4.42.
(d) HCl does not react with HE (both are acids), so no E⁻ forms. The mixture holds a weak acid and a strong acid, with no conjugate base to consume added acid.
| Point | What earns it |
|---|---|
| 1 | 50.0 mL, linked to [E⁻] = [HE] so that pH = pK_a |
| 1 | pH 8.65 from K_b of E⁻ |
| 1 | Not a buffer because all the weak acid has been used up |
| 1 | (c) still a buffer, pH 4.42 |
| 1 | (d) HCl cannot make the conjugate base E⁻ |
How did you do?
- Q1 or Q6(c) wrong: practise “moles first, total volume last” with Worked example 1 in the study guide.
- Q2, Q3 or Q7 wrong: use Figure 1 to decide what is left before choosing a method.
- Q4 wrong: revisit “Weak acid with weak base”.
- Q5 or Q6(a) wrong: re-work Worked examples 2 and 3.
Then tick off the topic checklist.