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: 25 °C, K_w = 1.0 × 10⁻¹⁴, so pK_a + pK_b = 14.00 for a conjugate pair. The acids HQ and HZ and the indicators J, K, L, M and P are fictional. Approximate colour-change ranges: bromocresol green 3.8–5.4 (yellow to blue); phenolphthalein 8.3–10.0 (colourless to pink).
Question 1 (multiple choice · foundation)
A weak acid HA has pK_a = 6.2. It is dissolved in a solution that is held at pH 8.0. Which statement is correct?
- (A) [A⁻] is greater than [HA].
- (B) [HA] is greater than [A⁻].
- (C) [HA] and [A⁻] are equal.
- (D) The predominant form cannot be found without the concentration of HA.
Answer and explanation
Answer: (A). pH 8.0 is above pK_a 6.2, so the deprotonated form predominates. In numbers, [A⁻]/[HA] = 10^(8.0 − 6.2) = 10^1.8 ≈ 63.
- (B) is the result for pH below pK_a; the comparison has been reversed.
- (C) is only true when pH = pK_a.
- (D) is wrong because, once the pH is fixed, the ratio [A⁻]/[HA] = K_a/[H₃O⁺] does not depend on how much acid is present.
Question 2 (multiple choice · core)
A weak base B has pK_b = 5.0. Which form predominates in a solution held at pH 7.0?
- (A) B, because the pH is greater than the pK_b.
- (B) BH⁺, because the pH is less than the pK_a of BH⁺.
- (C) B and BH⁺ are present at equal concentrations, because the solution is neutral.
- (D) B, because a base cannot be protonated in a neutral solution.
Answer and explanation
Answer: (B). Convert first: pK_a(BH⁺) = 14.0 − 5.0 = 9.0. The pH (7.0) is below 9.0, so the protonated form BH⁺ predominates. The ratio [B]/[BH⁺] = 10^(7.0 − 9.0) = 0.01.
- (A) compares pH with pK_b, which points the wrong way. pK_b is on the hydroxide side of the scale.
- (C) confuses “neutral solution” with “equal amounts of the two forms”. Equal amounts need pH = pK_a(BH⁺) = 9.0.
- (D) is false: H₃O⁺ is present at 10⁻⁷ M, and B has a strong enough attraction for protons to take them at this pH.
Question 3 (multiple choice · core)
A weak acid is titrated with sodium hydroxide. The pH at the equivalence point is 8.9. Which indicator is the best choice?
- (A) Indicator J, pK_a = 4.1
- (B) Indicator K, pK_a = 6.0
- (C) Indicator L, pK_a = 7.2
- (D) Indicator M, pK_a = 9.0
Answer and explanation
Answer: (D). An indicator’s colour changes over roughly pK_a ± 1, so the pK_a should be close to the equivalence-point pH. Indicator M (pK_a 9.0) changes colour from about pH 8 to 10, on the steep section around pH 8.9.
- (A) and (B) would change colour in the gently sloping region well before the equivalence point, giving a titre that is too small.
- (C) is the “pH 7” trap. A weak acid–strong base titration has a basic equivalence point because the conjugate base A⁻ is present, so pH 7.2 is still too early.
Question 4 (multiple choice · foundation)
Indicator P is a weak acid, HIn, with pK_a = 5.5. HIn is red and In⁻ is yellow. A few drops are added to a solution of pH 3.0. What colour is the solution?
- (A) Red
- (B) Yellow
- (C) Orange, because both forms are present at equal concentrations
- (D) Colourless, because the indicator is fully protonated
Answer and explanation
Answer: (A). pH 3.0 is 2.5 units below pK_a, so [In⁻]/[HIn] = 10^(3.0 − 5.5) ≈ 0.003. Almost all the indicator is HIn, which is red.
- (B) would need pH well above 5.5.
- (C) describes the colour near pH 5.5, where the two forms are similar in amount.
- (D) correctly says the indicator is mostly protonated, but the protonated form is red, not colourless.
Question 5 (calculation · core)
The weak acid HQ has pK_a = 3.85.
(a) Calculate the ratio [Q⁻]/[HQ] and the percentage of HQ in the acid form in a solution held at pH 2.50. (b) Repeat for pH 5.00, and state which form predominates. (c) Find the highest pH at which at least 90% of HQ is in the acid form.
Worked solution
(a) [Q⁻]/[HQ] = 10^(pH − pK_a) = 10^(2.50 − 3.85) = 10^(−1.35) = 0.045. Fraction as HQ = 1 / (1 + 0.0447) = 0.957, so 95.7% is HQ.
(b) [Q⁻]/[HQ] = 10^(5.00 − 3.85) = 10^1.15 = 14.1. Fraction as HQ = 1 / (1 + 14.1) = 0.066, so 6.6% is HQ. pH > pK_a, so Q⁻ predominates (93.4%).
(c) 90% HQ means [HQ]/[Q⁻] = 90/10 = 9, so [Q⁻]/[HQ] = 1/9. pH = pK_a + log(1/9) = 3.85 − 0.95 = 2.90. At pH 2.90 or below, at least 90% is HQ.
Suggested mark points (4): 1 for the ratio in (a); 1 for 95.7% (or 96%); 1 for the ratio, percentage and “Q⁻ predominates” in (b); 1 for pH 2.90 in (c). Accept “about one pH unit below pK_a” with a correct value of 2.9.
Common error: writing 10^(pK_a − pH) gives the inverse ratio, [HQ]/[Q⁻]. Check by asking “pH is below pK_a, so should the ratio [Q⁻]/[HQ] be less than 1?”
Question 6 (constructed response · core)
A student adds two drops of an unknown indicator to a series of buffer solutions and records the colour of each.
| Buffer pH | 3.0 | 4.0 | 4.5 | 5.0 | 5.5 | 6.0 | 7.0 |
|---|---|---|---|---|---|---|---|
| Colour | red | red | red-orange | orange | yellow-orange | yellow | yellow |
(a) Identify the colour of the protonated form, HIn, and estimate the pK_a of the indicator. Justify your estimate. (b) State how precise your estimate is, and suggest one change to the method that would make it more precise. (c) The student wants to use the indicator in two titrations: (i) aqueous ammonia with HCl, equivalence pH 5.28; (ii) acetic acid with NaOH, equivalence pH 8.72. For each, state whether the indicator is suitable and explain.
Worked solution and suggested Marlbridge rubric
(a) At low pH the indicator is mostly protonated, so HIn is red (and In⁻ is yellow). The colour is an even mixture (orange) at pH 5.0. An even mixture means [HIn] ≈ [In⁻], which happens when pH = pK_a. So pK_a ≈ 5.0.
(b) The buffers are 0.5 pH units apart and the colours are judged by eye. Red-orange at 4.5 and yellow-orange at 5.5 bracket the midpoint, so the estimate is about 5.0 ± 0.5. To improve it: use more buffers between pH 4.5 and 5.5 (for example every 0.1 unit), or measure the absorbance of each solution with a spectrophotometer instead of judging colour.
(c)(i) Suitable. The equivalence pH (5.28) is close to the indicator’s pK_a (about 5.0), so the colour changes on the steep part of the curve. (c)(ii) Not suitable. The equivalence pH (8.72) is far above the indicator’s range (about pH 4 to 6). The indicator would turn yellow while the solution is still in the gently rising region, well before the equivalence point, so the measured titre would be too small.
| Point | What earns it |
|---|---|
| 1 | HIn is red, with the reason (low pH → protonated form) |
| 1 | pK_a ≈ 5.0, linked to the even-mixture colour where [HIn] = [In⁻] |
| 1 | Precision of about ± 0.5 (or “between 4.5 and 5.5”) with a reason from the data |
| 1 | A valid improvement: closer-spaced buffers or an instrument such as a spectrophotometer |
| 1 | (i) suitable and (ii) not suitable, each with a comparison of equivalence pH with pK_a |
Do not award the last point for “suitable / not suitable” without the pH comparison.
Question 7 (explanation · stretch)
A weak acid HZ has pK_a = 5.80. A student titrates 20.00 mL of HZ solution, of concentration about 0.08 M, with 0.100 M NaOH. The pH at the equivalence point is 9.22. By mistake, the student uses bromocresol green as the indicator and records the volume at which the solution has fully turned blue: 4.60 mL.
(a) Explain why the student’s end point comes long before the equivalence point. (b) Calculate the concentration of HZ that the student would report, and state whether it is too high or too low. (c) Name a better indicator from the data at the top of this page and justify the choice.
Model answer and suggested Marlbridge rubric
(a) Bromocresol green changes colour over about pH 3.8 to 5.4, so its pK_a lies near the middle of that range, close to 5. The pH of the HZ solution passes 5.4 early in the titration, while most of the acid is still unreacted (5.4 is below the pK_a of HZ, 5.80, so HZ is still the main form). The equivalence point is at pH 9.22, far above the indicator’s range. So the indicator reaches its base colour long before enough NaOH has been added.
(b) Moles of NaOH = 0.100 mol L⁻¹ × 0.004 60 L = 4.60 × 10⁻⁴ mol = moles of HZ assumed. Concentration = 4.60 × 10⁻⁴ mol ÷ 0.020 00 L = 0.0230 M. This is too low: the real concentration is about 0.08 M, because the titre was far too small.
(c) Phenolphthalein (8.3–10.0). Its range contains the equivalence pH of 9.22, so the colour change (colourless to pink) falls on the steep section of the curve and the end point matches the equivalence point.
| Point | What earns it |
|---|---|
| 1 | The indicator’s range (or pK_a) is far below the equivalence pH |
| 1 | So the colour changes while most HZ is still unreacted / in the gently sloping region |
| 1 | 0.0230 M with working |
| 1 | States “too low” because the titre is too small |
| 1 | Phenolphthalein, justified by its range including (or pK_a close to) pH 9.22 |
How did you do?
- Q1, Q2 or Q5 wrong: re-read “The rule: compare pH with pK_a” and the weak-base section in the study guide.
- Q3, Q4 or Q6 wrong: revisit “Indicators are weak acids you can see” and Figure 2.
- Q7 incomplete: link the indicator choice to the volume you read and then to the concentration you calculate.
Then tick off the topic checklist.