★ Consolidation — Deepening L14–L16. Four students have been given the same titration curve and asked to identify the equivalence point, name a suitable indicator, and read the pKa. Only one student gets all three correct. Before reading on — can you identify who, and precisely diagnose what each of the others got wrong?
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Four students are each given a copy of the same titration curve. The curve shows pH on the y-axis (0–14) and volume of NaOH added on the x-axis (0–50 mL). The curve starts at pH 3.1, rises gradually through a plateau region, has a sharp jump between pH 7.5 and 11.5 centred at 25.00 mL, and levels off above pH 12 after 30 mL. At 12.50 mL, pH = 4.74.
Each student is asked: (1) identify the equivalence point and its pH; (2) identify a suitable indicator; (3) determine the pKa of the acid.
Student A: "The equivalence point is at pH 7. I would use bromothymol blue. The pKa is 4.74."
Student B: "The EP is at 25.00 mL, pH 9.5. I would use phenolphthalein. The pKa is 4.74."
Student C: "The EP is at 25.00 mL, pH 9.5. I would use methyl orange. The pKa is 7."
Student D: "The EP is at the maximum pH — above 12. Any indicator works. The pKa is the starting pH, 3.1."
Before reading on: Which student is correct on all three? Write a precise diagnosis of each incorrect student's errors — not just "they are wrong", but exactly which misconception is operating in each answer.
The four students represent the four most consequential errors in titration curve interpretation — and diagnosing each one with precision, rather than just identifying who is right, is the analytical rigour that Band 6 responses demonstrate.
Student B is correct on all three questions. The equivalence point is at 25.00 mL — the midpoint of the sharp jump on the volume axis — with EP pH ≈ 9.5 (midpoint of the pH range of the jump: (7.5 + 11.5)/2). Phenolphthalein (8.3–10.0) is appropriate — the EP pH of approximately 9.5 falls within its range. pKa = 4.74 — correctly read from the half-equivalence point at V_EP/2 = 12.50 mL.
Error 1 (EP): "The equivalence point is at pH 7." This is valid only for strong acid + strong base titrations. This curve shows a weak acid + strong base (buffer region before EP, EP pH above 7, starting pH ~3.1). The equivalence point is identified by volume — the midpoint of the steepest section — not by a target pH value.
Error 2 (Indicator): Bromothymol blue (6.0–7.6) does not encompass the EP pH of ~9.5. The student chose BTB because they incorrectly placed the EP at pH 7 — the two errors are causally linked. Phenolphthalein is required.
Error 3 (pKa): Student A actually identifies pKa = 4.74 correctly — this is the only answer Student A gets right.
Error 1 (Indicator): Methyl orange (3.1–4.4) transitions in the buffer region of this weak acid titration — far below the EP pH of ~9.5. Using methyl orange would give an endpoint when only ~10–20% of the acid had been neutralised, causing a catastrophic underestimate of acid concentration. The visual clarity of the colour change is irrelevant if the transition occurs at the wrong pH.
Error 2 (pKa): "pKa is 7 — the midpoint of the pH scale." pKa has nothing to do with pH 7 or the midpoint of the scale. pKa is the pH at the half-equivalence point of this specific weak acid — which is 4.74 for this curve. Student C correctly identifies the EP at 25.00 mL, pH 9.5.
Error 1 (EP): "The equivalence point is at the maximum pH on the curve — above 12." The highest pH on the curve is the post-equivalence plateau where excess NaOH dominates — this is not the equivalence point. The equivalence point is the midpoint of the steepest section at 25.00 mL, pH ~9.5.
Error 2 (Indicator): "All indicators work for all titrations." This is completely wrong — indicator selection depends on matching the transition range to the EP pH. Only phenolphthalein is appropriate here.
Error 3 (pKa): "pKa is the pH at the start of the curve — 3.1." The starting pH is determined by the initial concentration and Ka of the weak acid (pH = −log√(Ka × c)), not pKa directly. For this acid, starting pH 3.1 ≠ pKa 4.74.
| Student | EP identification | Indicator choice | pKa identification | Score |
|---|---|---|---|---|
| A | ✗ EP at pH 7 (strong/strong only) | ✗ BTB — below EP pH | ✓ pKa = 4.74 | 1/3 |
| B ✓ | ✓ 25.00 mL, pH ~9.5 | ✓ Phenolphthalein — covers EP pH | ✓ pKa = 4.74 from half-EP | 3/3 |
| C | ✓ 25.00 mL, pH 9.5 | ✗ Methyl orange — in buffer region | ✗ pKa = 7 (wrong) | 1/3 |
| D | ✗ Maximum pH point (excess base) | ✗ All indicators (wrong) | ✗ pKa = starting pH (wrong) | 0/3 |
Every titration curve question — whether sketching, interpreting, or calculating — can be answered systematically by working through six questions in order. Students who memorise this framework never need to rely on pattern-matching from memory.
What are the acid and base types? Determines curve shape, starting pH, and whether a buffer region exists. Strong acid list: HCl, H₂SO₄ (1st), HNO₃, HClO₄, HBr, HI — all others are weak.
What is the initial acid concentration and volume? Determines starting pH (weak acid: ICE table; strong acid: −log(c)) and total moles of acid (sets V_EP).
What volume of titrant reaches the equivalence point? V_EP = n(acid)/c(base). Marks the end of the buffer region and the centre of the jump.
What is the EP pH? Depends on salt hydrolysis: above 7 (weak acid/strong base); at 7 (strong/strong); below 7 (strong acid/weak base); gradual (weak/weak).
What is pKa? Read from pH at V_EP/2 — only meaningful for weak acid + strong base curves.
Which indicator is appropriate? Transition range must overlap EP pH: EP > 7 → phenolphthalein; EP ≈ 7 → any; EP < 7 → methyl orange or BTB.
A titration curve and a topographic map share an unexpected structural parallel — and once this analogy is established, the vocabulary of map reading transfers directly to curve reading, making the abstract features of a titration curve concrete and navigable.
Imagine a topographic map showing the ascent from sea level to a mountain summit. The horizontal axis represents distance travelled (like volume of titrant). The vertical axis represents altitude (like pH). The terrain has five characteristic zones:
Zone 1 — The coastal plain (starting region): nearly flat, slight upward slope. Corresponds to the starting pH of the weak acid — slowly rising as a small fraction of the acid is neutralised.
Zone 2 — The foothills (buffer region): gradual, steady ascent. You're climbing, but the gradient is modest. Corresponds to the buffer region where both HA and A⁻ coexist — pH rises gradually and predictably via Henderson-Hasselbalch.
Zone 3 — The cliff face (near the equivalence point): sudden, near-vertical ascent. One step takes you from valley floor to mountain ledge. Corresponds to the sharp pH jump at the equivalence point — a fraction of a drop of titrant produces a dramatic pH change.
Zone 4 — The mountain plateau (just past the EP): the sudden ascent gives way to a flatter region. Corresponds to levelling off just past the equivalence point where excess base dominates.
Zone 5 — The high plateau (post-equivalence): gently rising terrain at high altitude. Corresponds to the high-pH asymptote where excess NaOH dominates and pH changes become small again.
The equivalence point is the bottom of the cliff face — the point of maximum gradient (maximum dpH/dV). The half-equivalence point is a specific landmark in Zone 2 — the midpoint of the foothills — where the skilled hiker can read the exact altitude (pH = pKa) from the map.
Quick diagnostic: Is there a foothills region (buffer) before the cliff face (EP)? → Yes = weak acid. Is the cliff face above or below the 7-unit contour line (pH 7)? → Above = weak acid/strong base. This reading takes 10 seconds and correctly identifies the curve type before any calculation.
The mathematical inevitability of the sharp pH jump near the equivalence point — and why it is sharpest for strong/strong combinations — becomes intuitive through the analogy of a parking lot reaching capacity.
Imagine a parking lot with exactly 1000 spaces. Cars arrive at a steady rate. For the first 990 minutes, the lot is gradually filling and finding a space is easy — this is the buffer region, where plenty of both HA and A⁻ keep pH changes gradual. At 995 cars, only 5 spaces remain; at 999 cars, only 1 space. At 1000 cars, the lot is exactly full — this is the equivalence point.
Now: if the next car arrives (1001st car), there is absolutely no space — it must park on the street, completely changing the traffic pattern. This is the post-equivalence region — excess base dominates and pH rises sharply.
The critical insight: the transition from "nearly full" to "one car on the street" spans only a tiny volume range near the EP — because the buffering capacity (parking spaces) drops steeply near 100% occupancy. The pH jump is not abrupt because the reaction is faster near the EP — it is abrupt because of logarithm mathematics and stoichiometric depletion.
For a weak acid, the lot has a reservation system — even when main spaces are full, a small reserve (the weak acid ionisation equilibrium) provides a few extra spaces. This means the transition is slightly more gradual — a smaller pH jump. For a strong acid, there is no reservation system — the transition is abrupt and complete.
The five errors below reproduce themselves with remarkable consistency across HSC cohorts. Diagnosing each one explicitly — with the precise misconception at its root and a specific fix — is more effective preparation than additional practice without diagnosis.
Root cause: Students memorise "neutralisation → neutral → pH 7" without understanding that only a strong acid + strong base titration gives a neutral salt. Fix: Always apply the six-question framework; Question 4 forces salt hydrolysis reasoning before marking the EP. An EP at pH 7 for a weak acid + strong base is an automatic error.
Root cause: The terms are used interchangeably in casual conversation. Fix: Equivalence point = stoichiometric completion of the reaction (calculated from moles). Endpoint = observed indicator colour change (experimental detection). In ideal conditions these coincide; in practice the endpoint slightly precedes or follows the EP depending on indicator choice. Writing "I reached the equivalence point at the colour change" conflates the two and loses marks.
Root cause: Students know "buffer" is associated with a weak acid and guess its position. Fix: For weak acid + strong base, the buffer region is before the EP — HA and A⁻ coexist while HA is being consumed. After the EP, only excess NaOH dominates — no buffer character at all.
Root cause: Students confuse three separate pH quantities. Fix: Three distinct identities — Starting pH = −log√(Ka×c) [determined by c and Ka]; Half-EP pH = pKa [intrinsic property]; EP pH > 7 [determined by salt hydrolysis]. None of these three is the same as any other.
Root cause: Students learn "phenolphthalein for acid-base titrations" as a general rule without knowing it is only appropriate when EP pH > 7. Fix: Always use the three-step indicator rule: (1) state EP pH and explain why; (2) state indicator and its range; (3) confirm range covers EP pH. If the three-step justification cannot be completed, the selection is wrong.
| Error | Specific mistake | Root misconception | Fix |
|---|---|---|---|
| EP at pH 7 | Marking EP where curve crosses pH 7 | Neutralisation → neutral → pH 7 always | EP = midpoint of jump (volume axis); EP pH varies by type |
| Endpoint = equivalence | Using interchangeably in written response | Terms not distinguished | EP = calculated; endpoint = observed colour change |
| Buffer after EP | Drawing buffer plateau after the jump | Buffer ≈ weak acid/base type — position guessed | Buffer region is BEFORE EP for weak acid |
| pKa = starting pH or 7 | Reading pKa from wrong curve feature | Confusing three distinct pH quantities | pKa = pH at V_EP/2 only |
| Phenolphthalein always | Selecting Ph without justification | Generalised rule without EP pH reasoning | Three-step justification required; Ph only for EP > 7 |
A pharmaceutical quality control laboratory verifying the purity of acetylsalicylic acid (aspirin, Ka = 3.0 × 10⁻⁴) uses a weak acid + strong base titration. If a technician incorrectly selects methyl orange instead of phenolphthalein — drawn in by methyl orange's visually dramatic orange-to-yellow change — the reported titre would be obtained when only ~15–20% of the aspirin had been neutralised. The calculated purity would be approximately 15–20% of the true value. An entire production batch could be failed, or worse, passed with vastly insufficient active ingredient. The indicator selection error is not just a theoretical concern — it carries regulatory, financial, and patient safety consequences.
In winemaking, the titration curve is used to determine total titratable acidity (primarily tartaric acid, pKa1 = 2.98, pKa2 = 4.34). Oenologists read the half-equivalence point from the curve to verify pKa — confirming that the dominant acid matches the expected grape variety. A mismatch suggests adulteration or spoilage. The same curve-reading skills tested in HSC Module 6 are applied daily in regulated food laboratories.
"The equivalence point is where the curve crosses pH 7." — pH 7 is the EP only for strong acid + strong base. For any system involving a weak acid or base, the EP is above or below pH 7. The EP is identified by volume — the midpoint of the steepest section — regardless of the pH value at that point.
"pKa = 7 (the midpoint of the pH scale)." — pKa has no relationship to the midpoint of the pH scale. pKa is an intrinsic molecular property read from the half-equivalence point of the specific titration curve — it equals the pH when exactly half the weak acid has been neutralised.
"I chose methyl orange because it gives a very clear colour change." — Visual clarity of a colour change is irrelevant to indicator selection. The only criterion is whether the indicator's transition range encompasses the EP pH. Methyl orange (3.1–4.4) is completely unsuitable for any weak acid + strong base titration (EP pH > 7).
"The endpoint and equivalence point are the same thing." — The equivalence point is the calculated stoichiometric point; the endpoint is the experimentally observed colour change. A well-chosen indicator makes these coincide — but they are conceptually distinct, and a poorly chosen indicator separates them by many millilitres.
"The buffer region appears after the equivalence point for a weak acid." — The buffer region (where HA and A⁻ coexist) appears before the EP — it is the flat plateau visible in the early-to-middle section of the curve for a weak acid. After the EP, only excess NaOH dominates — there is no buffering.
A weak acid + strong base curve includes a buffer region before equivalence, with the half-equivalence point giving pKa and the equivalence point sitting above pH 7.
The four curve types can be distinguished by whether a buffer region appears, where the equivalence point lies, and whether a sharp enough pH jump exists for an indicator.
(a) Acid type — two features:
Feature 1: No buffer region before the equivalence point. A weak acid in the presence of partial neutralisation produces a buffer region (HA + A⁻ coexisting). The absence of any plateau is characteristic of a strong acid, which has no conjugate base buffer capacity.
Feature 2: The jump is centred near pH 7 (midpoint of 3.5–9.5 ≈ 6.5), consistent with a neutral salt from strong acid + strong base. For a weak acid, EP pH would be above 7.
Conclusion: HX is a strong acid.
(b) EP volume and pH:
EP volume = midpoint of steepest section = 20.00 mL (given as centre of the jump).
EP pH = midpoint of the jump = (3.5 + 9.5)/2 = 6.5 ≈ 7.0 — consistent with strong acid + strong base (neutral salt, no hydrolysis).
(c) Phenolphthalein suitability:
EP pH ≈ 7.0. The sharp jump spans approximately pH 3.5–9.5. Phenolphthalein's range (8.3–10.0) falls within this jump — phenolphthalein is suitable.
So are methyl orange (3.1–4.4) and BTB (6.0–7.6) — all three indicator ranges fall within the large jump of a strong/strong titration. Phenolphthalein gives the clearest visual change (colourless to pink) and is recommended.
(d) Concentration of HX:
At EP: n(NaOH) = n(HX)
n(NaOH) = c × V = 0.0500 × 0.02000 = 1.00 × 10⁻³ mol n(HX) = 1.00 × 10⁻³ mol c(HX) = n/V = 1.00 × 10⁻³ / 0.04000 = 0.0250 mol/LVerify strong acid: Starting pH = −log(0.0250) = 1.60. Observed 1.70 — within ±0.1 pH units of graphical reading precision. Consistent with strong acid classification.
Answer: (a) Strong acid — no buffer region before EP; jump centred near pH 7 consistent with neutral salt. (b) V_EP = 20.00 mL; EP pH ≈ 7.0 (midpoint of jump at 6.5). (c) Phenolphthalein is suitable — EP pH within the large jump (pH 3.5–9.5) that encompasses its range (8.3–10.0). (d) c(HX) = 0.0250 mol/L.
(a) EP identification: The large pH jump occurs centred near V = 25.00 mL (pH rises from 5.34 at 18.75 mL to 12.10 at 31.25 mL). EP volume = 25.00 mL. EP pH ≈ 8.57 (read directly at V_EP) or approximately (5.34 + 12.10)/2 = 8.72 — both confirm EP pH > 7, consistent with weak acid + strong base.
(b) pKa verification:
Half-EP volume = V_EP/2 = 25.00/2 = 12.50 mL At V = 12.50 mL, pH = 4.89 → pKa(experimental) = 4.89 pKa(theoretical) = −log(1.3 × 10⁻⁵) = 5 − log(1.3) = 5 − 0.114 = 4.89 ✓Perfect match — the half-EP pH correctly identifies the pKa of propanoic acid.
(c) Theoretical starting pH:
First find c(acid): at EP, n(NaOH) = n(acid) → c(acid) = (0.100 × 0.02500)/0.02500 = 0.100 mol/L.
Check: Ka/c = 1.3 × 10⁻⁵/0.100 = 1.3 × 10⁻⁴ < 0.0025 ✓ (assumption valid).
x = √(Ka × c) = √(1.3 × 10⁻⁵ × 0.100) = √(1.3 × 10⁻⁶) = 1.140 × 10⁻³ mol/L Theoretical starting pH = −log(1.140 × 10⁻³) = 2.94Observed: 3.03. Difference = 0.09 pH units — within ±0.1 pH units graphical precision ✓.
(d) Methyl orange error calculation:
Student stops at V = 6.25 mL of NaOH.
n(NaOH) at endpoint = 0.100 × 0.00625 = 6.25 × 10⁻⁴ mol Reported c(acid) = 6.25 × 10⁻⁴ / 0.02500 = 0.02500 mol/L True c(acid) = 0.1000 mol/L Percentage error = |0.02500 − 0.1000| / 0.1000 × 100% = 75.0% underestimationThe student stopped when only 6.25/25.00 = 25% of the propanoic acid had been neutralised — the reported concentration is only 25% of the true value. Methyl orange (3.1–4.4) transitions in the early buffer region of this titration (pKa = 4.89), far below the EP at pH 8.57.
(e) Three-step indicator justification:
Step 1 (EP pH and why): The equivalence point pH is approximately 8.6 — above neutral pH of 7.00. This occurs because the salt formed at equivalence is sodium propanoate (CH₃CH₂COONa), and the propanoate ion (CH₃CH₂COO⁻) is the conjugate base of a weak acid. It undergoes base hydrolysis: CH₃CH₂COO⁻ + H₂O ⇌ CH₃CH₂COOH + OH⁻, producing OH⁻ and raising pH above 7.
Step 2 (indicator and range): Phenolphthalein — transition range pH 8.3–10.0.
Step 3 (confirmation): The EP pH of 8.6 falls within phenolphthalein's range (8.3–10.0). As the last drops of NaOH are added at equivalence, the sharp pH jump passes through 8.3–10.0 — phenolphthalein changes from colourless to faint pink, giving a clear, reproducible endpoint coinciding with the equivalence point.
Answer: (a) V_EP = 25.00 mL; EP pH ≈ 8.57 (above 7, weak acid + strong base). (b) pKa from half-EP at 12.50 mL = 4.89, matching theoretical pKa = 4.89 ✓. (c) Theoretical starting pH = 2.94; observed 3.03 — within ±0.1 units of graphical precision. (d) Methyl orange endpoint → reported c = 0.0250 mol/L vs true 0.1000 mol/L → 75% underestimation. (e) EP pH ≈ 8.6 (basic — propanoate hydrolyses); phenolphthalein (8.3–10.0) covers EP pH; transitions within sharp jump at equivalence.
(a)(i) Starting pH — Solution P (HCl, strong acid):
HCl fully ionised: [H⁺] = 0.100 mol/L → pH = −log(0.100) = 1.00(a)(i) Starting pH — Solution Q (HF, weak acid):
Check: Ka/c = 6.8 × 10⁻⁴/0.100 = 6.8 × 10⁻³. x = √(6.8 × 10⁻⁴ × 0.100) = 8.25 × 10⁻³. Verify: 8.25 × 10⁻³/0.100 = 8.25% > 5% — assumption fails. Quadratic required:
x² + 6.8 × 10⁻⁴x − 6.8 × 10⁻⁵ = 0 x = (−6.8 × 10⁻⁴ + √(4.624 × 10⁻⁷ + 2.72 × 10⁻⁴))/2 = 7.91 × 10⁻³ mol/L Starting pH(HF) = −log(7.91 × 10⁻³) = 2.10(a)(ii) Half-equivalence point:
Solution P: The concept of a half-equivalence point (pH = pKa) applies only to weak acid curves because only weak acids have a buffer region where HA and A⁻ coexist before the EP. HCl has no Ka in the weak acid sense — it ionises completely and has no buffer region. There is no meaningful half-equivalence point for a strong acid titration.
Solution Q (HF): Both have the same V_EP = 20.00 mL. Half-EP at 10.00 mL:
pKa(HF) = −log(6.8 × 10⁻⁴) = 4 − log(6.8) = 4 − 0.833 = 3.17 Half-EP pH = pKa = 3.17(a)(iii) EP volume and pH:
Both: V_EP = n(acid)/c(base) = (0.100 × 0.02000)/0.100 = 20.00 mL.
HCl: HCl + NaOH → NaCl + H₂O. NaCl is a neutral salt (Na⁺ and Cl⁻ are neutral spectators). EP pH = 7.00.
HF: NaF is formed. F⁻ is the conjugate base of a weak acid → base hydrolysis: F⁻ + H₂O ⇌ HF + OH⁻.
Kb(F⁻) = Kw/Ka(HF) = 1.0 × 10⁻¹⁴ / 6.8 × 10⁻⁴ = 1.47 × 10⁻¹¹ [F⁻] at EP = (0.100 × 0.02000) / 0.04000 = 0.0500 mol/L [OH⁻] = √(1.47 × 10⁻¹¹ × 0.0500) = 8.57 × 10⁻⁷ → pOH = 6.07 → pH = 7.93EP pH(HF) = 7.93 (basic salt).
(b) Five key differences between the curves:
1. Starting pH: HCl starts at pH 1.00; HF starts at pH 2.10. HCl is fully ionised ([H⁺] = c); HF is only ~7.9% ionised at 0.100 mol/L → fewer H⁺ → higher starting pH.
2. Buffer region: HCl has no buffer region (no HA/A⁻ pair possible for a strong acid). HF has a significant buffer region before EP where HF and F⁻ coexist.
3. Jump size: HCl gives a larger jump (~7 units, pH ~4–11); HF gives a smaller jump (~4–5 units) because the buffer region before EP moderates the approach to equivalence.
4. EP pH: HCl → NaCl (neutral) → EP pH = 7.00. HF → NaF (basic) → EP pH = 7.93.
5. Half-EP: HCl has no meaningful half-EP (no buffer). HF half-EP at 10.00 mL gives pH = pKa = 3.17.
(c) Evaluating the claim:
The claim is incorrect. While HCl and HF are both monoprotic hydrogen halides, they have fundamentally different acid strengths: HCl is a strong acid (Ka → ∞, complete ionisation); HF is a weak acid (Ka = 6.8 × 10⁻⁴, ~7.9% ionised at 0.100 mol/L). Their titration curves differ in starting pH (1.00 vs 2.10), presence of buffer region (none vs significant), jump size (~7 units vs ~4–5 units), and EP pH (7.00 vs 7.93).
Quantitatively: at the half-equivalence point (10.00 mL NaOH), the HF curve shows pH = 3.17 — there is no equivalent feature on the HCl curve at all. The H–F bond is significantly stronger than H–Cl (bond enthalpy 570 vs 432 kJ/mol), making proton donation from HF far less favourable. Being hydrogen halides does not make acids equivalent — Ka determines curve shape, not halide identity.
(d) Indicator selection:
Solution P (HCl + NaOH, strong + strong, EP pH = 7.00): The sharp jump spans approximately pH 4–10. All three indicators are suitable (all transition ranges fall within the large jump). Best: phenolphthalein for visual clarity.
Solution Q (HF + NaOH, weak acid + strong base, EP pH = 7.93):
Step 1: EP pH = 7.93 — above 7 because F⁻ (conjugate base of weak acid HF) hydrolyses to give OH⁻.
Step 2: Phenolphthalein — range 8.3–10.0.
Step 3: EP pH 7.93 is just below phenolphthalein's lower limit (8.3), but the sharp jump extends from approximately pH 6.5 to pH 11.5 — phenolphthalein transitions within the upper part of this jump. Phenolphthalein is the most appropriate of the three available indicators. Methyl orange (3.1–4.4) is completely unsuitable (in buffer region, far below EP).
Answer: (a)(i) HCl: pH 1.00; HF: pH 2.10 (quadratic required). (a)(ii) HCl: no half-EP (strong acid, no buffer region); HF: half-EP at 10.00 mL, pH = pKa = 3.17. (a)(iii) Both: V_EP = 20.00 mL; HCl EP pH = 7.00; HF EP pH = 7.93. (b) Five differences: starting pH, buffer region, jump size, EP pH, half-EP — all explained by strong vs weak acid character. (c) Claim incorrect — HCl (Ka → ∞) and HF (Ka = 6.8 × 10⁻⁴) differ in ionisation, buffer capacity, jump size, and EP pH despite both being hydrogen halides; Ka determines curve shape. (d) HCl: any indicator (phenolphthalein recommended); HF: phenolphthalein (EP pH 7.93, transitions within jump pH 6.5–11.5; methyl orange completely unsuitable).
1 A student examining a titration curve for CH₃COOH + NaOH marks the equivalence point where the curve crosses pH 7. Diagnose this error.
2 A student reads the pKa of a weak acid from a titration curve as pH 9.4 — the pH at the equivalence point. Diagnose this error.
3 A student selects methyl orange for a titration of formic acid (HCOOH, Ka = 1.8 × 10⁻⁴) with NaOH because "it gives a vivid, easy-to-see colour change." Diagnose this error and calculate the percentage error in acid concentration if the methyl orange endpoint occurs at 20% of the true titre volume.
4 A student writes in their practical report: "I titrated until I reached the equivalence point, which I identified by the colour change of phenolphthalein from colourless to pink." Identify the conceptual error in this sentence and rewrite it correctly.
1. A student examines a titration curve: starting pH = 4.2; buffer plateau before EP; EP at V = 30.00 mL, pH = 9.8; at V = 15.00 mL, pH = 6.4. Which correctly extracts all key information?
2. Two students titrate 25.00 mL of 0.100 mol/L CH₃COOH with 0.100 mol/L NaOH. Student X uses phenolphthalein and records endpoint at 25.10 mL. Student Y uses methyl orange and records endpoint at 5.40 mL. Which student gives the more accurate acid concentration, and why?
3. A student examines a curve: sharp jump from pH 6.5 to 11.5; midpoint at V = 15.00 mL; pH at V = 7.50 mL reads 5.12. The student claims: "pKa = 5.12; Ka = 7.6 × 10⁻⁶; and since the EP is at 15.00 mL using 0.200 mol/L NaOH with 15.00 mL of acid, acid concentration = 0.200 mol/L." Evaluate this claim.
4. A titration curve is produced by adding 0.100 mol/L NaOH to 20.00 mL of a weak acid HA (Ka = 4.0 × 10⁻⁶). The equivalence point is at 20.00 mL of NaOH, with EP pH = 9.1. (a) State the concentration of HA. (b) Identify the appropriate indicator and write a complete three-step justification. (c) Identify the pH at the half-equivalence point and explain what this quantity represents. 5 MARKS
5. A student titrates 0.100 mol/L ammonia solution (NH₃, Kb = 1.8 × 10⁻⁵) with 0.100 mol/L HCl. (a) Identify the titration type and predict whether the EP pH is above, at, or below 7 — justify with a relevant equation. (b) Identify the appropriate indicator from methyl orange, BTB, or phenolphthalein — apply the three-step justification. (c) Explain why phenolphthalein would be unsuitable for this titration, with reference to the curve shape. 5 MARKS
6. Band 6 Extended Response (7 marks). A research laboratory has two 0.0500 mol/L acid solutions: Solution X is hydrochloric acid (HCl); Solution Y is lactic acid (CH₃CH(OH)COOH, Ka = 1.4 × 10⁻⁴). Both are titrated with 0.100 mol/L NaOH. (a) Calculate the starting pH of each solution. (b) Calculate the EP volume for each (assume 25.00 mL of each acid). (c) Describe four differences between the titration curves of X and Y, explaining why each difference arises with reference to acid strength. (d) Identify the appropriate indicator for each titration, applying the three-step justification for Solution Y. (e) A student claims the two curves should have identical shapes because they have the same concentration and are both titrated with the same NaOH concentration. Evaluate this claim quantitatively — use the half-equivalence point of Solution Y to support your argument. 7 MARKS
Go back to your Think First response at the top of this lesson. Now that you've worked through all five errors, both analogies, and three worked examples:
1. Error: Student marks EP at the pH 7 crossing point rather than the midpoint of the steepest section on the volume axis. Root misconception: "neutralisation → neutral → pH 7 always." Correct answer: for CH₃COOH + NaOH (weak acid + strong base), the EP pH is above 7 (~8.7) because acetate ion undergoes base hydrolysis: CH₃COO⁻ + H₂O ⇌ CH₃COOH + OH⁻. EP must be identified by volume at the point of inflection.
2. Error: pKa read from the EP pH (9.4) rather than the half-EP pH. Root misconception: confusing the three distinct pH quantities (starting pH, half-EP pH = pKa, EP pH). Correct method: half-EP volume = V_EP/2; read pH at that volume — that pH equals pKa.
3. Error: Indicator selected based on visual clarity, not EP pH. Why MO fails: EP pH for HCOOH + NaOH ≈ 8.7 (weak acid + strong base → EP above 7). Methyl orange (3.1–4.4) transitions in the early buffer region of the formic acid titration (pKa = 3.74), far below the EP. Percentage error: endpoint at 20% of true titre → reported c = 20% of true c → 80% underestimation.
4. Error: "I reached the equivalence point at the colour change" implies the colour change is the equivalence point. Corrected: "The endpoint (colour change of phenolphthalein from colourless to faint pink) was used to detect the equivalence point — the theoretical stoichiometric point at which all acetic acid had reacted with NaOH. In ideal conditions, the endpoint and equivalence point coincide."
Q1 — Acid type: Weak acid — starting pH 3.28 is above what a strong acid at this concentration would show; there is a buffer plateau visible in the gradual rise from 3.28 to 5.72 between V = 0 and V = 11.25 mL.
Q2 — EP: Large pH jump occurs at V = 15.00 mL (pH 8.20) — rising sharply from 5.72 at 11.25 mL to 12.05 at 18.75 mL. EP volume = 15.00 mL; EP pH ≈ 8.20 — above 7, consistent with weak acid + strong base.
Q3 — pKa: Half-EP at V_EP/2 = 7.50 mL; pH at 7.50 mL = 5.12 → pKa = 5.12.
Q4 — Ka: Ka = 10⁻⁵·¹² = 7.6 × 10⁻⁶.
Q5 — c(HA): n(NaOH) at EP = 0.200 × 0.01500 = 3.00 × 10⁻³ mol = n(HA). c(HA) = 3.00 × 10⁻³/0.01500 = 0.200 mol/L.
Q6 — Indicator: Step 1: EP pH = 8.20 — above 7 because A⁻ (conjugate base of weak acid HA) hydrolyses: A⁻ + H₂O ⇌ HA + OH⁻, producing OH⁻. Step 2: Phenolphthalein — range 8.3–10.0. Step 3: EP pH 8.20 is just below the lower limit of phenolphthalein (8.3), but the sharp jump extends from ~6 to 12 — phenolphthalein transitions within this jump. Phenolphthalein is most appropriate. Methyl orange is completely unsuitable.
Q7 — Equation: A⁻(aq) + H₂O(l) ⇌ HA(aq) + OH⁻(aq) — base hydrolysis of conjugate base produces OH⁻ → solution is basic at equivalence → EP pH above 7.
1. B — Starting pH 4.2, buffer plateau, EP pH 9.8 all confirm weak acid + strong base. Half-EP = 30.00/2 = 15.00 mL; pH at 15.00 mL = 6.4 → pKa = 6.4. Ka = 10⁻⁶·⁴ = 4.0 × 10⁻⁷. Phenolphthalein (8.3–10.0) covers EP pH 9.8. Option A identifies as strong/strong — wrong (buffer plateau and EP pH 9.8 rule this out). Option C reads pKa from EP pH — wrong. Option D reads pKa from starting pH — also wrong.
2. B — For CH₃COOH + NaOH, EP pH ≈ 8.7. Phenolphthalein (8.3–10.0) covers this EP → Student X's endpoint (25.10 mL) is accurate. Methyl orange (3.1–4.4) transitions in the buffer region — Student Y stops at 5.40/25.00 = 21.6% of the true EP volume → reports concentration at 21.6% of true value = 78.4% underestimation. Options A and C both incorrectly prioritise visual clarity over EP pH criterion. Option D attributes a 20 mL error to technique variation — impossible.
3. B — pKa = 5.12 (from half-EP) ✓; Ka = 10⁻⁵·¹² = 7.6 × 10⁻⁶ ✓. The concentration claim is numerically correct for these specific values (V(acid) = V_EP = 15.00 mL → equal volumes → equal concentrations at 0.200 mol/L) but the reasoning "equal concentrations → equal volumes → concentrations match" only holds when V(acid) = V_EP — the student's reasoning is incomplete, not wrong for this specific case.
Q4 (5 marks): (a) At EP: n(NaOH) = n(HA) → n(HA) = 0.100 × 0.02000 = 2.00 × 10⁻³ mol; c(HA) = 2.00 × 10⁻³/0.02000 = 0.100 mol/L. [1 mark.] (b) Step 1: EP pH = 9.1 — above 7 because A⁻ (conjugate base of weak acid HA) undergoes base hydrolysis: A⁻ + H₂O ⇌ HA + OH⁻, producing OH⁻. [1 mark.] Step 2: Phenolphthalein — transition range pH 8.3–10.0. [1 mark.] Step 3: EP pH 9.1 falls within phenolphthalein's range (8.3–10.0) — the indicator transitions within the sharp jump at equivalence, giving a sharp, reproducible endpoint. [1 mark.] (c) pH at half-EP = pKa = −log(4.0 × 10⁻⁶) = 5.40. Half-EP is at V_EP/2 = 10.00 mL. pKa is the intrinsic ionisation constant of the acid — it represents the pH at which half the acid has been neutralised, i.e. [HA] = [A⁻], and the Henderson-Hasselbalch log term = 0. [1 mark — 5 marks total.]
Q5 (5 marks): (a) Strong acid (HCl) + weak base (NH₃) titration. EP pH below 7 because the salt formed is NH₄Cl; NH₄⁺ is the conjugate acid of the weak base NH₃ and undergoes acid hydrolysis: NH₄⁺ + H₂O ⇌ NH₃ + H₃O⁺, producing H₃O⁺ and lowering pH below 7. [2 marks.] (b) Step 1: EP pH below 7 — NH₄⁺ hydrolyses to give H₃O⁺. Step 2: Methyl orange — range 3.1–4.4, or BTB — range 6.0–7.6. Step 3: The sharp jump for strong acid + weak base is centred below pH 7 (approximately pH 3–6.5 for a typical NH₃ + HCl system) — methyl orange or BTB both transition within this jump. [2 marks.] (c) Phenolphthalein (8.3–10.0) is unsuitable because its transition range is above the EP pH for this titration. In fact, the curve for strong acid + weak base levels off at a low pH above 7 in the post-EP region — phenolphthalein would not change colour at all during the titration, as the solution never reaches pH 8.3 in the normal titration range. [1 mark — 5 marks total.]
Q6 — Band 6 (7 marks): (a) X (HCl): pH = −log(0.0500) = 1.30. Y (lactic acid): Ka/c = 1.4 × 10⁻⁴/0.0500 = 2.8 × 10⁻³ > 0.0025 ✗. x = √(1.4 × 10⁻⁴ × 0.0500) = 2.65 × 10⁻³. Verify: 2.65 × 10⁻³/0.0500 = 5.3% > 5% — assumption invalid. Quadratic: x² + (1.4 × 10⁻⁴)x − (1.4 × 10⁻⁴)(0.0500) = 0 → x ≈ 2.61 × 10⁻³ mol/L → pH(Y) = −log(2.61 × 10⁻³) ≈ 2.58 (approximation still rounds to 2.58). [1 mark for both pH values.] (b) Both: n(acid) = 0.0500 × 0.02500 = 1.25 × 10⁻³ mol; V_EP = 1.25 × 10⁻³/0.100 = 12.50 mL for both. [1 mark.] (c) Four differences: (i) Starting pH: X = 1.30; Y = 2.58 — X lower because HCl fully ionised, lactic acid only partially ionised. (ii) Buffer region: X has none (strong acid, no HA/A⁻ pair); Y has a buffer plateau before EP (lactic acid + lactate ion coexist). (iii) Jump size: X has a larger jump (~7 units); Y has a smaller jump (~3–4 units) because buffer capacity moderates approach to EP. (iv) EP pH: X = 7.00 (NaCl neutral salt); Y above 7 (lactate ion hydrolyses: CH₃CH(OH)COO⁻ + H₂O ⇌ lactic acid + OH⁻). [2 marks — ½ mark each, min 4 differences.] (d) X: any indicator (large jump spans all three ranges); recommend BTB or phenolphthalein. Y — 3-step: Step 1: EP pH above 7 because lactate ion (conjugate base of weak acid) undergoes base hydrolysis, producing OH⁻. Step 2: Phenolphthalein — range 8.3–10.0. Step 3: EP pH for Y is above 7 (within the phenolphthalein range) — phenolphthalein transitions at equivalence. Methyl orange is completely unsuitable (in buffer region, ~3.5 pH units below EP). [2 marks.] (e) The claim is incorrect. While both solutions have the same concentration (0.0500 mol/L) and same V_EP (12.50 mL), their curve shapes are fundamentally different because Ka governs curve shape — not concentration alone. Quantitative support: the half-EP of Solution Y is at V = 6.25 mL; pH = pKa(lactic acid) = −log(1.4 × 10⁻⁴) = 3.85. There is no equivalent feature on Solution X's curve — HCl has no meaningful half-EP. At V = 6.25 mL for HCl, the pH is simply determined by excess H⁺ concentration, not by any buffer equilibrium. The two curves produce different pH values at every volume between 0 and V_EP, and the shapes (no buffer vs significant buffer, EP pH 7.00 vs >7) are entirely different — proving that concentration and titrant are not the only factors determining curve shape. Ka is the decisive variable. [1 mark — 7 marks total.]
Put your knowledge of titration curves, indicators and equivalence points to the test. Answer correctly to deal damage — get it wrong and the boss hits back. Pool: lessons 1–17.
Tick when you've finished all activities and checked your answers.