Three students have each been given a different acid solution and asked to calculate the pH. Each has chosen a different method. Only one method is correct for their acid. Before reading on — can you identify which student has chosen the right method, and precisely why each wrong method fails?
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Three students are each given a 0.100 mol/L solution of a different acid and asked to calculate the pH.
Student A (given HNO₃): "HNO₃ is a strong acid. I write HNO₃ → H⁺ + NO₃⁻. Since it ionises completely, [H⁺] = 0.100 mol/L. pH = −log(0.100) = 1.00."
Student B (given CH₃COOH, Ka = 1.8 × 10⁻⁵): "CH₃COOH is an acid. I write Ka = [H⁺][CH₃COO⁻]/[CH₃COOH]. I set [H⁺] = 0.100 mol/L because that is the concentration given, so Ka = (0.100)(0.100)/(0.100) = 0.100. pH = −log(0.100) = 1.00."
Student C (given HF, Ka = 6.8 × 10⁻⁴): "HF is a weak acid. I set up an ICE table: x = √(Ka × c) = √(6.8 × 10⁻⁵) = 8.25 × 10⁻³ mol/L. I check: 8.25% > 5% — assumption invalid. I solve the quadratic and get x = 7.91 × 10⁻³ mol/L. pH = −log(7.91 × 10⁻³) = 2.10."
Before reading on — which student used the correct method? Write a precise diagnosis of the specific error each incorrect student made. Then write what the correct pH should be for each acid.
The three students represent the three most consequential calculation errors in IQ2 — and diagnosing exactly what went wrong for each incorrect student, rather than simply knowing who is right, is the precision that separates Band 4 from Band 6 responses.
Student A is correct. HNO₃ is a strong acid — it ionises completely. The single arrow (→) is correct. [H⁺] = 0.100 mol/L exactly. pH = 1.00. This is the only case among the three where [H⁺] = c is a legitimate step.
Student B is incorrect — Error: applied the strong acid shortcut to a weak acid.
Student C is correct. Student C correctly identifies HF as weak, sets up the ICE table, uses x = √(Ka × c) as first estimate, checks assumption (8.25% > 5% — invalid), solves the quadratic (x = 7.91 × 10⁻³ mol/L), and obtains pH = 2.10. This is a methodologically complete solution.
| Student | Acid | Specific error | Correct pH |
|---|---|---|---|
| A | HNO₃ (strong) | None — correct ✓ | pH = 1.00 ✓ |
| B | CH₃COOH (Ka = 1.8 × 10⁻⁵) | Applied [H⁺] = c (strong acid shortcut) to a weak acid; [H⁺] is ~74× too high | pH = 2.87 (ICE table) |
| C | HF (Ka = 6.8 × 10⁻⁴) | None — correct ✓ (assumption correctly identified as invalid, quadratic applied) | pH = 2.10 ✓ |
The reason students apply the wrong calculation method is not a lack of mathematical ability — it is the absence of a systematic identification step before the calculation begins. A decision tree makes the identification automatic and the method selection unambiguous.
Every pH calculation in Module 6 begins with the same two questions: Is this an acid or a base? Is it strong or weak? The answers completely determine the calculation method.
| Scenario | Identification test | Method | Key formula |
|---|---|---|---|
| Strong acid alone | On the 6 strong acid list | Direct | [H⁺] = c → pH = −log[H⁺] |
| Weak acid alone | NOT on strong acid list | ICE table + assumption check | x = √(Ka×c) or quadratic → pH = −log(x) |
| Strong base alone | NaOH, KOH, Ca(OH)₂, Ba(OH)₂ | Direct | [OH⁻] = c×n(OH⁻) → pOH → pH = 14−pOH |
| Weak base alone | NOT on strong base list | ICE table (Kb) | [OH⁻] = √(Kb×c) → pOH → pH = 14−pOH |
| Strong acid + strong base mixed | Both strong | Moles → excess → c/V(total) | c(excess) = n(excess)/V(total) |
| Weak acid + strong base (before EP) | HA and A⁻ both present | Henderson-Hasselbalch | pH = pKa + log(n(A⁻)/n(HA)) |
The most critical branch is the strong/weak identification — which uses the six strong acid list from L05: HCl, H₂SO₄ (1st), HNO₃, HClO₄, HBr, HI. Everything else is weak.
The decision tree in Card 2 is the chemistry equivalent of a hospital triage system — and just as a doctor who gives every patient the same treatment regardless of diagnosis will harm some of them, a student who uses the same pH calculation method for every acid will get most problems wrong.
In a hospital emergency department, a triage nurse assesses every patient before any treatment begins. Two questions: How severe? What category? The answers direct the patient to the right team. The treatment only begins after the triage is complete.
A nurse who sends every patient to the cardiac team — regardless of actual condition — would be catastrophically wrong for most patients even if they applied the cardiac treatment perfectly. The pH equivalent: a student who applies [H⁺] = c for every acid is making the same categorical error. They are applying the right treatment (strong acid method) to the wrong patient (weak acid), and no matter how accurately they execute the arithmetic, the answer is fundamentally wrong because the method was selected before the diagnosis was made.
The triage step (strong or weak?) must always come first. The treatment step (which formula?) always comes second.
The most common arithmetic error in mixing calculations — using only one volume instead of the total combined volume — has an exact economic analogy that makes the error immediately obvious once the parallel is seen.
Imagine you have $50 Australian dollars and exchange $30 worth into Euros, leaving $20 AUD. You want to find the density of AUD in your wallet — AUD per unit of wallet space. Your wallet now holds both AUD and Euros: its total capacity is larger than when it only held AUD. To find the AUD density (concentration), divide $20 by the total wallet capacity — not by the original AUD-only capacity. Using the original (smaller) capacity gives a concentration that is too high.
The mixing equivalent: when excess H⁺ or OH⁻ remains after neutralisation, its concentration must be calculated using the total combined volume. The "wallet" is V(total) = V(acid) + V(base). The "AUD remaining" is n(excess species). The "AUD density" is c(excess) = n(excess)/V(total).
The five errors below are not random — they are the specific, predictable mistakes that appear most frequently in HSC Module 6 marking, and diagnosing each with a precise fix converts them from recurring traps into solved problems.
Error 1 — Using [H⁺] = c for a weak acid
Error 2 — Forgetting to check the simplifying assumption
Error 3 — Writing pOH as pH for a base
Error 4 — Using only V(acid) or V(base) in c(excess) calculation
Error 5 — Applying ICE table to a buffer mixture after partial neutralisation
"pH = pKa for any weak acid solution." — Incorrect. pH = pKa only at the half-equivalence point of a titration, when [A⁻] = [HA]. For a pure 0.100 mol/L weak acid, pH < pKa always.
"If Ka(acid) = Kb(base), they must be a conjugate pair." — Incorrect. Conjugate pairs are defined by a one-proton difference in formula, not by equal equilibrium constants. Ka(CH₃COOH) = Kb(NH₃) = 1.8 × 10⁻⁵ is a numerical coincidence — CH₃COOH and NH₃ are not conjugate.
"After partial neutralisation, I use the ICE table on the remaining HA." — Incorrect. The solution contains both HA and A⁻ — it is a buffer. ICE tables assume [A⁻] = 0 initially, which is violated. Henderson-Hasselbalch must be used.
"H₂SO₄ is monoprotic — [H⁺] = c(H₂SO₄)." — Incorrect. H₂SO₄ is diprotic (both ionisations are essentially complete in dilute solution) → [H⁺] = 2 × c(H₂SO₄).
(a) KOH — Identification: KOH is a strong base (on the strong base list). Method: direct. [OH⁻] = c(KOH) = 0.0400 mol/L (1 OH⁻ per formula unit). pOH = −log(0.0400) = 1.40. pH = 14.00 − 1.40 = 12.60. Sanity check: pH 12.60 > 7 ✓ (base).
(b) HBr — Identification: HBr is a strong acid (on the six strong acid list). Method: direct. [H⁺] = 0.0400 mol/L. pH = −log(0.0400) = 1.40. Sanity check: pH 1.40 < 7 ✓ (acid).
(c) HCN — Identification: HCN is NOT on the six strong acid list → weak acid. Ka = 6.2 × 10⁻¹⁰. Method: ICE table. Check assumption: Ka/c = 6.2 × 10⁻¹⁰/0.0400 = 1.55 × 10⁻⁸ << 0.0025 ✓. x = √(6.2 × 10⁻¹⁰ × 0.0400) = √(2.48 × 10⁻¹¹) = 4.98 × 10⁻⁶ mol/L. Verify: 4.98 × 10⁻⁶/0.0400 = 0.012% << 5% ✓. pH = −log(4.98 × 10⁻⁶) = 5.30. Sanity check: pH 5.30 < 7 ✓ (weak acid).
ANSWER: (a) pH = 12.60. (b) pH = 1.40. (c) pH = 5.30.
Both scenarios — common setup: n(OH⁻) from Ba(OH)₂ = 0.200 × 0.0350 × 2 = 1.40 × 10⁻² mol (Ba(OH)₂ is diprotic base → ×2). V(total) = 35.0 + 65.0 = 100.0 mL = 0.1000 L for both.
(a) HNO₃ at 0.150 mol/L: n(H⁺) = 0.150 × 0.0650 = 9.75 × 10⁻³ mol. Compare: n(OH⁻) = 0.01400 > n(H⁺) = 0.009750 → excess OH⁻. n(excess OH⁻) = 0.01400 − 0.009750 = 4.25 × 10⁻³ mol. c(OH⁻) = 4.25 × 10⁻³/0.1000 = 0.0425 mol/L. pOH = −log(0.0425) = 1.37. pH = 14.00 − 1.37 = 12.63. Sanity check: excess OH⁻ → pH > 7 ✓.
(b) HNO₃ at 0.0800 mol/L: n(H⁺) = 0.0800 × 0.0650 = 5.20 × 10⁻³ mol. n(OH⁻) = 0.01400 > n(H⁺) = 0.005200 → still excess OH⁻. n(excess) = 0.01400 − 0.005200 = 8.80 × 10⁻³ mol. c(OH⁻) = 8.80 × 10⁻³/0.1000 = 0.0880 mol/L. pOH = −log(0.0880) = 1.056. pH = 14.00 − 1.056 = 12.94.
ANSWER: (a) Excess OH⁻ = 4.25 × 10⁻³ mol; c(OH⁻) = 0.0425 mol/L; pH = 12.63. (b) Excess OH⁻ = 8.80 × 10⁻³ mol; c(OH⁻) = 0.0880 mol/L; pH = 12.94.
(a) pH of three solutions:
Solution 1 (HCl, strong acid): [H⁺] = 0.100 mol/L → pH = 1.00.
Solution 2 (CH₃COOH, weak acid): x = √(1.8 × 10⁻⁵ × 0.100) = 1.34 × 10⁻³ mol/L. Verify: 1.34% < 5% ✓. pH = 2.87.
Solution 3 (NH₃, weak base): [OH⁻] = √(1.8 × 10⁻⁵ × 0.100) = 1.34 × 10⁻³ mol/L. pOH = 2.87. pH = 14.00 − 2.87 = 11.13.
(b) Mixing HCl with NaOH: n(H⁺) = 0.100 × 0.0400 = 4.00 × 10⁻³ mol. n(OH⁻) = 0.100 × 0.0600 = 6.00 × 10⁻³ mol. Excess OH⁻ = 2.00 × 10⁻³ mol. V(total) = 100.0 mL = 0.1000 L. c(OH⁻) = 0.0200 mol/L. pOH = 1.70. pH = 12.30.
(c) Partial neutralisation of CH₃COOH: n(CH₃COOH) = 0.100 × 0.0250 = 2.50 × 10⁻³ mol. n(OH⁻) = 0.100 × 0.0250 = 2.50 × 10⁻³ mol. Moles are equal → equivalence point — all CH₃COOH converted to CH₃COO⁻. n(CH₃COO⁻) = 2.50 × 10⁻³ mol; n(CH₃COOH) = 0.
Solution type: salt solution of a weak acid's conjugate base (NOT a buffer — no HA remaining). CH₃COO⁻ hydrolyses: CH₃COO⁻ + H₂O ⇌ CH₃COOH + OH⁻. Kb(CH₃COO⁻) = Kw/Ka = 1.0 × 10⁻¹⁴/1.8 × 10⁻⁵ = 5.56 × 10⁻¹⁰. c(CH₃COO⁻) = 2.50 × 10⁻³/0.0500 = 0.0500 mol/L. [OH⁻] = √(5.56 × 10⁻¹⁰ × 0.0500) = 5.27 × 10⁻⁶ mol/L. pOH = 5.28. pH = 14.00 − 5.28 = 8.72 (basic salt solution).
(d) Ka × Kb and relative strengths: Ka(CH₃COOH) × Kb(NH₃) = 1.8 × 10⁻⁵ × 1.8 × 10⁻⁵ = 3.24 × 10⁻¹⁰ ≠ Kw (1.0 × 10⁻¹⁴). Therefore CH₃COOH and NH₃ are NOT a conjugate pair. The Ka × Kb = Kw relationship applies only within conjugate pairs. The equal Ka and Kb values are a numerical coincidence — CH₃COOH and NH₃ are equally strong in their respective roles (acid and base), which is confirmed by their pH values (2.87 and 11.13) being symmetric around 7. But this symmetry arises because Ka(CH₃COOH) = Kb(NH₃), not because they are conjugate.
ANSWER: (a) HCl: pH 1.00; CH₃COOH: pH 2.87; NH₃: pH 11.13. (b) Excess OH⁻ = 2.00 × 10⁻³ mol; pH = 12.30. (c) Equivalence point — basic salt solution; Kb(CH₃COO⁻) = 5.56 × 10⁻¹⁰; [OH⁻] = 5.27 × 10⁻⁶ mol/L; pH = 8.72. (d) Ka × Kb ≠ Kw → not conjugate; equal Ka and Kb is numerical coincidence; both equally strong in their respective roles; Ka × Kb = Kw applies only within conjugate pairs.
Minimum rules for Lesson 11 — consolidation checklist.
Each student response contains exactly one of the five IQ2 errors. Identify which error it is, state what is wrong, and write the corrected calculation.
For each of the following, state the identification (acid/base type), the method, and calculate pH. Show all working.
Look back at what you wrote in the Think First section. What has changed? What did you get right? What surprised you?
1. A student is asked to find the pH of 0.050 mol/L formic acid (HCOOH, Ka = 1.8 × 10⁻⁴) and writes: "[H⁺] = 0.050 mol/L. pH = −log(0.050) = 1.30." Which response correctly identifies all errors?
2. 20.0 mL of 0.300 mol/L HCl is mixed with 30.0 mL of 0.150 mol/L NaOH. Which correctly calculates the pH?
3. The pH of 0.100 mol/L solutions are: P = 1.00; Q = 11.13; R = 2.87; S = 12.70. A student claims Q and R must be a conjugate pair because their pH values sum to 14 (11.13 + 2.87 = 14.00). Evaluate this claim.
4. A student adds 25.0 mL of 0.100 mol/L NaOH to 25.0 mL of 0.100 mol/L CH₃COOH (Ka = 1.8 × 10⁻⁵). Which correctly calculates the pH?
5. Which of the following correctly distinguishes between using Henderson-Hasselbalch and using the ICE table method for calculating pH of a weak acid solution?
Question 6. Calculate the pH of each at 25°C. Show the identification step and full working for each. (a) 0.0250 mol/L HClO₄. (b) 0.0500 mol/L Ba(OH)₂. (c) 0.100 mol/L lactic acid (HC₃H₅O₃, Ka = 1.4 × 10⁻⁴). State whether the simplifying assumption is valid for each.
Question 7. 50.0 mL of 0.200 mol/L CH₃COOH (Ka = 1.8 × 10⁻⁵) is mixed with 20.0 mL of 0.200 mol/L NaOH. (a) Calculate the moles of CH₃COOH remaining and CH₃COO⁻ formed after the reaction. (b) Calculate the pH of the resulting mixture. (c) Explain why this pH is higher than the pH of the original pure 0.200 mol/L CH₃COOH solution (pH ≈ 2.57).
Question 8. A student has unlabelled solutions of 0.100 mol/L HNO₃ and 0.100 mol/L HNO₂ (Ka = 4.5 × 10⁻⁴). (a) Calculate the expected pH of each. (2 marks) (b) The student mixes 40.0 mL of each solution with 60.0 mL of 0.100 mol/L NaOH. Calculate the pH of each resulting mixture — show identification of what type of solution each forms. (3 marks) (c) Explain why the two mixing calculations require different methods (one requires simple excess calculation; one requires Henderson-Hasselbalch), despite both using "the same volume of the same NaOH." (2 marks)
Return to your diagnosis of the three students. The full analysis:
Two students were correct (A and C) — one for a strong acid, one for a weak acid requiring the quadratic. The only incorrect student (B) made the single most common Module 6 error: applying the strong acid shortcut to a weak acid.
Q1: C
HCOOH is a weak acid (not on the six strong acid list). [H⁺] = c is Error 1. ICE table required: x = √(1.8 × 10⁻⁴ × 0.050) = √(9.0 × 10⁻⁶) = 3.0 × 10⁻³. Check: 3.0 × 10⁻³/0.050 = 6.0% > 5% — assumption fails. Quadratic: x² + 1.8 × 10⁻⁴x − 9.0 × 10⁻⁶ = 0 → x ≈ 2.93 × 10⁻³ mol/L → pH ≈ 2.53. Option B confuses pKa with pH. Option D invents a non-existent formula.
Q2: B
n(H⁺) = 0.300 × 0.0200 = 6.00 × 10⁻³ mol. n(OH⁻) = 0.150 × 0.0300 = 4.50 × 10⁻³ mol. Excess H⁺ = 1.50 × 10⁻³ mol. V(total) = 20.0 + 30.0 = 50.0 mL = 0.0500 L. c(H⁺) = 1.50 × 10⁻³/0.0500 = 0.0300 mol/L. pH = 1.52. Option A uses V = 0.020 L (acid only — Error 4). Option C uses V = 0.030 L (base only — also Error 4). Option D incorrectly claims equal moles.
Q3: B
pH(R) + pH(Q) = 2.87 + 11.13 = 14.00 because Ka(R) = Kb(Q) = 1.8 × 10⁻⁵ — a numerical coincidence resulting in equal [H⁺] from R and [OH⁻] from Q. Conjugate pairs are defined by differing by exactly one H⁺: CH₃COOH/CH₃COO⁻ or NH₄⁺/NH₃ are conjugate pairs; CH₃COOH/NH₃ are not. Option A and C incorrectly redefine conjugate pairs using pH sums. Option D is wrong — pH values can sum to 14.
Q4: C
Equal moles of CH₃COOH and NaOH (0.100 × 0.0250 = 2.50 × 10⁻³ mol each) → equivalence point — all HA converted to A⁻. Solution contains CH₃COO⁻ (conjugate base of weak acid) → basic salt solution. Kb(CH₃COO⁻) = Kw/Ka = 5.56 × 10⁻¹⁰. c(CH₃COO⁻) = 2.50 × 10⁻³/0.0500 = 0.0500 mol/L. [OH⁻] = √(5.56 × 10⁻¹⁰ × 0.0500) = 5.27 × 10⁻⁶ mol/L. pOH = 5.28. pH = 8.72. Option A incorrectly identifies this as a buffer. Option B applies ICE table to post-equivalence solution — Error 5. Option D incorrectly assumes neutral salt.
Q5: B
The ICE table assumes [A⁻] = 0 initially — valid for a pure weak acid before any A⁻ has been produced. Henderson-Hasselbalch is required when significant A⁻ is present (from partial neutralisation), because the common ion effect of A⁻ suppresses ionisation of HA, making the simple ICE table method invalid. Option A gives wrong criteria. Option C is incorrect — H-H applies only to buffer solutions, not to pure weak acid. Option D is wrong — they give different answers when [A⁻] ≠ 0.
(a) HClO₄: Strong acid (on strong acid list) → [H⁺] = 0.0250 mol/L. pH = −log(0.0250) = 1.60.
(b) Ba(OH)₂: Strong base (on strong base list), diprotic → [OH⁻] = 2 × 0.0500 = 0.100 mol/L. pOH = −log(0.100) = 1.00. pH = 14.00 − 1.00 = 13.00. Sanity check: pH 13.00 > 7 ✓.
(c) Lactic acid: NOT on strong acid list → weak acid. Ka/c = 1.4 × 10⁻⁴/0.100 = 1.4 × 10⁻³ < 0.0025 ✓. x = √(1.4 × 10⁻⁴ × 0.100) = √(1.4 × 10⁻⁵) = 3.742 × 10⁻³ mol/L. Verify: 3.742 × 10⁻³/0.100 = 3.74% < 5% ✓ — assumption valid. pH = −log(3.742 × 10⁻³) = 2.43. Sanity check: pH 2.43 < 7 ✓.
(a) Moles after reaction: n(CH₃COOH) = 0.200 × 0.0500 = 0.01000 mol. n(OH⁻) = 0.200 × 0.0200 = 4.00 × 10⁻³ mol. CH₃COOH + OH⁻ → CH₃COO⁻ + H₂O (goes to completion). n(CH₃COOH) remaining = 0.01000 − 4.00 × 10⁻³ = 6.00 × 10⁻³ mol. n(CH₃COO⁻) formed = 4.00 × 10⁻³ mol.
(b) pH of buffer: pKa = −log(1.8 × 10⁻⁵) = 4.744. pH = pKa + log(n(A⁻)/n(HA)) = 4.744 + log(4.00 × 10⁻³/6.00 × 10⁻³) = 4.744 + log(0.667) = 4.744 − 0.176 = 4.57.
(c) Why pH is higher than pure 0.200 mol/L CH₃COOH: In the original pure acid solution, no CH₃COO⁻ was present initially — the equilibrium CH₃COOH ⇌ H⁺ + CH₃COO⁻ produced [H⁺] = 1.90 × 10⁻³ mol/L (pH = 2.72). After adding NaOH, the solution is a buffer containing both CH₃COOH (6.00 × 10⁻³ mol) and CH₃COO⁻ (4.00 × 10⁻³ mol). The CH₃COO⁻ already present suppresses further ionisation of CH₃COOH (common ion effect — the product of the equilibrium is already present, shifting equilibrium left by Le Chatelier). This reduces [H⁺] well below what the pure acid would produce, raising pH from ≈2.72 to 4.57. Henderson-Hasselbalch accounts for this suppression mathematically through the log(n(A⁻)/n(HA)) term.
(a) Expected pH: HNO₃ (strong acid): [H⁺] = 0.100 mol/L → pH = 1.00. HNO₂ (weak acid, Ka = 4.5 × 10⁻⁴): Check: Ka/c = 4.5 × 10⁻⁴/0.100 = 4.5 × 10⁻³ > 0.0025 ✗. x = √(4.5 × 10⁻⁴ × 0.100) = √(4.5 × 10⁻⁵) = 6.71 × 10⁻³ mol/L. Verify: 6.71% > 5% ✗ — assumption fails. Quadratic: x² + 4.5 × 10⁻⁴x − 4.5 × 10⁻⁵ = 0 → x = (−4.5 × 10⁻⁴ + √(2.025 × 10⁻⁷ + 1.8 × 10⁻⁴))/2 = (−4.5 × 10⁻⁴ + 0.01345)/2 = 6.50 × 10⁻³ mol/L → pH = −log(6.50 × 10⁻³) = 2.19.
(b) Mixing with NaOH: n(NaOH) = 0.100 × 0.0600 = 6.00 × 10⁻³ mol for both. n(HNO₃) = n(HNO₂) = 0.100 × 0.0400 = 4.00 × 10⁻³ mol for both. V(total) = 100.0 mL = 0.1000 L for both.
HNO₃ + NaOH: n(OH⁻) > n(H⁺) → excess OH⁻ = 6.00 × 10⁻³ − 4.00 × 10⁻³ = 2.00 × 10⁻³ mol. c(OH⁻) = 2.00 × 10⁻³/0.1000 = 0.0200 mol/L. pOH = 1.70. pH = 12.30 (excess strong base solution).
HNO₂ + NaOH: n(OH⁻) = 6.00 × 10⁻³ mol > n(HNO₂) = 4.00 × 10⁻³ mol. All HNO₂ is neutralised and there is excess NaOH: excess OH⁻ = 6.00 × 10⁻³ − 4.00 × 10⁻³ = 2.00 × 10⁻³ mol. c(OH⁻) = 0.0200 mol/L. pOH = 1.70. pH = 12.30 (same excess NaOH calculation — beyond equivalence point).
(c) Why different methods: For HNO₃ + NaOH: HNO₃ is strong — it is fully ionised before mixing, with H⁺ freely available. When NaOH is added in excess beyond equivalence, there is simply more OH⁻ than H⁺ — a straightforward excess calculation using c(excess OH⁻)/V(total). No equilibrium is involved after the reaction. For HNO₂ + NaOH with NaOH in excess: same logic applies — both are treated as "simple excess" calculations beyond the equivalence point. However, if NaOH were insufficient (e.g. only 20.0 mL NaOH), the HNO₂ calculation would differ: HNO₂ is weak — partial neutralisation would leave both HNO₂ and NO₂⁻ in solution simultaneously (a buffer). The common ion effect of NO₂⁻ suppresses further ionisation of HNO₂, making the simple ICE table method invalid and requiring Henderson-Hasselbalch. This is the fundamental distinction: strong acid partial neutralisation still gives a simple excess calculation (all H⁺ from the strong acid already "counts"); weak acid partial neutralisation creates a buffer that requires H-H because both HA and A⁻ are significant species at equilibrium.
Climb platforms, hit checkpoints, and answer questions on pH calculations, mixing acids and bases, and Band 6 explanations. Quick recall from lessons 1–11.