Year 12 Chemistry Module 6, Acid/Base Reactions ★ Consolidation ⏱ ~45 min Lesson 11 of 19 IQ2 Mastery

IQ2 Mastery, pH Calculations, Mixing & Method Selection

Three error patterns recur in IQ2 pH calculation questions, applying [H⁺] = c to weak acids, forgetting to verify the 5% assumption, and using the wrong total volume in mixing calculations. Three students attempt the same three pH problems below. Two are correct; one makes the most common HSC error. Before reading on, can you identify which one?

Optional consolidation. This lesson introduces no new syllabus dot points, it consolidates material taught earlier in the module. It is valuable practice, and it is not required to complete the core route.

Today's hook: Three specific pH calculation errors account for most lost marks on this topic. This lesson diagnoses all three, and gives you the exact fixes. Can you spot the error before reading on?

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Warm up first

Three quick questions from earlier lessons. Pulling old material back to mind before you learn something new makes the new material stick better, so this is not busywork.

Worksheets

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Four printable worksheets that build from the foundations up to exam-style questions, start at whatever level suits you.

Build Foundations & guided practice Apply Application practice Master Mastery challenge Exam Exam-style questions

★ CONSOLIDATION, NO NEW SYLLABUS DOT POINTS

This lesson deepens and integrates content from L08–L10: strong/weak pH calculations, ICE tables, mixing problems, Henderson-Hasselbalch, and ΔHn comparisons. The goal is to eliminate the five highest-frequency IQ2 calculation errors through explicit diagnosis, decision-tree thinking, and high-mark extended response practice.

Think First

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.

🌳 Key Relationships, This Lesson

Step 1: Acid or base? → Step 2: Strong or weak?

Strong acid (HCl, HNO₃, H₂SO₄, HClO₄, HBr, HI): [H⁺] = c × n(H⁺) → pH = −log[H⁺] Weak acid (everything else): ICE table → Ka = x²/(c−x) → x = √(Ka×c) or quadratic → pH = −log(x) Strong base (NaOH, KOH, Ca(OH)₂, Ba(OH)₂): [OH⁻] = c × n(OH⁻) → pOH → pH = 14 − pOH Weak base: ICE table (Kb) → [OH⁻] = x → pOH → pH = 14 − pOH

Mixing strong acid + strong base:

n(H⁺) = c(acid) × V(acid) | n(OH⁻) = c(base) × V(base) × n(OH⁻ per unit) Excess → c(excess) = n(excess) / V(total) | V(total) = V(acid) + V(base) always

Partial neutralisation (weak acid + strong base, before EP):

HA + OH⁻ → A⁻ + H₂O (goes to completion) pH = pKa + log(n(A⁻)/n(HA)) | At half-EP: n(A⁻) = n(HA) → pH = pKa

Five-error checklist (run after every multi-step pH calculation):

(1) Did I identify acid/base type? (2) Did I check the 5% assumption? (3) For base: pH = 14 − pOH? (4) In mixing: V(total) used? (5) If A⁻ present from neutralisation: used H-H, not ICE table?

Learning Intentions

Know

Key facts

The decision tree: strong/weak → acid/base → given concentration or Ka → select method
Common error patterns in IQ2 calculations and their diagnoses
The six strong acids: HCl, H₂SO₄ (1st), HNO₃, HClO₄, HBr, HI

Understand

Concepts

Why method selection depends on what information is given and what is asked
Why V(total) is always required when mixing two solutions
How to diagnose errors by working backwards from incorrect answers

Can do

Skills

Select the correct calculation method for any pH scenario
Diagnose and fix the five most common IQ2 calculation errors
Justify method selection using chemical reasoning in high-mark responses

Key Terms

pH of mixed solutions

When two solutions are mixed, calculate moles of H⁺ and OH⁻; find excess moles; divide by total volume.

Neutralisation point

When moles H⁺ = moles OH⁻; [H⁺] = [OH⁻] = 10⁻⁷ mol L⁻¹; pH = 7 for strong acid/base.

Excess acid calculation

After mixing, n(excess H⁺) = n(H⁺) − n(OH⁻); then [H⁺] = n(excess)/V(total).

Excess base calculation

After mixing, n(excess OH⁻) = n(OH⁻) − n(H⁺); then [OH⁻] = n(excess)/V(total); find pH via pOH.

high-mark explanation

Requires correct identification of species, justification of pH change, and linking to Kw and equilibrium.

Dilution effect on pH

Diluting a strong acid reduces [H⁺] proportionally; pH approaches 7 but never exceeds it for an acid.

Core Content

Cross-lesson links: This mastery lesson consolidates all IQ2 pH skills from L07–L10, Kw, ICE tables, Ka/Kb, dilution, and mixing. The five error types diagnosed here also appear in buffer calculations (L12–L13) and every titration curve region (L14–L17). Passing this lesson means you can answer any IQ2 question on the HSC.

Who is correct? Three students

Card 1, Who Is Correct? The Three Students Analysed

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 high-mark 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.

Error Diagnosis

Student B writes [H⁺] = 0.100 mol/L for CH₃COOH, treating it as if it were completely ionised. But CH₃COOH is a weak acid with Ka = 1.8 × 10⁻⁵. At equilibrium, only ~1.3% of CH₃COOH molecules have ionised. [H⁺] = 1.34 × 10⁻³ mol/L, not 0.100 mol/L. The stated pH = 1.00 is approximately 1.87 units too low, overestimating [H⁺] by a factor of 74. Student B also reveals a secondary error: calculating Ka = 0.100 from the assumed concentrations treats Ka as a calculated result, when it is a fixed constant (1.8 × 10⁻⁵), not something derived from assumed [H⁺].

Fix

Identify acid type first: CH₃COOH is NOT on the six strong acid list → weak acid → ICE table required. Check assumption: Ka/c = 1.8 × 10⁻⁵/0.100 = 1.8 × 10⁻⁴ << 0.0025 ✓. x = √(1.8 × 10⁻⁵ × 0.100) = 1.34 × 10⁻³ mol/L. Verify: 1.34% < 5% ✓. Correct pH = 2.87.

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 ✓

Must Do

Before writing a single equation or number in any pH calculation, identify the acid or base type. Write this explicitly at the start of your working: "HNO₃ is a strong acid, complete ionisation" or "CH₃COOH is a weak acid, ICE table required." This five-second identification step prevents the most common category of error in the entire module and is required for full method marks.

Common Error

Student B's error, using [H⁺] = c for a weak acid, is the most frequently penalised calculation error in Module 6 HSC marking. It typically costs 3–4 marks across a multi-part question because every subsequent calculation (Ka, ΔHn comparison, buffer pH) built on this wrong [H⁺] is also wrong.

Student A (HNO₃, strong) is correct: [H⁺] = c, pH = 1.00. Student B (CH₃COOH) error: applied [H⁺] = c to a weak acid, overestimates [H⁺] by 74×; correct pH = 2.87 from ICE table. Student C (HF, Ka = 6.8 × 10⁻⁴) correct: assumption fails (8.25% > 5%), uses quadratic → pH = 2.10. Always identify acid type FIRST.

Pause, copy the highlighted definition into your book before moving on.

Student B's fundamental error in calculating pH of CH₃COOH was:

The method decision tree

Card 2, The Decision Tree: Selecting the Right Method Every Time

We just saw that Student B's error was skipping acid identification before selecting the method. That raises a question: What is the complete decision tree that covers every IQ2 scenario so you never pick the wrong method? This card answers it → six scenarios keyed to identification questions: strong/weak acid, strong/weak base, mixing, partial neutralisation, each has a unique method.

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.

Must Do

Laminate this decision tree in your working memory before the HSC. In every exam question involving pH, write the identification step first, then select the method. The decision tree does not slow you down, it prevents the 3–4 mark error of using the wrong method, which always costs more time to recover from than the 5 seconds the identification step takes.

Common Error

Students treat H₂SO₄ as monoprotic and calculate [H⁺] = c(H₂SO₄). H₂SO₄ is diprotic, in dilute solution both protons are donated → [H⁺] = 2 × c(H₂SO₄). A 0.050 mol/L H₂SO₄ solution gives [H⁺] = 0.100 mol/L → pH = 1.00, not pH = 1.30. This is an error on the strong acid branch, equally common and equally costly as the strong/weak confusion.

Six-scenario decision tree: (1) Strong acid → [H⁺] = c → pH; (2) Weak acid → ICE table → x = √(Ka × c) → check < 5% → pH; (3) Strong base → [OH⁻] = c × n → pOH → pH = 14 − pOH; (4) Weak base → ICE Kb table → [OH⁻] → pOH → pH; (5) Mixing SA+SB → moles → excess → n/V(total); (6) Partial neutralisation → Henderson-Hasselbalch.

Add the highlighted point to your notes before the check below.

0.050 mol/L H₂SO₄ gives [H⁺] of:

Deep dive, Extension. The decision tree itself is useful core revision. The ICE-table, weak-base Kb and Henderson–Hasselbalch branches within it are Extension, and are scored separately from the basic core route.

Analogy: medical triage

Card 3, Analogy 1: The Medical Triage System

We just saw the six-scenario decision tree, identification before calculation. That raises a question: Is there an analogy that makes the "identify first, calculate second" principle unforgettable? This card answers it → hospital triage: diagnose the patient (acid type) before prescribing treatment (formula); applying the right treatment to the wrong patient always fails.

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.

Insight

Where this analogy breaks down: A patient's condition can be ambiguous and require further testing. In chemistry, the acid type identification is unambiguous once the six strong acid list is memorised, HCl, H₂SO₄ (1st), HNO₃, HClO₄, HBr, HI are always strong, everything else is always weak, with no grey area. The analogy correctly captures the sequential diagnosis-then-treatment logic but overstates the ambiguity in the chemistry diagnosis step.

Must Do

Use the triage analogy in study practice. Before each pH calculation, physically write or say aloud: "Triage: this acid is [strong/weak], method is [direct/ICE table]." This verbal triage step takes five seconds and hard-wires the identification habit before the habit of jumping straight to calculation can override it.

Triage principle: acid type identification (strong/weak) must come BEFORE formula selection, identical to hospital triage (diagnose before treating). Wrong method + perfect arithmetic = wrong answer every time. The six strong acid list (HCl, HNO₃, H₂SO₄, HClO₄, HBr, HI) makes the diagnosis unambiguous; everything else is weak with no grey area.

Pause, write the highlighted definition into your book.

The triage analogy is limited because acid type identification requires further testing to confirm, just like medical diagnosis.

Analogy: currency exchange (total volume)

Card 4, Analogy 2: The Currency Exchange (Why V(total) Is Required in Mixing)

We just saw the triage analogy for acid identification. That raises a question: What about mixing errors, why do students keep forgetting V(total) and using only one volume? This card answers it → currency exchange analogy: the excess species "occupies" the total wallet (total volume); using only V(acid) artificially inflates the concentration.

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).

Common Error

Using V(acid) alone as the total volume after mixing always overestimates c(excess) by a factor of V(total)/V(acid). For a 50 mL + 50 mL mix, this factor is 2, the error doubles [H⁺] or [OH⁻], reducing pH by log(2) = 0.30 units for excess acid. This 0.30 pH unit error is large enough to move a correct answer into the wrong mark band. Write "V(total) = V₁ + V₂ = ___" explicitly before calculating c(excess).

Insight

Where this analogy breaks down: Currency exchange involves a physical exchange, AUD is gone, replaced by Euros. In neutralisation, H⁺ and OH⁻ do not exchange, they react and are destroyed, forming H₂O. The remaining species simply occupies a larger volume. The analogy correctly captures the volume-expansion aspect but misrepresents the chemistry of what happens to the reacting species.

Mixing volume rule: c(excess) = n(excess) / V(total) where V(total) = V(acid) + V(base). Using only V(acid) doubles [H⁺] for equal volumes, causing a 0.30 pH unit error. Write "V(total) = V₁ + V₂ = ___" explicitly before calculating c(excess), this one line prevents the most common arithmetic error in mixing calculations.

Add the highlighted point to your notes before the check below.

After mixing 30 mL of acid with 20 mL of base, the total volume used in c(excess) = n/V is:

Five IQ2 errors and misconceptions

Beyond the syllabus. The method-selection decision tree is useful revision. Items relying on Henderson–Hasselbalch or weak-base Kb calculations are extension and sit outside the core score — for the exam, the strong acid/base and Ka pathways are what you need.

Card 5, The Five Most Common IQ2 Calculation Errors: Diagnosed with Fixes

We just saw the currency exchange analogy for V(total) in mixing calculations. That raises a question: What are all five of the highest-frequency IQ2 calculation errors in one consolidated reference? This card answers it → five errors: wrong method for weak acid, unchecked assumption, pOH reported as pH, wrong volume in mixing, ICE table applied to a buffer, all sharing the same root cause.

A student writes: "pH = −log(1.8 × 10⁻⁵) = 4.74" for a 0.1 mol/L acetic acid solution, confusing pKa with pH. Another calculates [H⁺] = 0.1 mol/L for 0.1 mol/L CH₃COOH, applying the strong acid shortcut to a weak acid. These two are the most common errors in IQ2 pH calculation responses. Here are all five, diagnosed with exact fixes.

Error 1, Using [H⁺] = c for a weak acid

Diagnosis

Applied strong acid shortcut without identifying acid type

Fix

Identify acid type first (six strong acid list); if weak, set up ICE table

Error 2, Forgetting to check the simplifying assumption

Diagnosis

Used x = √(Ka × c) without verifying x/c < 5%

Fix

After calculating x, always compute (x/c) × 100% before accepting the result. If ≥ 5%, apply quadratic

Error 3, Writing pOH as pH for a base

Diagnosis

Calculated pOH correctly but wrote it as the final answer without subtracting from 14

Fix

For any base, the final step is always pH = 14 − pOH. Sanity check: pH must be > 7 for a base

Error 4, Using only V(acid) or V(base) in c(excess) calculation

Diagnosis

Used the original component volume instead of the combined total volume

Fix

V(total) = V(acid) + V(base) always. Write this line explicitly before calculating c(excess)

Error 5, Applying ICE table to a buffer mixture after partial neutralisation

Diagnosis

Used x = √(Ka × c) for a solution containing both HA and A⁻ produced by neutralisation

Fix

If A⁻ is present in significant amounts, the solution is a buffer, use Henderson-Hasselbalch: pH = pKa + log(n(A⁻)/n(HA))

Must Do

Run the five-error checklist at the end of every multi-step pH calculation: (1) Did I identify acid/base type before calculating? (2) Did I check the assumption if I used the square root shortcut? (3) For a base, did I subtract pOH from 14? (4) In a mixing problem, did I use V(total)? (5) If A⁻ was formed by neutralisation, did I use Henderson-Hasselbalch? Five checks, 30 seconds, eliminates the five highest-frequency errors in one pass.

Five IQ2 errors checklist: (1) [H⁺] = c for weak acid, identify type first, use ICE table; (2) simplifying assumption skipped, verify x/c × 100% < 5%; (3) pOH reported as pH for base, always pH = 14 − pOH; (4) V(acid) used instead of V(total), write V₁ + V₂ explicitly; (5) ICE table applied to buffer after partial neutralisation, use Henderson-Hasselbalch when both HA and A⁻ present. Root cause of all five: method selected before system identified.

Pause, write the highlighted checklist into your book before the check below.

After partial neutralisation of CH₃COOH with NaOH, leaving both CH₃COOH and CH₃COO⁻ present, the correct method to find pH is:

Misconceptions to Fix

✗ Wrong: "pH = pKa for any weak acid solution."

✓ Right: 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.

✗ Wrong: "If Ka(acid) = Kb(base), they must be a conjugate pair."

✓ Right: 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.

✗ Wrong: "After partial neutralisation, I use the ICE table on the remaining HA."

✓ Right: 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.

✗ Wrong: "H₂SO₄ is monoprotic, [H⁺] = c(H₂SO₄)."

✓ Right: H₂SO₄ is diprotic (both ionisations are essentially complete in dilute solution) → [H⁺] = 2 × c(H₂SO₄).

Worked examples and activities

Calculate pH of each at 25°C: (a) 0.0400 mol/L KOH; (b) 0.0400 mol/L HBr; (c) 0.0400 mol/L HCN (Ka = 6.2 × 10⁻¹⁰)

(a) KOH, Identification: KOH is a strong base. Method: direct. [OH⁻] = c(KOH) = 0.0400 mol/L. 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. 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. Check assumption: Ka/c = 6.2 × 10⁻¹⁰/0.0400 = 1.55 × 10⁻⁸ << 0.0025 ✓. x = √(6.2 × 10⁻¹⁰ × 0.0400) = 4.98 × 10⁻⁶ mol/L. Verify: 0.012% << 5% ✓. pH = −log(4.98 × 10⁻⁶) = 5.30.

ANSWERS: (a) pH = 12.60. (b) pH = 1.40. (c) pH = 5.30.

(a) 35.0 mL of 0.200 mol/L Ba(OH)₂ + 65.0 mL of 0.150 mol/L HNO₃ → pH? (b) Repeat with 0.0800 mol/L HNO₃.

Common setup: n(OH⁻) from Ba(OH)₂ = 0.200 × 0.0350 × 2 = 1.40 × 10⁻² mol. V(total) = 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. Excess OH⁻ = 0.01400 − 0.009750 = 4.25 × 10⁻³ mol. c(OH⁻) = 4.25 × 10⁻³/0.1000 = 0.0425 mol/L. pOH = 1.37. pH = 12.63.

(b) HNO₃ at 0.0800 mol/L: n(H⁺) = 0.0800 × 0.0650 = 5.20 × 10⁻³ mol. Excess OH⁻ = 0.01400 − 0.005200 = 8.80 × 10⁻³ mol. c(OH⁻) = 0.0880 mol/L. pOH = 1.056. pH = 12.94.

ANSWERS: (a) pH = 12.63. (b) pH = 12.94.

0.100 mol/L solutions: Sol 1 = HCl; Sol 2 = CH₃COOH (Ka = 1.8×10⁻⁵); Sol 3 = NH₃ (Kb = 1.8×10⁻⁵). (a) pH each. (b) Mix 40.0 mL Sol 1 + 60.0 mL 0.100 mol/L NaOH. (c) Add 25.0 mL 0.100 mol/L NaOH to 25.0 mL Sol 2. (d) Explain Ka × Kb relationship.

(a) pH: Sol 1 (HCl, strong): pH = 1.00. Sol 2 (CH₃COOH): x = √(1.8×10⁻⁵×0.100)=1.34×10⁻³; 1.34% <5% ✓. pH = 2.87. Sol 3 (NH₃): [OH⁻]=1.34×10⁻³; pOH=2.87. pH = 11.13.

(b) HCl + NaOH: n(H⁺) = 4.00×10⁻³ mol. n(OH⁻) = 6.00×10⁻³ mol. Excess OH⁻ = 2.00×10⁻³ mol. V(total) = 0.1000 L. c(OH⁻) = 0.0200 mol/L. pOH = 1.70. pH = 12.30.

(c) Equal moles → equivalence point: All CH₃COOH → CH₃COO⁻. n(CH₃COO⁻) = 2.50×10⁻³ mol. c = 0.0500 mol/L. Kb = 5.56×10⁻¹⁰. [OH⁻] = 5.27×10⁻⁶ mol/L. pOH = 5.28. pH = 8.72 (basic salt solution).

(d) Ka × Kb relationship: Ka(CH₃COOH) × Kb(NH₃) = 3.24×10⁻¹⁰ ≠ Kw (1.0×10⁻¹⁴). Therefore they are NOT a conjugate pair. Equal Ka and Kb is a numerical coincidence. pH symmetry (2.87 + 11.13 = 14) arises from equal equilibrium constants, not conjugate relationship.

ANSWERS: (a) 1.00 / 2.87 / 11.13. (b) pH = 12.30. (c) EP; pH = 8.72. (d) Ka×Kb ≠ Kw → not conjugate; equal Ka and Kb is coincidence.

📓 Lesson 11 Consolidation Checklist

Always identify acid/base type before selecting calculation method, write it explicitly
Strong acids: HCl, H₂SO₄ (1st), HNO₃, HClO₄, HBr, HI, everything else is weak
H₂SO₄ is diprotic: [H⁺] = 2 × c(H₂SO₄) in dilute solution
Always check x/c × 100% < 5% after using the simplifying assumption, if ≥ 5%, apply quadratic
For bases: pH = 14 − pOH always; sanity check pH > 7
Mixing: V(total) = V₁ + V₂, write this before calculating c(excess)
Buffer (HA + A⁻ both present): use Henderson-Hasselbalch, NOT ICE table
Ka × Kb = Kw applies only within conjugate pairs, not for any random acid and base

✏️ Activities

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.

Student P (0.050 mol/L HCOOH, Ka = 1.8 × 10⁻⁴): "[H⁺] = 0.050 mol/L; pH = −log(0.050) = 1.30." Error type: ___ Correct pH: ___
Student Q (0.100 mol/L NH₃, Kb = 1.8 × 10⁻⁵): "[OH⁻] = √(1.8 × 10⁻⁵ × 0.100) = 1.34 × 10⁻³; pH = −log(1.34 × 10⁻³) = 2.87." Error type: ___ Correct pH: ___
Student R (mixing: 30.0 mL of 0.200 mol/L HCl + 20.0 mL of 0.100 mol/L NaOH): "excess H⁺ = 4.0 × 10⁻³ mol; c(H⁺) = 4.0 × 10⁻³/0.030 = 0.133 mol/L; pH = 0.875." Error type: ___ Correct pH: ___
Student S (HF, Ka = 6.8 × 10⁻⁴, c = 0.100 mol/L): "x = √(6.8 × 10⁻⁴ × 0.100) = 8.25 × 10⁻³ mol/L. pH = −log(8.25 × 10⁻³) = 2.08." Error type: ___ Correct pH: ___
Student T (25.0 mL 0.100 mol/L CH₃COOH + 12.5 mL 0.100 mol/L NaOH added; Ka = 1.8 × 10⁻⁵): "I use the ICE table: x = √(Ka × 0.100) = 1.34 × 10⁻³; pH = 2.87." Error type: ___ Correct pH: ___

For each of the following, state the identification (acid/base type), the method, and calculate pH. Show all working.

0.200 mol/L H₂SO₄ (treat both ionisations as complete in dilute solution)
0.0500 mol/L Ca(OH)₂
0.150 mol/L CH₃NH₂ (methylamine, Kb = 4.4 × 10⁻⁴)
40.0 mL of 0.100 mol/L HCl mixed with 10.0 mL of 0.100 mol/L NaOH
30.0 mL of 0.200 mol/L CH₃COOH (Ka = 1.8 × 10⁻⁵) mixed with 10.0 mL of 0.200 mol/L NaOH

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50 mL of 0.2 mol/L HCl is mixed with 30 mL of 0.2 mol/L NaOH. Before calculating, predict: (a) will the final solution be acidic, basic, or neutral? (b) which reagent is in excess?

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Use the calculator. A solution has [OH⁻] = 1.0 × 10⁻⁵ mol/L at 25 °C. What is its pH?

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Practice questions

Multiple Choice

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Short Answer

Question 6. (4 marks) Calculate the pH of each at 25°C. Show the identification step and full working. (a) 0.0250 mol/L HClO₄. (b) 0.0500 mol/L Ba(OH)₂. (c) 0.100 mol/L lactic acid (Ka = 1.4 × 10⁻⁴). State whether the simplifying assumption is valid for each.

Question 7. (4 marks) 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. (b) Calculate the pH. (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. (7 marks, Band 6) A student has unlabelled solutions of 0.100 mol/L HNO₃ and 0.100 mol/L HNO₂ (Ka = 4.5 × 10⁻⁴). (a) Calculate expected pH of each. (2 marks) (b) Mix 40.0 mL of each with 60.0 mL of 0.100 mol/L NaOH, calculate pH showing solution type identification. (3 marks) (c) Explain why the two mixing calculations require different methods despite using "the same NaOH." (2 marks)

Show All Answers

Weak-acid [H⁺] error. HCOOH is weak; [H⁺] = c is Error 1. ICE table: x = 3.0×10⁻³; 6.0% > 5% → quadratic; pH ≈ 2.53.

Using the combined volume. V(total) = 50.0 mL; c(H⁺) = 0.0300 mol/L; pH = 1.52.

When pH + pOH sums to 14. pH sum = 14 because Ka = Kb (coincidence), not because they are conjugate. Conjugate pairs require 1-proton difference in formula.

Equivalence-point method. Equal moles → EP; Kb(CH₃COO⁻) = 5.56×10⁻¹⁰; pH = 8.72.

ICE vs Henderson-Hasselbalch. ICE table for pure weak acid; H-H when A⁻ present from partial neutralisation.

Q6 Sample: (a) HClO₄: strong → pH = −log(0.0250) = 1.60. (b) Ba(OH)₂: diprotic → [OH⁻] = 0.100 → pOH = 1.00 → pH = 13.00. (c) Lactic acid: weak; Ka/c = 1.4×10⁻³ < 0.0025 ✓; x = 3.742×10⁻³; 3.74% < 5% ✓; pH = 2.43.

Q7 Sample: (a) n(HA) = 0.01000 mol; n(OH⁻) = 4.00×10⁻³; n(HA)remaining = 6.00×10⁻³; n(A⁻) = 4.00×10⁻³. (b) pKa = 4.744; pH = 4.744 + log(4/6) = 4.57. (c) CH₃COO⁻ already present suppresses ionisation (common ion effect, Le Chatelier shifts left); H-H accounts for this suppression.

Q8 Sample: (a) HNO₃: pH = 1.00. HNO₂: assumption check 6.71% > 5% → quadratic → x = 6.50×10⁻³ → pH = 2.19. (b) Both give excess NaOH = 2.00×10⁻³ mol; c(OH⁻) = 0.0200; pOH = 1.70; pH = 12.30 for both. (c) If NaOH insufficient for HNO₂: partial neutralisation creates buffer (HA + A⁻ both present), requiring H-H; for HNO₃: strong acid fully ionised, no buffer possible, simple excess calculation always.

Calculation practice

Written calculation questions in the style and mark range of the HSC paper. Set your working out in full, because method marks are awarded for the steps as well as the final value.

AnalyseBand 5(3 marks) A 40.00 mL sample of 0.100 mol/L hydrochloric acid is mixed with 60.00 mL of 0.0500 mol/L sodium hydroxide solution at 25 °C. Calculate the pH of the resulting mixture.

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Retrieve and reflect

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🏆 Review

What is the first step in any pH calculation?

Why is [H⁺] = c wrong for CH₃COOH?

What volume must you use when calculating c(excess) after mixing acid and base?

When is Henderson-Hasselbalch used instead of an ICE table?

Does Ka(CH₃COOH) = Kb(NH₃) mean they are a conjugate pair?

Revisit Think First

Return to your diagnosis of the three students. The 2023 HSC marking notes flagged exactly these errors. The full analysis:

Student A (HNO₃): Correct. HNO₃ is a strong acid → [H⁺] = c = 0.100 mol/L → pH = 1.00.
Student B (CH₃COOH): Incorrect. Applied [H⁺] = c to a weak acid, Error 1. CH₃COOH is ~1.3% ionised at 0.100 mol/L; [H⁺] = 1.34 × 10⁻³ mol/L. Correct pH = 2.87. Student B also misused Ka, treating it as a calculated result rather than a fixed constant.
Student C (HF): Correct. Correctly identified as weak, used ICE table, caught the failed assumption (8.25% > 5%), applied quadratic, obtained pH = 2.10.

Two students were correct (A and C). The only incorrect student (B) made the single most common Module 6 error.

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