QCE Chemistry EA strategy: 2026 guide

QCEChemistryStudy guide15 min readUpdated 7 Oct 2026

Quick answer

QCE Chemistry EA rewards systematic calculation skill (equilibrium, pH, redox, electrolysis, thermochemistry), confident organic chemistry (naming plus reactions plus mechanism), integrated spectroscopy (IR plus NMR plus MS), and a six-week preparation routine that drills past papers under exam conditions.

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How the EA is structured
Topic frequency analysis
Calculation patterns
Strategy by section
Worked example: extended response
Common student errors
Six-week preparation routine
QCAA marking criteria
Check your knowledge

How the EA is structured

QCAA Chemistry EA is two papers sat in the external assessment block (26 October to 17 November 2026). Each paper is 90 minutes plus 5 minutes of perusal. QCAA sets the mark allocation and question types, so check the current syllabus and QCAA's assessment pages for the exact format.

Expect a mix of multiple choice, short response and data analysis items across the papers, with extended items integrating Unit 4 organic chemistry and analytical techniques. Use the released papers to see how items have been laid out, and remember the 2026 papers follow the 2025 syllabus.

Calculator-active. QCAA data booklet provided.

Cumulative across Unit 3 (energy and reactions) and Unit 4 (organic chemistry and analytical techniques).

Topic frequency analysis

Year on year, the following topics dominate the EA.

From Unit 3:

Equilibrium and Kc calculations (every year).
Le Chatelier predictions (every year).
pH for strong and weak acids; buffer pH (every year).
Titration curves and equivalence point (most years).
Redox half-equations and standard reduction potentials (every year).
Electrolysis with Faraday's law calculation (most years).
Rate of reaction with activation energy (most years).
Thermochemistry calorimetry calculation (most years).

From Unit 4:

IUPAC naming of organic compounds (every year).
Predicting reaction products from a named reaction class (every year).
Multi-step synthesis pathway (most years).
Integrated spectroscopy structural determination (most years).
IR + NMR + MS combined interpretation (most years).

The Great Barrier Reef stimulus turns up most years as the Le Chatelier vehicle: a CO₂ pulse acidifies the ocean carbonate equilibrium and the system shifts to dissolve carbonate.

Acidification of the Great Barrier Reef surface waters: an atmospheric CO₂ pulse drives the carbonate equilibrium toward bicarbonate and H⁺, depleting the carbonate that corals need for

\text{CaCO}_3

skeleton growth.

A strong-acid plus strong-base titration is the canonical Paper 2 figure for indicator selection: the equivalence jump straddles pH 7 and bromothymol blue (pH 6.0 to 7.6) sits cleanly inside the jump.

Strong-acid strong-base titration of 25 mL of 0.10 M HCl with 0.10 M NaOH; bromothymol blue's pH range sits entirely inside the vertical jump, so the endpoint matches equivalence within one drop.

For weak-acid buffers, Henderson-Hasselbalch plots a straight line of slope 1 against $\log([A^-]/[HA])$ and the effective region is $\text{p}K_a \pm 1$ .

Henderson-Hasselbalch line of slope 1 crossing

\text{pH} = \text{p}K_a

at the half-buffer point; QCAA buffer-design questions live inside the shaded

\text{p}K_a \pm 1

region.

Calculation patterns

Equilibrium. ICE table. Substitute into Kc expression. Solve for unknown (often a quadratic; use the small-x approximation only when x is less than 5 percent of initial concentration).

pH (strong acid). $pH = -\log_{10}[H^+]$ where [H+] equals the acid concentration.

pH (weak acid). $K_a = \frac{[H^+][A^-]}{[HA]}$ . Approximate $[H^+] = \sqrt{K_a c}$ when $c >> K_a$ , otherwise solve the quadratic.

Buffer pH. Henderson-Hasselbalch: $pH = pK_a + \log_{10}([A^-]/[HA])$ .

Titration. $c_1 V_1 / n_1 = c_2 V_2 / n_2$ for stoichiometric ratio $n_1 : n_2$ .

Redox. Balance half-equations: balance atoms first (except H and O), then O with water, then H with H+, then charge with electrons. Combine half-equations matching electron count.

Faraday's law. $Q = It$ . Moles deposited $= Q / (zF)$ where z is electrons per ion and F = 96500 C/mol.

Thermochemistry. $q = mc\Delta T$ for solution calorimetry. $\Delta H = -q / n$ (per mole of reaction).

Bond energy. $\Delta H = \sum E(\text{bonds broken}) - \sum E(\text{bonds formed})$ .

Strategy by section

Multiple choice. Read the stem carefully. Eliminate two implausible options before deciding between the remaining two. Mark and move on if uncertain.

Short response. 2 to 4 marks. Identify the chemistry concept. Apply the formula. Show working. State the answer with units and sig fig.

Extended response (8 to 10 marks). Plan first (1 to 2 minutes). Identify the components. Structure the response with subheadings or clear paragraphs. Use chemical equations as part of the argument.

Worked example: extended response

"Outline a synthesis of ethyl ethanoate from ethanol as the only organic starting material. Justify the choice of conditions and predict possible side products."

Plan. Convert some ethanol to ethanoic acid (oxidation). Esterify the remaining ethanol with the ethanoic acid.

Step 1. Oxidation of ethanol to ethanoic acid using acidified potassium dichromate (orange to green colour change). Heat under reflux to prevent loss of volatile aldehyde intermediate; allow full oxidation to the acid.

CH_3CH_2OH \xrightarrow[H^+, \Delta, \text{reflux}]{K_2Cr_2O_7} CH_3COOH

Step 2. Esterification of ethanoic acid with remaining ethanol using concentrated H2SO4 catalyst.

CH_3COOH + CH_3CH_2OH \rightleftharpoons CH_3COOCH_2CH_3 + H_2O

Conditions. Excess ethanol shifts equilibrium toward product (Le Chatelier). Heat under reflux but distil ester off as it forms (driving equilibrium further toward product). Concentrated H2SO4 acts as both catalyst and dehydrating agent.

Side products. Incomplete oxidation gives ethanal. Dehydration of ethanol (with H2SO4) gives ethene. Distillation purifies.

Common student errors

Significant figures. 3 sig fig unless data specifies otherwise. pH to 2 decimal places.

Units missing. Every numerical answer needs units.

Markovnikov mistakes. In alkene plus HX, the H adds to the carbon with more hydrogens.

Wrong functional group identification in spectroscopy. Always check the data booklet ranges before assigning.

Catastrophic error compounding. If a result is implausible (negative concentration, pH above 14), recheck.

Calculator-style data interpretation. Always interpret numerically and chemically.

Six-week preparation routine

Weeks 1-2. Key knowledge review. QCAA Chemistry syllabus as checklist. Map each subject matter point to your notes. Identify gaps.

Weeks 3-4. Calculation drills. 30 minutes per day rotating through equilibrium, pH, redox, electrolysis, thermochemistry, organic naming. One past Paper 1 per week.

Week 5. Extended response and integrated spectroscopy. Practice 8 to 10 mark items. Build a personal cheat sheet of named reactions and their conditions. One past Paper 2 per week.

Week 6. Full timed exam pairs. Two Paper 1 plus Paper 2 pairs over the week, marked against QCAA reports. Identify topics with persistent errors and revisit.

QCAA marking criteria

The EA awards marks for:

Correct chemistry (right concept, right equation, right product).
Show working (method marks even if arithmetic slips).
Significant figures (3 sig fig unless specified).
Units throughout.
Clear scientific communication.

Top band requires consistency across all five.

Check your knowledge

Six EA-style multi-part questions in the QCE Chemistry Paper 2 voice. Mark allocations follow QCAA conventions and the ISMG criterion that each part principally rewards is signposted in the solution. Attempt under exam conditions (1 minute per mark), then check against the solutions block. Three significant figures, units throughout, and explicit show-working.

A buffer is prepared by dissolving 4.10 g of sodium ethanoate ( $CH_3COONa$ , $M_r = 82.03$ ) in 250 mL of 0.200 mol L $^{-1}$ ethanoic acid ( $K_a = 1.74 \times 10^{-5}$ ). (a) Calculate the pH of the buffer. (b) Calculate the new pH after 5.00 mL of 1.00 mol L $^{-1}$ NaOH is added. (c) State, with justification, whether this buffer better resists added acid or added base. (6 marks)
Methanol is synthesised at a Gladstone gas-to-liquids plant by $CO_{(g)} + 2H_{2(g)} \rightleftharpoons CH_3OH_{(g)}$ , $\Delta H = -91 \ \text{kJ mol}^{-1}$ . (a) Predict and justify the effect on the equilibrium yield of methanol of (i) increasing temperature, (ii) decreasing volume, (iii) removing methanol as it forms. (b) The plant operates at 250 degrees C and 100 atm with a copper-zinc oxide catalyst rather than at 25 degrees C and 1 atm. Justify the choice of operating conditions in terms of yield, rate, and economics. (7 marks)
A 25.00 mL sample of vinegar is diluted to 250.0 mL. A 25.00 mL aliquot of this diluted vinegar is titrated against 0.1023 mol L $^{-1}$ NaOH; the mean titre is 24.85 mL. (a) Calculate the concentration of ethanoic acid in the original vinegar in g L $^{-1}$ ( $M_r(CH_3COOH) = 60.05$ ). (b) State the most appropriate indicator from phenolphthalein ( $pK_a = 9.4$ ), methyl orange ( $pK_a = 3.7$ ), or bromothymol blue ( $pK_a = 7.0$ ), and justify your choice. (c) The vinegar bottle claims 4.0 percent w/v ethanoic acid. Determine, with calculation, whether the measured concentration is consistent with the label. (6 marks)
(a, 3) Predict the products and write the balanced equation for electrolysis of concentrated aqueous sodium chloride (the chlor-alkali process) using inert electrodes. Justify your prediction of the cathode product over the alternative. (b, 3) Calculate the mass of chlorine gas produced when a current of 25.0 kA is passed for 1.00 hour at 95.0 percent current efficiency ( $F = 96\,500 \ \text{C mol}^{-1}$ , $M_r(Cl_2) = 70.90$ ). (6 marks)
An organic compound has molecular formula $C_4H_8O_2$ . Its IR spectrum shows a strong absorption at 1735 cm $^{-1}$ and no broad absorption near 3000 cm $^{-1}$ . The $^1$ H NMR shows three signals: a triplet at 1.0 ppm (3H), a singlet at 2.0 ppm (3H), and a quartet at 4.1 ppm (2H). The mass spectrum shows a parent ion at $m/z = 88$ and a base peak at $m/z = 43$ . (a) Identify the compound and draw its structure. (b) Assign every spectroscopic feature listed (IR band, each NMR signal with multiplicity, and the $m/z = 43$ fragment). (7 marks)
The Great Barrier Reef is acidifying. The partial pressure of atmospheric $CO_2$ has risen from 280 ppm pre-industrial to 425 ppm today; in surface seawater this equilibrates by $CO_{2(g)} + H_2O \rightleftharpoons H_2CO_{3(aq)} \rightleftharpoons H^+_{(aq)} + HCO_3^-_{(aq)}$ , with downstream equilibrium $CO_3^{2-} + H^+ \rightleftharpoons HCO_3^-$ controlling carbonate availability for coral aragonite. (a) Use Le Chatelier to explain qualitatively how rising atmospheric $CO_2$ shifts these equilibria and lowers carbonate ion concentration. (b) If surface seawater pH has fallen from 8.20 to 8.05 since pre-industrial times, calculate the percentage increase in $H^+$ concentration. (c) Discuss, with reference to the equilibria, why this seemingly small pH change is biologically significant for coral calcification. (8 marks)

Solutions

Q1: (a) Henderson-Hasselbalch: $pH = pK_a + \log([A^-]/[HA])$ . $n(CH_3COONa) = 4.10 / 82.03 = 0.0500 \ \text{mol}$ ; $[A^-] = 0.0500 / 0.250 = 0.200 \ \text{mol L}^{-1}$ . $pK_a = -\log(1.74 \times 10^{-5}) = 4.76$ . $pH = 4.76 + \log(0.200/0.200) = 4.76$ . (b) Added NaOH: $n = 0.00500 \times 1.00 = 5.00 \times 10^{-3} \ \text{mol}$ . New $n(HA) = 0.0500 - 0.00500 = 0.0450 \ \text{mol}$ ; new $n(A^-) = 0.0500 + 0.00500 = 0.0550 \ \text{mol}$ . New volume $= 0.255 \ \text{L}$ (but cancels). $pH = 4.76 + \log(0.0550/0.0450) = 4.76 + 0.087 = 4.85$ . (c) Equal initial $[HA]$ and $[A^-]$ at 0.200 mol L $^{-1}$ , so buffer capacity is symmetric; with equal moles either side, this buffer resists added acid and added base equally. This question principally tests application of equilibrium (application criterion) and judgement of buffer behaviour (judgement criterion).
The original buffer pH is 4.76 and rises to 4.85 after addition; capacity is symmetric.
Q2: (a) (i) Forward reaction is exothermic ( $\Delta H < 0$ ); increasing temperature shifts equilibrium reverse, decreasing methanol yield and $K_c$ . (ii) Decreasing volume increases all partial pressures; equilibrium shifts to the side with fewer moles of gas (3 mol reactants $\rightarrow$ 1 mol product), increasing methanol yield ( $K_c$ unchanged). (iii) Removing methanol shifts equilibrium forward, increasing conversion (Le Chatelier). (b) 250 degrees C and 100 atm is a rate-yield-cost compromise: at 25 degrees C the rate is uneconomic even with catalyst; higher temperature drops equilibrium yield but increases rate, and the copper-zinc catalyst raises the rate further without shifting equilibrium. 100 atm increases yield without the construction cost of a 300+ atm plant. Unconverted CO and $H_2$ are recycled, so a moderate per-pass yield at acceptable rate maximises plant throughput per dollar. This question rewards application of Le Chatelier (application criterion) and judgement against industrial constraints (judgement criterion).
Temperature decreases yield, pressure increases yield, product removal drives conversion; industrial conditions optimise rate-yield-economics.
Q3: (a) Moles NaOH per titration $= 0.1023 \times 0.02485 = 2.542 \times 10^{-3} \ \text{mol}$ . By 1:1 stoichiometry, moles ethanoic acid in 25.00 mL of dilute $= 2.542 \times 10^{-3} \ \text{mol}$ . Scaling to 250.0 mL of dilute: $\times 10 = 2.542 \times 10^{-2} \ \text{mol}$ , which came from 25.00 mL of original. Concentration in original $= 2.542 \times 10^{-2} / 0.02500 = 1.017 \ \text{mol L}^{-1}$ . In g L $^{-1}$ : $1.017 \times 60.05 = 61.1 \ \text{g L}^{-1}$ . (b) Weak acid versus strong base equivalence point lies in the basic region (pH around 8.7); phenolphthalein ( $pK_a = 9.4$ , transition 8.2 to 10.0) brackets this. Methyl orange would change colour well before equivalence; bromothymol blue is marginal. (c) Label 4.0 percent w/v $= 40 \ \text{g L}^{-1}$ ; measured 61.1 g L $^{-1}$ is 53 percent higher, well outside any reasonable measurement uncertainty, so the vinegar is not consistent with the label (this would be a regulatory non-conformance). This question principally tests knowledge and analysis of titration (knowledge and analysis criterion) and judgement against a labelled value (judgement criterion).
The vinegar is 61.1 g L $^{-1}$ ethanoic acid; phenolphthalein is the correct indicator; the result is inconsistent with the 4.0 percent label.
Q4: (a) At the cathode, both $H_2O$ ( $E^{\circ} = -0.83 \ \text{V}$ ) and $Na^+$ ( $E^{\circ} = -2.71 \ \text{V}$ ) compete; water has the higher reduction potential and is reduced preferentially, giving $H_2$ . At the anode, both $Cl^-$ ( $E^{\circ} = +1.36 \ \text{V}$ ) and $H_2O$ ( $E^{\circ} = +1.23 \ \text{V}$ ) compete; on thermodynamics alone water should oxidise first, but overpotential at most electrodes makes $Cl_2$ evolution kinetically favoured, and in concentrated brine the chloride concentration further favours $Cl_2$ . Overall: $2NaCl_{(aq)} + 2H_2O_{(l)} \rightarrow 2NaOH_{(aq)} + H_{2(g)} + Cl_{2(g)}$ . (b) Charge $Q = It = 25\,000 \times 3600 = 9.00 \times 10^{7} \ \text{C}$ . Effective charge at 95 percent: $0.95 \times 9.00 \times 10^{7} = 8.55 \times 10^{7} \ \text{C}$ . Moles electrons $= 8.55 \times 10^{7} / 96\,500 = 886 \ \text{mol}$ . Moles $Cl_2 = 886 / 2 = 443 \ \text{mol}$ (two electrons per $Cl_2$ ). Mass $= 443 \times 70.90 = 3.14 \times 10^{4} \ \text{g} = 31.4 \ \text{kg}$ . This question tests application of electrochemistry (application criterion) and justification of kinetic over thermodynamic prediction (judgement criterion).
The cathode produces $H_2$ and the anode produces $Cl_2$ with 31.4 kg of $Cl_2$ generated per hour at 25.0 kA and 95 percent efficiency.
Q5: (a) Compound is ethyl ethanoate, $CH_3COOCH_2CH_3$ . (b) IR 1735 cm $^{-1}$ is the C=O stretch of an ester; absence of broad band near 3000 cm $^{-1}$ rules out carboxylic acid O-H or alcohol O-H. $^1$ H NMR: 1.0 ppm triplet (3H) is the $OCH_2\underline{CH_3}$ methyl coupled to the adjacent $CH_2$ (n+1 = 3); 2.0 ppm singlet (3H) is the acetyl methyl $\underline{CH_3}CO$ with no neighbouring H; 4.1 ppm quartet (2H) is the $O\underline{CH_2}CH_3$ deshielded by the ester oxygen and coupled to the methyl (n+1 = 4). Mass spectrum: $M^+ = 88$ matches $C_4H_8O_2$ ; base peak at $m/z = 43$ is the acylium cation $CH_3CO^+$ formed by alpha-cleavage of the C-O bond adjacent to the carbonyl. This question rewards application of spectroscopic rules (application criterion) and synthesis across IR, NMR, and MS (knowledge and analysis criterion).
The compound is ethyl ethanoate; every spectroscopic feature is consistent with this structure.
Q6: (a) Higher atmospheric $p(CO_2)$ shifts the first equilibrium forward, increasing dissolved $H_2CO_3$ . This shifts the second equilibrium forward, increasing $[H^+]$ and $[HCO_3^-]$ (lowering pH). Increased $[H^+]$ shifts the third equilibrium ( $CO_3^{2-} + H^+ \rightleftharpoons HCO_3^-$ ) to the right, lowering $[CO_3^{2-}]$ . (b) Pre-industrial $[H^+] = 10^{-8.20} = 6.31 \times 10^{-9} \ \text{mol L}^{-1}$ ; today $[H^+] = 10^{-8.05} = 8.91 \times 10^{-9} \ \text{mol L}^{-1}$ . Percentage increase $= (8.91 - 6.31) / 6.31 \times 100 = 41 \ \text{percent}$ . (c) Coral aragonite ( $CaCO_3$ ) precipitation requires saturation of $Ca^{2+}$ and $CO_3^{2-}$ . The 41 percent rise in $[H^+]$ depletes $[CO_3^{2-}]$ via the third equilibrium, reducing the saturation state $\Omega_{ar}$ and slowing calcification rates; below $\Omega_{ar} = 3.0$ , branching corals on the GBR begin to lose net calcification, which is why a 0.15 pH unit drop has outsized ecological significance. This question tests application of Le Chatelier (application criterion), knowledge and analysis of pH calculation (knowledge and analysis criterion), and judgement of ecological significance (judgement criterion); the discussion in (c) directly addresses the communication criterion through structured paragraph linkage of chemistry to biology.
Atmospheric $CO_2$ rise drives all three equilibria toward reducing $[CO_3^{2-}]$ , raises $[H^+]$ by 41 percent, and depletes carbonate available for coral aragonite formation.

Sources & how we know this

Chemistry General senior syllabus 2025 — QCAA