WACE Chemistry Unit 3 deep dive: equilibrium, acids and bases, and redox for the 2026 exam

WACEChemistryStudy guide19 min readUpdated 23 Sep 2026

Revision deep dive for WACE Chemistry Unit 3: dynamic equilibrium and Le Chatelier's principle, equilibrium constants, solubility, acids and bases, pH and Kw, buffers, indicators and titrations, oxidation numbers and half-equations, galvanic and electrolytic cells, and corrosion, with worked examples and links to every Unit 3 dot point.

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How Unit 3 fits the exam
1. Dynamic equilibrium and Le Chatelier's principle
2. Equilibrium constants
3. Solubility equilibria
4. Acids, bases and pH
5. Buffers, indicators and titrations
6. Redox: oxidation numbers and half-equations
7. Galvanic cells and standard electrode potentials
8. Electrolysis
9. Corrosion
Common mistakes
Check your knowledge

How Unit 3 fits the exam

Unit 3 (equilibrium, acids and bases, and redox reactions) is half of the WACE Chemistry exam content, and it supplies most of the calculations. This deep dive runs through every Unit 3 dot point on the site in a sensible revision order. For format, timing and conventions such as observations and state symbols, see the exam strategy guide. Unit 4 is covered in the organic chemistry and synthesis deep dive.

1. Dynamic equilibrium and Le Chatelier's principle

Dot point: chemical equilibrium and Le Chatelier's principle.

In a closed system, a reversible reaction reaches dynamic equilibrium when the forward and reverse rates are equal. Both reactions continue; the concentrations stay constant.

Change	Effect on rates	Shift	Kc
Add a reactant	Forward rate increases	Right	Unchanged
Halve the volume (gases)	Rate of the side with more gas particles increases more	Towards fewer gas particles	Unchanged
Increase temperature	Both rates increase, the endothermic direction more	Endothermic direction	Changes
Add a catalyst	Both rates increase equally	None	Unchanged
Add an inert gas at constant volume	No change in concentrations	None	Unchanged

Reading a concentration-time graph

For $2\text{NO}_2(g) \rightleftharpoons \text{N}_2\text{O}_4(g)$ (brown to colourless), the volume is suddenly halved.

On the graph, both concentrations jump up at the instant of the change (same moles, half the volume). Then $[\text{NO}_2]$ gradually falls and $[\text{N}_2\text{O}_4]$ gradually rises as the system shifts towards fewer gas particles.

Observation: the gas mixture darkens immediately, then fades partly, ending darker than it was originally because $[\text{NO}_2]$ is still higher than before the change.

2. Equilibrium constants

Dot points: equilibrium constants and calculations, the reaction quotient and predicting direction.

Equilibrium law

For $a\text{A} + b\text{B} \rightleftharpoons c\text{C} + d\text{D}$ :

K_c = \frac{[\text{C}]^c[\text{D}]^d}{[\text{A}]^a[\text{B}]^b}

Pure solids and liquids (including water as the solvent) are left out.

Calculating and using Kc

For $\text{H}_2(g) + \text{I}_2(g) \rightleftharpoons 2\text{HI}(g)$ at a certain temperature, the equilibrium concentrations are $[\text{H}_2] = 0.10$ , $[\text{I}_2] = 0.20$ and $[\text{HI}] = 0.80$ mol L $^{-1}$ .

K_c = \frac{(0.80)^2}{(0.10)(0.20)} = 32.

A large Kc means products are favoured at equilibrium. If a different mixture at the same temperature has a reaction quotient $Q = 10$ , then $Q < K_c$ and the reaction proceeds in the forward direction until $Q = K_c$ .

3. Solubility equilibria

Dot point: solubility equilibria and Ksp.

A sparingly soluble salt in contact with its saturated solution is at equilibrium, for example $\text{AgCl}(s) \rightleftharpoons \text{Ag}^+(aq) + \text{Cl}^-(aq)$ , with $K_{sp} = [\text{Ag}^+][\text{Cl}^-]$ . A precipitate forms when the ionic product exceeds $K_{sp}$ . Adding a common ion (such as $\text{Cl}^-$ from NaCl) shifts the equilibrium left and lowers the solubility.

4. Acids, bases and pH

Dot points: acids, bases and pH, conjugate pairs and amphiprotic species, strong and weak acids and bases, Ka and Kb, polyprotic acids, self-ionisation of water and Kw.

Brønsted-Lowry: an acid donates a proton, a base accepts one. Each acid has a conjugate base that differs by one $\text{H}^+$ .
Amphiprotic species can do both: $\text{H}_2\text{O}$ , $\text{HCO}_3^-$ , $\text{HSO}_4^-$ , $\text{H}_2\text{PO}_4^-$ .
Strong acids ( $\text{HCl}$ , $\text{HNO}_3$ , $\text{H}_2\text{SO}_4$ first ionisation) ionise almost completely. Weak acids (ethanoic acid) ionise partially; a larger $K_a$ means a stronger acid.
Polyprotic acids ionise in steps, each step with a smaller $K_a$ , because removing $\text{H}^+$ from an increasingly negative ion is harder.

Water and pH

K_w = [\text{H}^+][\text{OH}^-] = 1.0 \times 10^{-14} \text{ at } 25\,°\text{C} \qquad \text{pH} = -\log[\text{H}^+]

Neutral does not always mean pH 7

Self-ionisation of water is endothermic, so $K_w$ increases with temperature. At 50 °C, $K_w \approx 5.5 \times 10^{-14}$ , so in pure water $[\text{H}^+] = \sqrt{5.5 \times 10^{-14}}$ and the pH is about 6.6.

The water is still neutral because $[\text{H}^+] = [\text{OH}^-]$ . Neutral means equal concentrations, not pH 7.

Dilution by a factor of 10 changes the pH of a strong acid by 1 unit: a pH 2.0 solution diluted 100 times has pH 4.0.

5. Buffers, indicators and titrations

Dot points: buffers, acid-base indicators, volumetric analysis and titration curves.

A buffer contains a weak acid and its conjugate base in similar, significant amounts, for example ethanoic acid and sodium ethanoate.

How a buffer resists pH change

$\text{CH}_3\text{COOH}(aq) + \text{H}_2\text{O}(l) \rightleftharpoons \text{CH}_3\text{COO}^-(aq) + \text{H}_3\text{O}^+(aq)$

Add a little acid: the added $\text{H}_3\text{O}^+$ increases the reverse rate, so the equilibrium shifts left and the large reservoir of $\text{CH}_3\text{COO}^-$ consumes most of the added $\text{H}_3\text{O}^+$ . The pH falls only slightly.

Add a little base: $\text{OH}^-$ reacts with $\text{H}_3\text{O}^+$ , lowering its concentration. The forward rate is now greater, so more $\text{CH}_3\text{COOH}$ ionises and replaces most of the lost $\text{H}_3\text{O}^+$ .

Indicators are weak acids whose acid and conjugate-base forms have different colours. Choose one whose colour change range includes the pH at the equivalence point: phenolphthalein (about pH 8 to 10) for a weak acid with a strong base; methyl orange (about pH 3 to 4.5) for a strong acid with a weak base.

Titration curves. Strong acid with strong base: equivalence at pH 7 with a long steep section. Weak acid with strong base: equivalence above 7. Strong acid with weak base: equivalence below 7.

6. Redox: oxidation numbers and half-equations

Dot points: oxidation numbers and half-equations, redox and electrochemistry.

Oxidation is loss of electrons (oxidation number increases); reduction is gain (oxidation number decreases).

Balancing a half-equation in acidic solution

Permanganate is reduced to manganese(II): $\text{MnO}_4^- \rightarrow \text{Mn}^{2+}$ .

Balance atoms other than O and H: already balanced (1 Mn each side).
Balance O with water: $\text{MnO}_4^- \rightarrow \text{Mn}^{2+} + 4\text{H}_2\text{O}$ .
Balance H with $\text{H}^+$ : $\text{MnO}_4^- + 8\text{H}^+ \rightarrow \text{Mn}^{2+} + 4\text{H}_2\text{O}$ .
Balance charge with electrons: left is $-1 + 8 = +7$ , right is $+2$ , so add 5 electrons to the left.

\text{MnO}_4^-(aq) + 8\text{H}^+(aq) + 5e^- \rightarrow \text{Mn}^{2+}(aq) + 4\text{H}_2\text{O}(l)

Check: Mn goes from +7 to +2, a gain of 5 electrons.

7. Galvanic cells and standard electrode potentials

Dot points: galvanic cells, standard electrode potentials.

Predicting a cell

A cell is built from $\text{Zn}|\text{Zn}^{2+}$ ( $E° = -0.76$ V) and $\text{Ag}|\text{Ag}^+$ ( $E° = +0.80$ V).

The half-cell with the higher reduction potential is reduced: $\text{Ag}^+ + e^- \rightarrow \text{Ag}$ at the cathode (positive). Zinc is oxidised at the anode (negative): $\text{Zn} \rightarrow \text{Zn}^{2+} + 2e^-$ .

$E°_\text{cell} = E°_\text{cathode} - E°_\text{anode} = 0.80 - (-0.76) = 1.56$ V.

Electrons flow through the wire from zinc to silver. In the salt bridge, anions migrate towards the anode half-cell and cations towards the cathode half-cell.

A positive $E°_\text{cell}$ predicts a spontaneous reaction under standard conditions, but it says nothing about the rate.

8. Electrolysis

Dot points: electrolytic cells and electrolysis, quantitative electrolysis and Faraday's laws.

In an electrolytic cell, a power supply drives a non-spontaneous reaction. The cathode is negative (reduction) and the anode is positive (oxidation). Predict products by choosing the strongest oxidant present to be reduced at the cathode and the strongest reductant to be oxidised at the anode, remembering that water itself can react.

Electrolysis of copper(II) sulfate with inert electrodes

Cathode: $\text{Cu}^{2+}(aq) + 2e^- \rightarrow \text{Cu}(s)$ , observed as a pink-brown solid coating the electrode.

Anode: sulfate is very hard to oxidise, so water is oxidised: $2\text{H}_2\text{O}(l) \rightarrow \text{O}_2(g) + 4\text{H}^+(aq) + 4e^-$ , observed as bubbles of colourless gas.

Over time the blue colour fades and the solution becomes more acidic.

Quantity. A current of 2.00 A for 30.0 minutes passes $Q = It = 3600$ C, which is $\dfrac{3600}{96\,485} = 0.0373$ mol of electrons, depositing $0.0187$ mol of copper, about 1.19 g.

9. Corrosion

Dot point: corrosion of iron and its prevention.

Rusting is an electrochemical process: iron is oxidised ( $\text{Fe} \rightarrow \text{Fe}^{2+} + 2e^-$ ) and oxygen is reduced in the presence of water ( $\text{O}_2 + 2\text{H}_2\text{O} + 4e^- \rightarrow 4\text{OH}^-$ ). Salt water speeds it up by improving ionic conduction. Prevention works by excluding water and oxygen (paint, oil, plastic coating), by sacrificial protection (a more reactive metal such as zinc or magnesium is oxidised instead), or by cathodic protection (an external supply makes the iron the cathode).

Common mistakes

Where Unit 3 marks go missing

Saying that Kc changes when a concentration changes. Only temperature changes Kc.
Including water or solids in an equilibrium expression.
Mixing up the signs of electrodes between galvanic and electrolytic cells.
Forgetting electrons in half-equations, or balancing atoms but not charge.
Claiming a buffer keeps the pH exactly constant. It resists change; the pH still moves slightly.

Check your knowledge

For $\text{CO}(g) + 2\text{H}_2(g) \rightleftharpoons \text{CH}_3\text{OH}(g)$ , predict the shift when the volume is increased. (Answer: left, towards more gas particles.)
What is the conjugate base of $\text{H}_2\text{PO}_4^-$ ? (Answer: $\text{HPO}_4^{2-}$ .)
Find the oxidation number of Cr in $\text{Cr}_2\text{O}_7^{2-}$ . (Answer: +6.)

Then try the Unit 3 practice quiz.

Sources & how we know this

Chemistry ATAR course Year 12 syllabus — School Curriculum and Standards Authority (SCSA)
Chemistry past ATAR course examinations — School Curriculum and Standards Authority (SCSA)
Chemistry ATAR course: syllabus and support materials — SCSA

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