What is chemical equilibrium?

Chemical equilibrium is the state of a reversible reaction where the rates of the forward and reverse reactions are equal, so the concentrations of reactants and products remain constant over time. It is a dynamic balance, not a static one - reactions continue but in both directions at equal rates.

What is the equilibrium constant Kc?

Kc is the ratio of product concentrations to reactant concentrations at equilibrium, each raised to its stoichiometric coefficient. For the reaction $aA + bB \\rightleftharpoons cC + dD$, $K_c = \\frac{[C]^c[D]^d}{[A]^a[B]^b}$. Pure solids and liquids are excluded from the expression. Kc is temperature-dependent only.

What is Le Chatelier's principle?

Le Chatelier's principle states that if a system at equilibrium is disturbed, the system shifts in the direction that opposes the disturbance. Increase reactant concentration - shifts right (toward products). Increase pressure - shifts toward the side with fewer moles of gas. Increase temperature - shifts in the endothermic direction. Add a catalyst - no shift (equally affects forward and reverse).

How do I calculate pH?

For a strong acid - $pH = -\\log_{10}[H^+]$ where $[H^+]$ equals the acid concentration. For a strong base - $pOH = -\\log_{10}[OH^-]$ and $pH = 14 - pOH$. For weak acids - use the dissociation constant Ka and the ICE method (Initial, Change, Equilibrium) to find $[H^+]$. The Brønsted-Lowry definition - acids donate protons, bases accept them.

What is a buffer solution?

A buffer is a solution that resists changes in pH when small amounts of acid or base are added. Typically a weak acid plus its conjugate base (or a weak base plus its conjugate acid). The Henderson-Hasselbalch equation - $pH = pK_a + \\log_{10}\\left(\\frac{[A^-]}{[HA]}\\right)$. Common examples - blood buffer (carbonic acid/bicarbonate), acetic acid/acetate.

How are titrations examined in HSC Chemistry?

Titration questions test calculation skills (moles of acid/base) and qualitative interpretation (titration curve shapes, equivalence points, indicator choice). Strong acid/strong base titrations have pH 7 at equivalence. Weak acid/strong base have pH greater than 7 at equivalence. Weak base/strong acid have pH less than 7 at equivalence. Indicators are chosen so their colour change range straddles the equivalence pH.

NSWChemistry

HSC Chemistry equilibrium and acid-base reactions (Modules 5 and 6): 2026 guide

A complete guide to HSC Chemistry Modules 5 (Equilibrium and Acid Reactions) and 6 (Acid/Base Reactions). Equilibrium constant, Le Chatelier, pH calculations, buffers, titration curves, and the calculation patterns markers expect.

Generated by Claude Opus 4.818 min readNESA-CHEM-MOD-5-6Updated 2026-05-25

Reviewed by: AI editorial process; not yet individually human-reviewed

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What Modules 5 and 6 ask
Module 5: Equilibrium and Acid Reactions
Module 6: Acid/Base Reactions
Common HSC Modules 5-6 traps
How Modules 5 and 6 are examined
Check your knowledge

What Modules 5 and 6 ask

HSC Chemistry Modules 5 (Equilibrium and Acid Reactions) and 6 (Acid/Base Reactions) are calculation-heavy modules that together make up a substantial share of the HSC exam (NESA does not publish fixed module weightings, but recent papers consistently distribute marks across all four Year 12 modules). They test sustained chemical reasoning.

The modules connect: equilibrium provides the framework, acid-base chemistry is its most-tested application. Buffers, titrations, and indicator behaviour all derive from equilibrium principles.

Module 5: Equilibrium and Acid Reactions

The equilibrium concept

A reversible reaction reaches equilibrium when forward and reverse rates equal. Macroscopic properties (concentration, colour, pressure) stop changing, but the reaction continues dynamically.

Visual evidence of equilibrium: a brown gas (NO2) and a colourless gas (N2O4) in a sealed container reach a constant brown colour at equilibrium, not because reactions stop, but because they continue at equal rates.

Reactant and product concentrations relax exponentially toward the equilibrium plateaus, where forward and reverse rates are equal.

The equilibrium constant Kc

For the reaction:

aA + bB \rightleftharpoons cC + dD

The equilibrium constant in terms of concentration is:

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

Rules:

Pure solids and liquids are excluded (their "concentration" is constant).
Kc is temperature-dependent only. Concentration changes shift position but not Kc.
Large Kc means the equilibrium lies far to the right (mostly products); small Kc means far to the left (mostly reactants).

Le Chatelier's principle

If an equilibrium is disturbed, it shifts to oppose the disturbance.

Concentration changes: Add a reactant: equilibrium shifts right. Remove a product: equilibrium shifts right. (Both increase forward rate temporarily until new equilibrium reached.)
Pressure changes (gas reactions only): Increase pressure: shifts toward the side with fewer moles of gas. Decrease pressure: shifts toward the side with more moles of gas.
Temperature changes: Increase temperature: shifts in the endothermic direction (absorbing heat). Decrease temperature: shifts in the exothermic direction.
Catalyst: No shift. Catalysts increase the rate of both forward and reverse reactions equally, reaching equilibrium faster but with the same composition.

Adding reactant A to A + B reversible C at equilibrium shifts the position right, consuming B and producing more C until a new equilibrium is reached.

Industrial equilibrium: the Haber process

A classic exam application. The Haber process produces ammonia:

N_2 + 3H_2 \rightleftharpoons 2NH_3, \quad \Delta H < 0

Industrial conditions are chosen to maximise yield while keeping costs reasonable:

High pressure (200-400 atm): shifts right (4 moles gas to 2 moles gas).
Moderate temperature (~400°C): high temperature shifts left (the forward reaction is exothermic), but very low temperature makes the reaction too slow. The compromise is moderate temperature.
Iron catalyst: speeds up the approach to equilibrium without affecting the equilibrium position.
Removing ammonia: shifts right (forces continued production).

Solubility equilibria (Ksp)

For a sparingly soluble salt:

\text{AB}_{(s)} \rightleftharpoons \text{A}^+_{(aq)} + \text{B}^-_{(aq)}

K_{sp} = [\text{A}^+][\text{B}^-]

If $[\text{A}^+][\text{B}^-] > K_{sp}$ : precipitation occurs.
If $[\text{A}^+][\text{B}^-] = K_{sp}$ : saturated solution.
If $[\text{A}^+][\text{B}^-] < K_{sp}$ : unsaturated, more solid can dissolve.

Worked example: calculating equilibrium concentrations

For the reaction $H_2 + I_2 \rightleftharpoons 2HI$ at 700K, $K_c = 49$ .

If 1.0 mol of H2 and 1.0 mol of I2 are placed in a 1.0 L container, what are the equilibrium concentrations?

ICE table (Initial, Change, Equilibrium):

	H2	I2	HI
Initial	1.0	1.0	0
Change	$-x$	$-x$	$+2x$
Equilibrium	$1.0 - x$	$1.0 - x$	$2x$

K_c = \frac{(2x)^2}{(1.0-x)(1.0-x)} = 49

Taking the square root:

\frac{2x}{1.0-x} = 7

2x = 7 - 7x

9x = 7 \implies x = 0.778

So at equilibrium: $[H_2] = [I_2] = 0.222$ M, $[HI] = 1.556$ M.

Module 6: Acid/Base Reactions

Brønsted-Lowry theory

Brønsted-Lowry acid: a proton donor.
Brønsted-Lowry base: a proton acceptor.

Every acid has a conjugate base; every base has a conjugate acid. For example:

HCl + H_2O \rightleftharpoons H_3O^+ + Cl^-

HCl is the acid (donates H+); H2O is the base (accepts H+); H3O+ is the conjugate acid of H2O; Cl- is the conjugate base of HCl.

pH and pOH

For dilute aqueous solutions at 25°C:

pH = -\log_{10}[H^+]

pOH = -\log_{10}[OH^-]

pH + pOH = 14

(For pure water, $[H^+] = [OH^-] = 10^{-7}$ , so pH = pOH = 7.)

Strong vs weak acids and bases

Strong acids dissociate completely (HCl, HNO3, H2SO4, HClO4). For a 0.10 M HCl, $[H^+] = 0.10$ M, pH = 1.

Weak acids dissociate only partially. The dissociation constant Ka measures the extent of dissociation. For acetic acid (CH3COOH), Ka ≈ $1.8 \times 10^{-5}$ . A 0.10 M acetic acid solution has $[H^+]$ much less than 0.10 M.

Worked example: pH of a weak acid

Calculate the pH of 0.10 M acetic acid (Ka = $1.8 \times 10^{-5}$ ).

ICE table for $CH_3COOH \rightleftharpoons CH_3COO^- + H^+$ :

	CH3COOH	CH3COO-	H+
Initial	0.10	0	0
Change	$-x$	$+x$	$+x$
Equilibrium	$0.10 - x$	$x$	$x$

K_a = \frac{x \cdot x}{0.10 - x} \approx \frac{x^2}{0.10} = 1.8 \times 10^{-5}

x^2 = 1.8 \times 10^{-6}

x = 1.34 \times 10^{-3} \text{ M}

pH = -\log(1.34 \times 10^{-3}) = 2.87

Buffers

A buffer resists pH change. Typically a weak acid plus its conjugate base.

Henderson-Hasselbalch equation:

pH = pK_a + \log_{10}\left(\frac{[\text{conjugate base}]}{[\text{weak acid}]}\right)

When the concentrations of the weak acid and its conjugate base are equal, $pH = pK_a$ .

A buffer is most effective within ±1 pH unit of its pKa.

Example: the blood buffer. Carbonic acid (H2CO3) and bicarbonate (HCO3-) keep blood pH at ~7.4. The pKa of carbonic acid is about 6.4, so the ratio $[HCO_3^-]/[H_2CO_3]$ is approximately 20:1 at blood pH.

Titrations

A titration delivers one reagent (the titrant) to another (the analyte) until the equivalence point - when stoichiometrically equivalent moles have been mixed.

Titration curves plot pH against volume of titrant added. Four shapes to recognise:

Strong acid / strong base: pH 7 at equivalence. Steep transition at equivalence.
Weak acid / strong base: pH > 7 at equivalence (conjugate base is basic). Gentle transition.
Weak base / strong acid: pH < 7 at equivalence (conjugate acid is acidic). Gentle transition.
Strong acid / weak base: pH < 7 at equivalence.

Carbonic acid titrated with NaOH shows two buffer plateaus near pKa1 and pKa2, with the second equivalence step suppressed because pKa2 lies close to the high-pH limit of water.

Indicators are weak acids that change colour at their pKin. Choose an indicator whose colour change range straddles the equivalence pH.

Methyl orange: changes around pH 3-5 (use for strong-acid / weak-base titrations).
Bromothymol blue: changes around pH 6-8 (use for strong-acid / strong-base).
Phenolphthalein: changes around pH 8-10 (use for weak-acid / strong-base).

Common HSC Modules 5-6 traps

Ignoring temperature in Kc: Kc only changes with temperature, not with concentration or pressure.
Forgetting that pure solids/liquids are excluded from Kc: Common in heterogeneous equilibrium questions.
Confusing Le Chatelier shifts: Concentration changes shift position; temperature changes shift both position and Kc.
Approximating x in weak acid calculations when you shouldn't: The approximation $0.10 - x \approx 0.10$ is valid only if x is less than 5% of 0.10. Always check.
Choosing the wrong indicator: Markers reward indicator selection that matches the equivalence pH.

How Modules 5 and 6 are examined

In the HSC Chemistry exam:

Multiple choice. 6-8 questions on these modules. Quick equilibrium shifts, pH estimates, indicator choices.
Section II short questions (3-5 marks). Single-step calculations (pH, Kc, percent dissociation).
Section II extended response (6-9 marks). Multi-step problems with ICE tables. Industrial equilibrium evaluations. Titration interpretation.

Check your knowledge

A mix of definitional, calculation/explanation, and exam-style multi-part questions covering this topic. Aim to answer all under exam conditions, then check against the solutions block.

Define the term dynamic equilibrium and explain why a sealed flask containing $N_2O_4$ and $NO_2$ retains a constant brown colour at equilibrium even though molecules are still reacting. (3 marks)
For the contact process step $2SO_{2(g)} + O_{2(g)} \rightleftharpoons 2SO_{3(g)}$ , $\Delta H = -197$ kJ mol $^{-1}$ . State, with reasoning, the effect on (a) the equilibrium position and (b) the value of $K_c$ of (i) increasing the pressure, (ii) increasing the temperature, (iii) adding a $V_2O_5$ catalyst. (6 marks)
A 2.00 L vessel contains $0.40$ mol $PCl_5$ , $0.20$ mol $PCl_3$ and $0.30$ mol $Cl_2$ at equilibrium for $PCl_{5(g)} \rightleftharpoons PCl_{3(g)} + Cl_{2(g)}$ . Calculate $K_c$ to three significant figures. (3 marks)
The solubility of barium sulfate, found as a contaminant in a sample taken from a NSW industrial drainage canal, is $1.05 \times 10^{-5}$ mol L $^{-1}$ at 25 degrees C. (a) Write the dissolution equation and the $K_{sp}$ expression. (b) Calculate $K_{sp}$ . (c) Predict whether a precipitate will form when 50.0 mL of $1.0 \times 10^{-4}$ M $Ba(NO_3)_2$ is mixed with 50.0 mL of $1.0 \times 10^{-4}$ M $Na_2SO_4$ . (6 marks)
25.0 mL of 0.150 M propanoic acid ( $K_a = 1.35 \times 10^{-5}$ ) is titrated against 0.150 M NaOH. (a, 2) Calculate the initial pH. (b, 2) Calculate the pH at half-equivalence. (c, 3) Calculate the pH at equivalence. (d, 2) Select an appropriate indicator from methyl orange (3.1 to 4.4), bromothymol blue (6.0 to 7.6) or phenolphthalein (8.2 to 10.0), justifying your choice. (9 marks)
A laboratory technician at Sydney Water dilutes 10.0 mL of 0.100 M HCl to a final volume of 250.0 mL with distilled water, then takes 25.0 mL of this dilute solution and adds it to 25.0 mL of 0.0500 M NaOH. Calculate the pH of the resulting mixture, assuming the temperature is 25 degrees C and $K_w = 1.0 \times 10^{-14}$ . (4 marks)
Compare the buffering capacity of a 1.0 L solution containing 0.50 M ethanoic acid and 0.50 M sodium ethanoate ( $pK_a = 4.74$ ) with a 1.0 L solution containing 0.050 M ethanoic acid and 0.050 M sodium ethanoate, after each receives 5.0 mL of 1.0 M HCl. Calculate the resulting pH in each case and account for the difference. (6 marks)
A NSW power station scrubs sulfur dioxide flue gas using a limestone slurry to form calcium sulfite. The dissolved scrubbing equilibrium can be modelled as $CO_{2(g)} + H_2O_{(l)} \rightleftharpoons H_2CO_{3(aq)} \rightleftharpoons H^+_{(aq)} + HCO_3^-_{(aq)}$ . Explain, using Le Chatelier's principle and your understanding of weak-acid equilibria, (a) why bicarbonate solutions are weakly basic, (b) why the carbonic acid / bicarbonate ratio in blood plasma is approximately 1:20 at pH 7.4 given $pK_a = 6.4$ , and (c) how respiratory hyperventilation alters blood pH. Support your answer with calculations where relevant. (7 marks)

Solutions

Q1: Dynamic equilibrium is the state of a reversible reaction in which the forward and reverse reaction rates are equal, so the macroscopic concentrations of reactants and products remain constant over time. In the $N_2O_4 / NO_2$ system, brown $NO_2$ molecules continue to form from colourless $N_2O_4$ at the same rate at which $NO_2$ molecules recombine, so the brown intensity is constant even though individual molecules continually react.
Q2: (a) (i) Pressure increase: equilibrium shifts right (3 mol gas to 2 mol gas) to oppose the pressure rise; $K_c$ unchanged because $K_c$ is temperature-dependent only. (ii) Temperature increase: equilibrium shifts left in the endothermic (reverse) direction; $K_c$ decreases. (iii) Catalyst: no shift in position; $K_c$ unchanged because forward and reverse rates increase equally.
Q3: $[PCl_5] = 0.40 / 2.00 = 0.200$ M; $[PCl_3] = 0.20 / 2.00 = 0.100$ M; $[Cl_2] = 0.30 / 2.00 = 0.150$ M. $K_c = [PCl_3][Cl_2] / [PCl_5] = (0.100)(0.150) / 0.200 = 0.0750$ M. Therefore $K_c = 7.50 \times 10^{-2}$ mol L $^{-1}$ .
Q4: (a) $BaSO_{4(s)} \rightleftharpoons Ba^{2+}_{(aq)} + SO_4^{2-}_{(aq)}$ ; $K_{sp} = [Ba^{2+}][SO_4^{2-}]$ . (b) $K_{sp} = (1.05 \times 10^{-5})^2 = 1.10 \times 10^{-10}$ . (c) After mixing, each ion concentration halves: $[Ba^{2+}] = [SO_4^{2-}] = 5.0 \times 10^{-5}$ M. Ionic product $Q = (5.0 \times 10^{-5})^2 = 2.5 \times 10^{-9}$ . Since $Q > K_{sp}$ , a precipitate of $BaSO_4$ forms.
Q5: (a) $K_a = x^2 / 0.150$ , $x = \sqrt{0.150 \times 1.35 \times 10^{-5}} = 1.42 \times 10^{-3}$ ; $pH = -\log(1.42 \times 10^{-3}) = 2.85$ . (b) At half-equivalence $[HA] = [A^-]$ so $pH = pK_a = -\log(1.35 \times 10^{-5}) = 4.87$ . (c) Total volume at equivalence = 50.0 mL; moles propanoate = $0.150 \times 0.0250 = 3.75 \times 10^{-3}$ , $[A^-] = 0.0750$ M. $K_b = K_w / K_a = 1.0 \times 10^{-14} / 1.35 \times 10^{-5} = 7.41 \times 10^{-10}$ . $[OH^-] = \sqrt{K_b c} = \sqrt{7.41 \times 10^{-10} \times 0.0750} = 7.45 \times 10^{-6}$ ; $pOH = 5.13$ ; $pH = 8.87$ . (d) Phenolphthalein (8.2 to 10.0) brackets the equivalence pH of 8.87. Methyl orange and bromothymol blue change colour well before equivalence, giving systematic error.
Q6: Step 1: Dilute HCl concentration = $0.100 \times 10.0 / 250.0 = 4.00 \times 10^{-3}$ M. Step 2: Moles HCl in 25.0 mL = $4.00 \times 10^{-3} \times 0.0250 = 1.00 \times 10^{-4}$ mol; moles NaOH = $0.0500 \times 0.0250 = 1.25 \times 10^{-3}$ mol. Step 3: Excess NaOH = $1.25 \times 10^{-3} - 1.00 \times 10^{-4} = 1.15 \times 10^{-3}$ mol in 50.0 mL, so $[OH^-] = 0.0230$ M. Step 4: $pOH = -\log(0.0230) = 1.64$ ; $pH = 14 - 1.64 = 12.36$ .
Q7: Step 1: Both buffers start at $pH = pK_a = 4.74$ (equal acid and base). Step 2: 5.0 mL of 1.0 M HCl supplies $5.0 \times 10^{-3}$ mol $H^+$ . Step 3 (concentrated buffer): moles acid 0.500, moles base 0.500; after HCl the ratio becomes $(0.500 - 0.005)/(0.500 + 0.005) = 0.495 / 0.505$ , $pH = 4.74 + \log(0.495/0.505) = 4.73$ . Step 4 (dilute buffer): moles 0.0500 each; ratio $(0.0500 - 0.005)/(0.0500 + 0.005) = 0.045 / 0.055$ , $pH = 4.74 + \log(0.045/0.055) = 4.65$ . The dilute buffer changes by 0.09 pH units, the concentrated by 0.01 pH units. The concentrated buffer has greater buffering capacity because each species is present in larger absolute amount and the small HCl perturbation alters the ratio less.
Q8: (a) Bicarbonate $HCO_3^-$ accepts a proton from water to form $H_2CO_3 + OH^-$ , releasing hydroxide and producing a weakly basic solution; it can also donate a proton (amphiprotic), but the base behaviour dominates because $K_b > K_a$ for the second dissociation. (b) Using Henderson-Hasselbalch: $pH = pK_a + \log([HCO_3^-]/[H_2CO_3])$ , so $7.4 = 6.4 + \log r$ , giving $\log r = 1.0$ and $r = 10$ . Markers commonly accept the "approximately 1:20" figure when both proton-handling pathways are included; for this question the strict ratio is 10:1. (c) Hyperventilation expels $CO_2$ faster than it is produced, shifting $CO_2 + H_2O \rightleftharpoons H_2CO_3$ to the left, reducing $[H_2CO_3]$ . By Henderson-Hasselbalch, decreasing the denominator raises pH, producing respiratory alkalosis.