SACE Stage 2 Chemistry common mistakes
SACE Stage 2 Chemistry · Understand the traps and how to avoid them.
Calculating the pH of a weak acid by substituting the acid concentration directly as [H+] or by using [H+] = sqrt(Ka × c) without checking whether the approximation is valid.
Candidates memorise strong-acid formulas and assume every acid ionises completely, so they skip the equilibrium expression and the ICE table.
What to do instead
Write the Ka expression and an ICE table, let x = [H+], solve the equilibrium equation, and only use the square-root shortcut when the initial concentration is much larger than Ka; check that x is less than about 5% of the initial concentration.
Assuming the equivalence point of every titration has pH 7 and choosing an indicator by colour memory rather than by matching its transition range to the steep part of the titration curve.
Candidates conflate equivalence with neutrality and do not consider hydrolysis of the salt formed at equivalence.
What to do instead
Sketch the titration curve, calculate or reason about the pH at equivalence from the conjugate acid or base, and select an indicator whose transition range lies on the steep section of that curve.
Balancing redox half-equations by counting atoms only, leaving charge unbalanced or using the wrong number of electrons when combining the half-equations.
Candidates focus on balancing oxygen and hydrogen and forget that electron loss and gain must also balance charge.
What to do instead
Balance each half-equation separately: balance atoms other than O and H, balance O with H2O, balance H with H+, then balance charge with electrons; multiply each half-equation so the electrons cancel before adding.
Predicting that increasing pressure always shifts an equilibrium to the right, or that adding a catalyst increases the equilibrium yield.
Candidates apply Le Chatelier’s principle as a set of slogans and do not count only gaseous species or distinguish rate effects from position-of-equilibrium effects.
What to do instead
Count the total moles of gas on each side, including only gaseous species, and compare them before predicting a pressure shift; state explicitly that a catalyst changes the rate of approach to equilibrium but not the equilibrium constant or yield.
Describing a catalyst as increasing the yield in an industrial equilibrium process, or drawing rate–time graphs that show the catalysed reaction reaching a different final amount of product.
Candidates blur the distinction between kinetics and equilibrium because both topics use collision theory and energy profiles.
What to do instead
Separate the ideas: a catalyst lowers activation energy and increases both forward and reverse rates equally, so equilibrium is reached faster but the final composition is unchanged; on a rate–time graph the catalysed curve should level off at the same value.
Calculating theoretical yield or atom economy by using masses directly without converting to moles, or by ignoring the limiting reagent in a multistep calculation.
Candidates treat mass as if it were proportional to particles and often use the first reactant listed as though it must be limiting.
What to do instead
Convert all given masses to moles, divide by stoichiometric coefficients to identify the limiting reagent, use the limiting reagent’s mole ratio to find the theoretical yield, and distinguish percentage yield from atom economy by using the balanced equation and molar masses.
Naming organic compounds by choosing a parent chain that is not the longest continuous carbon chain, or by numbering from the end that gives the first substituent rather than the lowest set of locants.
Candidates scan from left to right and stop at the first obvious chain, and they confuse “first substituent” with “lowest locant set.”
What to do instead
Find every possible continuous chain, choose the longest one containing the principal functional group, number to give the lowest locant set at the first point of difference, and then list substituents alphabetically with correct prefixes.
Predicting that oxidation of a primary alcohol always stops at the aldehyde, or that a tertiary alcohol is easily oxidised to a ketone.
Candidates memorise “alcohol oxidises to aldehyde or ketone” without linking the product to the number of hydrogens on the carbinol carbon.
What to do instead
Classify the alcohol as primary, secondary or tertiary, then state the controlled product: primary alcohol to aldehyde with distillation and to carboxylic acid with excess oxidant under reflux, secondary alcohol to ketone, and tertiary alcohol resistant to oxidation under normal conditions.
Drawing the repeat unit of an addition polymer with a double bond still present, or drawing a condensation polymer without removing the small molecule such as water or HCl.
Candidates copy the monomer structure and do not track which bonds break and which atoms leave during polymerisation.
What to do instead
For addition polymerisation, break the C=C double bond and show two single bonds forming the backbone; for condensation polymerisation, remove one small molecule per amide or ester linkage and show the correct repeat unit with the linkage intact.
Assigning the anode and cathode incorrectly in an electrochemical cell, reversing the sign of Ecell, or calculating Ecell by adding the two reduction potentials without changing the sign of the oxidation half-cell.
Candidates memorise “anode is positive” from electrolysis and mix it with galvanic-cell conventions, and they do not consistently apply Ecell = Ecathode − Eanode.
What to do instead
Use a fixed convention: oxidation always occurs at the anode and reduction at the cathode; for a galvanic cell the cathode is positive, and Ecell = E(reduction at cathode) − E(reduction at anode); for electrolysis, the external supply reverses the spontaneity but the electrode reaction definitions stay the same.