Loav · VCE Chemistry · Unit 4 / Area of Study 2
U4A2-02 · Expanded original 2026 teaching guide · Study guide
Count reactions before calculating concentrations
A titration uses a measured volume of a solution with known concentration to determine an unknown amount. A redox titration works because the analyte and titrant exchange electrons in a predictable stoichiometric ratio. The equivalence point is the theoretical point at which the reactants have been supplied in the required ratio. The endpoint is the observed signal used to estimate that point. They should be close, but they are not the same definition.
Write and balance the reaction before using numbers. “One mole reacts with one mole” is not a general titration rule. Electron balance determines the coefficients; the coefficients determine the mole relationship. A concentration ratio is not automatically a mole ratio, because solution volumes may differ. Track each physical sample: the original solution, a diluted volumetric-flask solution, and the aliquot actually titrated are distinct quantities.
Iodine measures reactive carbon–carbon unsaturation
In the stipulated ideal addition model, one mole of I₂ adds across one mole of C=C bonds. For a molecule with two reactive alkene bonds, one mole of molecules consumes two moles of iodine when addition is complete. The iodine-to-molecule mole ratio therefore measures the number of reactive C=C bonds per molecule for a pure compound, or an average for a specified mixture. Count C=C bonds, not C=O bonds in acid or ester groups. A saturated fatty acid still has a carbonyl bond in its carboxyl group.
The course's iodine-based degree of unsaturation is not a generic index of hydrogen deficiency. That broader calculation also counts rings and carbonyl bonds and would answer a different question. State the model's assumptions: complete selective addition, known sample amount, and no competing iodine-consuming reactions. Real analytical procedures may supply specialised reagents; follow the given chemistry rather than inventing a procedure. These examples use supplied amounts and no back-titration method.
Worked example 1: C=C bonds per molecule
An original idealised dataset states that 0.564 g of a pure fatty acid of molar mass 282 g mol⁻¹ consumes 0.5076 g iodine, M(I₂) = 253.8 g mol⁻¹. Amount of acid = 0.564/282 = 0.00200 mol. Amount of iodine = 0.5076/253.8 = 0.002000 mol. The ratio n(I₂)/n(acid) = 1.00, consistent with one reactive C=C bond per molecule. This does not locate the double bond or distinguish geometrical isomers. Its COOH carbonyl must not be added to this count.
If iodine value is defined in a question as grams I₂ consumed per 100 g sample, this example gives (0.5076/0.564) × 100 = 90.0 g I₂ per 100 g. Define that quantity before using it; it is not a dimensionless bond count.
Direct permanganate titration: explain the colour
In acidic solution, permanganate can oxidise Fe²⁺ to Fe³⁺. The half-equations are Fe²⁺ → Fe³⁺ + e⁻ and MnO₄⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H₂O. Multiply the iron equation by five and add: MnO₄⁻ + 8H⁺ + 5Fe²⁺ → Mn²⁺ + 4H₂O + 5Fe³⁺. Both sides have total charge +17. One mole of permanganate therefore reacts with five moles of Fe²⁺.
Permanganate is purple. Before equivalence, each small addition is consumed while the reducing analyte remains. The endpoint is the first persistent very faint colour from a slight excess of MnO₄⁻; describe the species and persistence, rather than memorising an isolated colour word. Mn²⁺ is effectively colourless when dilute and very pale pink at higher concentration according to the 2026 data book. It is not the source of the endpoint's persistent permanganate colour.
Worked example 2: follow the aliquot
A 10.00 mL original Fe²⁺ solution is diluted to 100.0 mL. A 25.00 mL aliquot of the diluted solution requires 20.00 mL of 0.02000 mol L⁻¹ MnO₄⁻ in excess acid. n(MnO₄⁻) = 0.02000 × 0.02000 = 0.0004000 mol. The aliquot contains 5 × 0.0004000 = 0.002000 mol Fe²⁺. Its concentration is 0.002000/0.02500 = 0.08000 mol L⁻¹. The original was diluted tenfold, so its concentration was 0.8000 mol L⁻¹.
Independent check: four aliquots make the 100.0 mL flask, containing 0.008000 mol Fe²⁺ in total. Those moles came from 0.01000 L original solution: 0.008000/0.01000 = 0.8000 mol L⁻¹. Dilution changes concentration and volume, not the amount transferred.
Excess and limiting reactants
For any specified mixture, compare n/coefficient for each reacting species. The smallest value identifies the limiting reagent, assuming the reaction proceeds as stated. Multiply that reaction extent by coefficients to find consumption; subtract consumption from the initial amount to find excess remaining. For 0.00100 mol MnO₄⁻ and 0.00400 mol Fe²⁺ in excess acid, Fe²⁺ limits: 0.00400/5 is smaller than 0.00100/1. Only 0.000800 mol permanganate reacts, leaving 0.000200 mol. This is a direct reaction calculation, not a back titration.
Measurement choices that matter
Use concordant titres as specified by the school method, and do not average a rough trial indiscriminately with precise trials. A suitable pipette measures the aliquot, while a volumetric flask defines dilution volume. A titre is final burette reading minus initial reading. Under VCAA conventions, a burette graduated at 0.1 mL supports readings at marks or half-marks, such as 10.50 or 10.55 mL, rather than an invented 10.53 mL. Retain guard digits during calculation and round the final result appropriately.
Practice with worked answers
1. A pure compound requires 0.00600 mol I₂ for 0.00200 mol compound. Interpret the result.
The ratio is 3.00 mol I₂ per mol compound, consistent with three reactive C=C bonds per molecule under complete selective addition. It gives no double-bond positions.
2. A 25.00 mL Fe²⁺ aliquot uses 15.00 mL of 0.01000 M permanganate in excess acid. Calculate [Fe²⁺].
n(MnO₄⁻) = 0.01000 × 0.01500 = 0.0001500 mol. Multiply by five: n(Fe²⁺) = 0.0007500 mol. Divide by 0.02500 L: [Fe²⁺] = 0.03000 mol L⁻¹.
3. Mix 0.00200 mol MnO₄⁻ and 0.00600 mol Fe²⁺ in excess acid. What remains?
Fe²⁺ is limiting because 0.00600/5 = 0.00120, less than 0.00200/1. Permanganate consumed = 0.00120 mol; remaining = 0.000800 mol. All Fe²⁺ is consumed in the stated ideal model.
4. Why does overshooting the endpoint usually overestimate the analyte concentration?
The recorded titrant volume exceeds the stoichiometric requirement. Treating all of it as reacted with analyte overestimates titrant moles consumed, then analyte moles and concentration. Repeating without overshooting targets this error; extra decimal places do not fix it.
Safety: all practical work is school-supervised with approved chemical-specific controls. Acidified oxidants and iodine are not home-experiment reagents. These calculations are original simulated teaching examples, not laboratory instructions or student primary data.
Go deeper: justify the ratio and track the physical sample
Each oxidation half-equation measures electrons released per analyte molecule; the reduction half-equation measures electrons accepted per titrant ion. Equalise those electron totals before cancelling them. This explains why the same permanganate titrant can have different mole ratios with different analytes. Fe²⁺ releases one electron, whereas oxidation of a primary alcohol all the way to a carboxylic acid releases four per molecule. A memorised five-to-one ratio from iron cannot be transferred to an alcohol.
A titration also assumes sufficiently complete, selective reaction and suitable reaction speed under the specified conditions. If several reducing compounds are present, the titre may measure their combined consumption rather than one named analyte. A stoichiometrically correct calculation does not fix a non-selective experiment. State assumptions explicitly when interpreting a mixture.
Worked example 3: direct organic titration with a changed mole ratio
In an original idealised model, ethanol is quantitatively oxidised to ethanoic acid. The oxidation half-equation is C₂H₅OH + H₂O → CH₃COOH + 4H⁺ + 4e⁻. Combined with acidic permanganate reduction, the equation is 5C₂H₅OH + 4MnO₄⁻ + 12H⁺ → 5CH₃COOH + 4Mn²⁺ + 11H₂O. Both sides contain ten carbon atoms, forty-two hydrogen atoms, twenty-one oxygen atoms and four manganese atoms; each side has total charge +8.
A 20.00 mL sample is diluted to 200.0 mL. Its 25.00 mL aliquot uses 16.00 mL of 0.03000 mol L⁻¹ permanganate in excess acid. Titrant amount = 0.03000 × 0.01600 = 0.0004800 mol. Ethanol amount in the aliquot = (5/4) × 0.0004800 = 0.0006000 mol. Diluted concentration = 0.0006000/0.02500 = 0.02400 mol L⁻¹. Original concentration = 0.02400 × (200.0/20.00) = 0.2400 mol L⁻¹.
Check by the whole-flask route: eight aliquots contain 0.004800 mol, originating from 0.02000 L. Their ratio again gives 0.2400 mol L⁻¹. Neither route multiplies by the flask volume and the dilution factor independently; doing both would double-count the same scaling. This is a direct titration, not a back titration.
Understand why procedural errors change an answer
Adding pure water to the conical flask after transferring an aliquot changes concentration but not analyte moles; with otherwise suitable reaction conditions, the equivalence volume is unchanged. Accidentally diluting the titrant is different: more volume is needed to supply the required titrant moles. If the calculation still uses the old, higher labelled concentration, it overestimates analyte amount. Trace the actual physical change first, then the measured titre, then the calculated result. Do not automatically label every extra-water error “too low”.
Transfer practice
5. Using the ethanol equation above, mix 0.00300 mol ethanol with 0.00200 mol permanganate in excess acid. Find the limiting reagent and ethanol remaining.
Compare 0.00300/5 = 0.000600 with 0.00200/4 = 0.000500. Permanganate limits. Ethanol consumed = 5 × 0.000500 = 0.00250 mol; ethanol remaining = 0.000500 mol. Smaller initial moles alone was not the justification: the coefficients were essential.
6. A pure model compound has two reactive C=C bonds per molecule. A 0.00150 mol sample is supplied with 0.00400 mol iodine. Is iodine sufficient, and how much remains in the ideal model?
Required iodine = 2 × 0.00150 = 0.00300 mol. It is sufficient, leaving 0.00100 mol. This calculation concerns supplied reactant quantities and complete addition, not measurement of an excess by back titration.
7. An endpoint is reached after 18.50 mL, but a student continues to 19.00 mL. Assuming the first endpoint was correct, what is the approximate percentage overestimate in analyte amount?
Amount inferred is proportional to recorded titre, so excess fraction = (19.00 − 18.50)/18.50 = 0.0270, approximately 2.7%. This is systematic overshoot for this trial. Averaging several similarly overshot titres does not remove that bias; better endpoint control is needed.
