SACE Board SACE 2022-style practice: A 25.00\ mL sample of a solution containing iron(II) ions was acidified and titrated with 0.0200\ mol L^-1 potassium permanganate. The average titre was 21.30\ mL. The reaction is MnO_4^- + 5Fe^2+ + 8H^+ → Mn^2+ + 5Fe^3+ + 4H_2O. Calculate the concentration of Fe^2+ in the sample.

Step 1: n(MnO_4^-) = cV = 0.0200 × 0.02130 = 4.260 × 10^-4\ mol. (1 mark) Step 2: the equation gives a 1:5 ratio of MnO_4^- to Fe^2+, so n(Fe^2+) = 5 × 4.260 × 10^-4 = 2.130 × 10^-3\ mol. (2 marks) Step 3: c(Fe^2+) = nV = 2.130 × 10^-30.02500 = 0.0852\ mol L^-1. (2 marks)

SACE Board SACE 2020-style practice: The dissolved oxygen in a water sample was measured by a Winkler-type method in which the liberated iodine was titrated with sodium thiosulfate: I_2 + 2S_2O_3^2- → 2I^- + S_4O_6^2-. A 50.0\ mL sample required 8.40\ mL of 0.0100\ mol L^-1 thiosulfate. Given that O_2 + 4I^- + 4H^+ → 2I_2 + 2H_2O, calculate the concentration of dissolved oxygen in mol L^-1.

Step 1: n(S_2O_3^2-) = 0.0100 × 0.00840 = 8.40 × 10^-5\ mol. (1 mark) Step 2: I_2 : S_2O_3^2- = 1:2, so n(I_2) = 12(8.40 × 10^-5) = 4.20 × 10^-5\ mol. (1 mark) Step 3: from O_2 → 2I_2, the ratio is 1:2, so n(O_2) = 12(4.20 × 10^-5) = 2.10 × 10^-5\ mol. (1 mark) Step 4: c(O_2) = 2.10 × 10^-50.0500 = 4.20 × 10^-4\ mol L^-1. (2 marks)

§-Syllabus dot point

SAChemistrySyllabus dot point

How can redox titrations measure the concentration of oxidising or reducing species in a sample?

Use redox titrations, including permanganate titrations, to determine the concentration of an analyte from balanced half-equations and stoichiometry.

Permanganate and iodine/thiosulfate redox titrations: combining half-equations to find the mole ratio, self-indicating endpoints, and fully worked SACE-style calculations from titre to analyte concentration.

Generated by Claude Opus 4.89 min answerUpdated 2026-06-02

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

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What this dot point is asking
Lead worked calculation
Self-indicating endpoints
Building the overall equation
Iodine/thiosulfate (iodometric) titrations
Why it matters for monitoring

What this dot point is asking

SACE expects you to construct the balanced overall equation from half-equations, read off the correct (often non $1:1$ ) ratio, and complete the stoichiometric calculation, including multi-step iodine/thiosulfate analyses.

Lead worked calculation

Permanganate titration of oxalate

A $0.200\ \text{g}$ sample of impure sodium oxalate was dissolved, acidified, and titrated with $0.0150\ \text{mol L}^{-1}$ $\text{KMnO}_4$ ; the average titre was $28.20\ \text{mL}$ . Find the percentage purity. ( $M(\text{Na}_2\text{C}_2\text{O}_4) = 134.00\ \text{g mol}^{-1}$ .)

The key challenge in a redox titration calculation is finding the correct mole ratio before doing any arithmetic. We derive the ratio from the half-equations, then work through the stoichiometry.

Step 1: Combine the half-equations to find the mole ratio

Write both half-equations with the electrons transferred:

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

\text{C}_2\text{O}_4^{2-} \rightarrow 2\text{CO}_2 + 2e^-

Electrons must balance: multiply the reduction half by $2$ and the oxidation half by $5$ , giving the overall equation:

2\text{MnO}_4^- + 5\text{C}_2\text{O}_4^{2-} + 16\text{H}^+ \rightarrow 2\text{Mn}^{2+} + 10\text{CO}_2 + 8\text{H}_2\text{O.}

The mole ratio is $\text{MnO}_4^- : \text{C}_2\text{O}_4^{2-} = 2:5$ .

Step 2: Calculate moles of permanganate from the titre

Use $n = cV$ , converting the titre to litres:

n(\text{MnO}_4^-) = 0.0150 \times 0.02820 = 4.230 \times 10^{-4}\ \text{mol.}

Step 3: Apply the mole ratio to find moles of oxalate

For every 2 moles of $\text{MnO}_4^-$ there are 5 moles of $\text{C}_2\text{O}_4^{2-}$ , so multiply by $\dfrac{5}{2}$ :

n(\text{C}_2\text{O}_4^{2-}) = \frac{5}{2} \times 4.230 \times 10^{-4} = 1.058 \times 10^{-3}\ \text{mol.}

Step 4: Convert moles to mass and find percentage purity

Multiply by the molar mass of sodium oxalate, then divide by the sample mass and convert to a percentage:

m = nM = 1.058 \times 10^{-3} \times 134.00 = 0.1417\ \text{g.}

\%\text{ purity} = \frac{0.1417}{0.200} \times 100 = 70.9\%.

Final answer: The sample is $70.9\%$ pure sodium oxalate.

Self-indicating endpoints

Potassium permanganate is its own indicator. $\text{MnO}_4^-$ is intensely purple, while the product $\text{Mn}^{2+}$ is almost colourless. While reductant remains, each drop is decolourised; at the endpoint the first tiny excess of $\text{MnO}_4^-$ gives the solution a faint permanent pink. No separate indicator is needed. Iodine titrations are nearly self-indicating (the brown $\text{I}_2$ fades), but a starch indicator is added near the endpoint to give a sharp blue-black to colourless change.

Building the overall equation

The crux of redox titration calculations is the ratio, which is rarely $1:1$ .

Write each half-equation with its electrons.
Multiply each so the electrons lost equal the electrons gained.
Add and cancel electrons, $\text{H}^+$ and $\text{H}_2\text{O}$ where possible.
Read the mole ratio of analyte to titrant directly from the coefficients.

Common ratios to know: $\text{MnO}_4^- : \text{Fe}^{2+} = 1:5$ ; $\text{MnO}_4^- : \text{C}_2\text{O}_4^{2-} = 2:5$ ; $\text{Cr}_2\text{O}_7^{2-} : \text{Fe}^{2+} = 1:6$ ; $\text{I}_2 : \text{S}_2\text{O}_3^{2-} = 1:2$ .

Iodine/thiosulfate (iodometric) titrations

Many oxidisers are measured indirectly. The oxidiser first liberates iodine from excess iodide ( $\text{oxidiser} + \text{I}^- \rightarrow \text{I}_2$ ), and the liberated $\text{I}_2$ is then titrated with standard thiosulfate. Because two steps are chained, you must apply two ratios in sequence, as in the dissolved-oxygen example above. This indirect approach lets a single standard (thiosulfate) measure many different oxidising analytes.

Why it matters for monitoring

Redox titrations quantify species that acid-base methods cannot, such as dissolved oxygen, chemical oxygen demand, iron content in water and ores, and the concentration of disinfectants. The Winkler dissolved-oxygen analysis and permanganate-based oxygen-demand tests are standard tools in environmental water monitoring.

Exam-style practice questions

Practice questions written in the style of SACE Board exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

SACE 20225 marksA

25.00\ \text{mL}

sample of a solution containing iron(II) ions was acidified and titrated with

0.0200\ \text{mol L}^{-1}

potassium permanganate. The average titre was

21.30\ \text{mL}

. The reaction is

\text{MnO}_4^- + 5\text{Fe}^{2+} + 8\text{H}^+ \rightarrow \text{Mn}^{2+} + 5\text{Fe}^{3+} + 4\text{H}_2\text{O}

. Calculate the concentration of

\text{Fe}^{2+}

in the sample.

Show worked answer →

Step 1: $n(\text{MnO}_4^-) = cV = 0.0200 \times 0.02130 = 4.260 \times 10^{-4}\ \text{mol}$ . (1 mark)

Step 2: the equation gives a $1:5$ ratio of $\text{MnO}_4^-$ to $\text{Fe}^{2+}$ , so $n(\text{Fe}^{2+}) = 5 \times 4.260 \times 10^{-4} = 2.130 \times 10^{-3}\ \text{mol}$ . (2 marks)

Step 3: $c(\text{Fe}^{2+}) = \dfrac{n}{V} = \dfrac{2.130 \times 10^{-3}}{0.02500} = 0.0852\ \text{mol L}^{-1}$ . (2 marks)

SACE 20205 marksThe dissolved oxygen in a water sample was measured by a Winkler-type method in which the liberated iodine was titrated with sodium thiosulfate:

\text{I}_2 + 2\text{S}_2\text{O}_3^{2-} \rightarrow 2\text{I}^- + \text{S}_4\text{O}_6^{2-}

. A

50.0\ \text{mL}

sample required

8.40\ \text{mL}

0.0100\ \text{mol L}^{-1}

thiosulfate. Given that

\text{O}_2 + 4\text{I}^- + 4\text{H}^+ \rightarrow 2\text{I}_2 + 2\text{H}_2\text{O}

, calculate the concentration of dissolved oxygen in

\text{mol L}^{-1}

Show worked answer →

Step 1: $n(\text{S}_2\text{O}_3^{2-}) = 0.0100 \times 0.00840 = 8.40 \times 10^{-5}\ \text{mol}$ . (1 mark)

Step 2: $\text{I}_2 : \text{S}_2\text{O}_3^{2-} = 1:2$ , so $n(\text{I}_2) = \tfrac{1}{2}(8.40 \times 10^{-5}) = 4.20 \times 10^{-5}\ \text{mol}$ . (1 mark)

Step 3: from $\text{O}_2 \rightarrow 2\text{I}_2$ , the ratio is $1:2$ , so $n(\text{O}_2) = \tfrac{1}{2}(4.20 \times 10^{-5}) = 2.10 \times 10^{-5}\ \text{mol}$ . (1 mark)

Step 4: $c(\text{O}_2) = \dfrac{2.10 \times 10^{-5}}{0.0500} = 4.20 \times 10^{-4}\ \text{mol L}^{-1}$ . (2 marks)