Carry out volumetric analysis calculations using moles, concentration and stoichiometry.

Using titration data to find an unknown concentration, the mole-ratio method, standard solutions, and back titration.

Generated by Claude Opus 4.77 min answerUpdated 2026-05-30

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

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What this dot point is asking

You must reliably convert between volume, concentration and moles, and use the mole ratio from a balanced equation.

The core relationships

The amount of substance in solution is found from:

A standard solution is one of accurately known concentration, often made from a primary standard such as anhydrous sodium carbonate, weighed precisely and dissolved to a known volume.

The mole-ratio method

The reliable four-step routine is:

Find the moles of the substance you know fully (known concentration and volume).
Use the balanced equation to find moles of the unknown via the mole ratio.
Divide those moles by the volume of the unknown to get its concentration.
Check units and significant figures.

example

Finding an unknown acid concentration

$25.0\ \text{mL}$ of hydrochloric acid is titrated with $0.100\ \text{mol L}^{-1}$ sodium hydroxide. The average titre is $22.4\ \text{mL}$ . The equation is $\text{HCl} + \text{NaOH} \rightarrow \text{NaCl} + \text{H}_2\text{O}$ . Find the acid concentration.

Step 1: $n(\text{NaOH}) = c \times V = 0.100 \times 0.0224 = 2.24 \times 10^{-3}\ \text{mol}$ .

Step 2: The mole ratio of HCl to NaOH is 1 to 1, so $n(\text{HCl}) = 2.24 \times 10^{-3}\ \text{mol}$ .

Step 3: $c(\text{HCl}) = n / V = \dfrac{2.24 \times 10^{-3}}{0.0250} = 0.0896\ \text{mol L}^{-1}$ .

So the acid is about $0.0896\ \text{mol L}^{-1}$ .

Back titration

When a substance reacts slowly or is insoluble, a known excess of reagent is added, then the unreacted excess is titrated. The amount that reacted is found by subtracting the leftover moles from the total added. This is useful for analysing antacids and impure carbonates.

In the exam, write the balanced equation, calculate moles of the fully known species, apply the mole ratio, then divide by the unknown's volume in litres, quoting a sensible number of significant figures.

Exam-style practice questions

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

2021 TASC3 marksThe concentration of a hydrochloric acid solution was determined by titration: 2HCl(aq) + Na2CO3(aq) -> 2NaCl(aq) + H2O(l) + CO2(g). 25.0 mL of hydrochloric acid was placed in a conical flask and 23.35 mL of sodium carbonate solution of concentration 0.0729 mol L-1 was required for neutralisation. Calculate the concentration of the hydrochloric acid.

Show worked answer →

Moles of sodium carbonate: n(Na2CO3) = c x V = 0.0729 x 0.02335 = 1.702 x 10-3 mol. (1 mark)

Mole ratio: from the equation, 2 mol HCl react with 1 mol Na2CO3, so n(HCl) = 2 x 1.702 x 10-3 = 3.405 x 10-3 mol. (1 mark)

Concentration of HCl: c = n / V = 3.405 x 10-3 / 0.0250 = 0.136 mol L-1. (1 mark)

2022 TASC2 marksThe concentration of Fe2+ in river water was measured. Three 20.00 mL samples of diluted river water were titrated with 0.01053 mol L-1 acidified potassium permanganate, with titres of 24.44, 24.40 and 24.42 mL. The reaction is MnO4-(aq) + 5Fe2+(aq) + 8H+(aq) -> 5Fe3+(aq) + Mn2+(aq) + 4H2O(l). Show that the average number of moles of MnO4- delivered from the burette was about 2.6 x 10-4 moles.

Show worked answer →

Average titre: (24.44 + 24.40 + 24.42) / 3 = 24.42 mL = 0.02442 L. (The three titres are concordant, so all are used.)

Moles of permanganate: n(MnO4-) = c x V = 0.01053 x 0.02442 = 2.57 x 10-4 mol.

This is approximately 2.6 x 10-4 moles, as required. (2 marks: average titre and moles.)

2023 TASC4 marksTo find the concentration of Cu2+ in a 100.0 mL copper(II) sulfate solution, 25.0 mL is pipetted and mixed with excess potassium iodide: 2Cu2+(aq) + 4I-(aq) -> 2CuI(s) + I2(aq). The iodine is titrated with 0.0500 mol L-1 sodium thiosulfate: I2(aq) + 2S2O3^2-(aq) -> 2I-(aq) + S4O6^2-(aq). 40.0 mL of thiosulfate is used. What was the concentration of Cu2+ in the original copper(II) sulfate solution?

Show worked answer →

Moles of thiosulfate: n = 0.0500 x 0.0400 = 2.00 x 10-3 mol.

From the titration ratio (2 thiosulfate : 1 iodine), n(I2) = 2.00 x 10-3 / 2 = 1.00 x 10-3 mol.

From the first reaction (2 Cu2+ : 1 I2), n(Cu2+) = 2 x 1.00 x 10-3 = 2.00 x 10-3 mol in the 25.0 mL aliquot.

Concentration of Cu2+ = n / V = 2.00 x 10-3 / 0.0250 = 0.0800 mol L-1. Because the 25.0 mL aliquot was taken from the original undiluted solution, the original concentration is also 0.0800 mol L-1. (4 marks: thiosulfate moles, iodine moles, copper moles, concentration.)

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