TASC TCE 2021-style practice: The concentration of a hydrochloric acid solution was determined by titration: 2HCl(aq) + Na_2CO_3(aq) → 2NaCl(aq) + H_2O(l) + CO_2(g). 25.0\ mL of hydrochloric acid required 23.35\ mL of sodium carbonate solution of concentration 0.0729\ mol L^-1 for neutralisation. Calculate the concentration of the hydrochloric acid.

Moles of sodium carbonate: n(Na_2CO_3) = cV = 0.0729 × 0.02335 = 1.702 × 10^-3\ mol. (1 mark) Mole ratio: 2\ mol of HCl react with 1\ mol of Na_2CO_3, so n(HCl) = 2 × 1.702 × 10^-3 = 3.405 × 10^-3\ mol. (1 mark) Concentration: c(HCl) = n/V = 3.405 × 10^-3 / 0.0250 = 0.136\ mol L^-1. (1 mark)

TASC TCE 2023-style practice: To find [Cu^2+] in a copper(II) sulfate solution, 25.0\ mL is mixed with excess potassium iodide: 2Cu^2+(aq) + 4I^-(aq) → 2CuI(s) + I_2(aq). The liberated iodine is titrated with 0.0500\ mol L^-1 sodium thiosulfate: I_2(aq) + 2S_2O_3^2-(aq) → 2I^-(aq) + S_4O_6^2-(aq). 40.0\ mL of thiosulfate is used. Calculate [Cu^2+] in the original solution.

Moles of thiosulfate: n = 0.0500 × 0.0400 = 2.00 × 10^-3\ mol. (1 mark) Titration ratio is 2\ S_2O_3^2- : 1\ I_2, so n(I_2) = 2.00 × 10^-3 / 2 = 1.00 × 10^-3\ mol. (1 mark) From the first reaction (2\ Cu^2+ : 1\ I_2), n(Cu^2+) = 2 × 1.00 × 10^-3 = 2.00 × 10^-3\ mol in the 25.0\ mL aliquot. (1 mark) c(Cu^2+) = n/V = 2.00 × 10^-3 / 0.0250 = 0.0800\ mol L^-1. (1 mark)

§-Syllabus dot point

TASChemistrySyllabus dot point

How do we calculate an unknown concentration from titration data?

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

Using titration data to find an unknown concentration, the mole-ratio method, standard solutions, dilution, back titration and double (redox) titrations, with fully worked TASC-style examples.

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

TASC expects you to convert reliably between volume, concentration and moles, apply the mole ratio from a balanced equation, and handle dilution, back titrations and two-stage redox (double) titrations.

The core relationships

A standard solution has accurately known concentration, usually made by weighing a primary standard (a pure, stable, non-hygroscopic solid such as anhydrous sodium carbonate) precisely and dissolving it to an exact volume in a volumetric flask. Concordant titres agree to within about $0.10\ \text{mL}$ ; average only the concordant runs and discard rough trials.

The mole-ratio method

The reliable routine is:

Write the balanced equation.
Find the moles of the species you know completely (known $c$ and $V$ ): the titrant.
Use the balanced equation's coefficient ratio to find moles of the unknown.
Divide those moles by the volume of the unknown to get its concentration.

Finding an unknown acid concentration

Find the concentration of a hydrochloric acid solution given that $25.0\ \text{mL}$ aliquots were titrated with $0.100\ \text{mol L}^{-1}$ sodium hydroxide, yielding concordant titres of $22.45$ , $22.40$ and $22.35\ \text{mL}$ . The reaction is $\text{HCl}(aq) + \text{NaOH}(aq) \rightarrow \text{NaCl}(aq) + \text{H}_2\text{O}(l)$ .

Step 1: Average the concordant titres

All three titres agree to within $0.10\ \text{mL}$ , so they are all concordant and should all be included in the average. Averaging reduces the effect of small random errors in reading the burette:

V(\text{NaOH}) = \frac{22.45 + 22.40 + 22.35}{3} = 22.40\ \text{mL} = 0.02240\ \text{L}.

Step 2: Calculate the moles of the titrant (NaOH)

The concentration and volume of sodium hydroxide are both known, so the moles can be calculated directly using $n = cV$ . Remember to use the volume in litres:

n(\text{NaOH}) = cV = 0.100 \times 0.02240 = 2.24 \times 10^{-3}\ \text{mol}.

Step 3: Apply the stoichiometric mole ratio

The balanced equation shows a $1:1$ ratio between $\text{HCl}$ and $\text{NaOH}$ . At the equivalence point the moles of acid neutralised equal the moles of base added:

n(\text{HCl}) = n(\text{NaOH}) = 2.24 \times 10^{-3}\ \text{mol}.

Step 4: Calculate the concentration of the unknown acid

Divide the moles of hydrochloric acid by the volume of acid used (the pipette volume, converted to litres):

c(\text{HCl}) = \frac{n}{V} = \frac{2.24 \times 10^{-3}}{0.0250} = 0.0896\ \text{mol L}^{-1}.

Final answer: $c(\text{HCl}) = 0.0896\ \text{mol L}^{-1}$

Dilution

If a stock solution is diluted before titrating, account for the dilution factor. The amount of solute is conserved on dilution, so:

Back titration

When a substance reacts slowly or is insoluble (an antacid tablet, impure limestone), add a known excess of standard reagent, let it react fully, then titrate the unreacted excess. The amount that reacted with the sample equals total added minus leftover. This avoids slow or poorly defined end points.

Apparatus and good technique

Accurate volumetric analysis depends on the right glassware and technique. A pipette delivers a fixed, accurately known volume of the analyte into the conical flask; a burette delivers a variable, precisely measured volume of titrant; and a volumetric flask makes up a standard solution to an exact volume. Rinse the burette with the titrant and the pipette with the analyte (not water) so residual water does not dilute them, but rinse the conical flask with water only, because any extra water there does not change the moles of analyte present. Read the bottom of the meniscus at eye level, and swirl continuously near the end point, adding dropwise once the indicator begins to flicker.

Redox and double titrations

Not every titration is acid-base. In a redox titration, an oxidant such as acidified permanganate or dichromate is titrated against a reductant such as iron(II), and a coloured reagent like permanganate can act as its own indicator (the first persistent pink marks the end point). A double (two-stage) titration, as in the copper-iodine-thiosulfate example above, links two reactions through a shared species, so you carry the moles from the titration step back through the first reaction to reach the analyte. Always combine the two mole ratios carefully.

In the exam, write the balanced equation, average the concordant titres, calculate moles of the fully known species, apply the mole ratio, undo any dilution, 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.

TCE 20213 marksThe concentration of a hydrochloric acid solution was determined by titration:

2\text{HCl}(aq) + \text{Na}_2\text{CO}_3(aq) \rightarrow 2\text{NaCl}(aq) + \text{H}_2\text{O}(l) + \text{CO}_2(g)

25.0\ \text{mL}

of hydrochloric acid required

23.35\ \text{mL}

of sodium carbonate solution of concentration

0.0729\ \text{mol L}^{-1}

for neutralisation. Calculate the concentration of the hydrochloric acid.

Show worked answer →

Moles of sodium carbonate: $n(\text{Na}_2\text{CO}_3) = cV = 0.0729 \times 0.02335 = 1.702 \times 10^{-3}\ \text{mol}$ . (1 mark)

Mole ratio: $2\ \text{mol}$ of $\text{HCl}$ react with $1\ \text{mol}$ of $\text{Na}_2\text{CO}_3$ , so $n(\text{HCl}) = 2 \times 1.702 \times 10^{-3} = 3.405 \times 10^{-3}\ \text{mol}$ . (1 mark)

Concentration: $c(\text{HCl}) = n/V = 3.405 \times 10^{-3} / 0.0250 = 0.136\ \text{mol L}^{-1}$ . (1 mark)

TCE 20234 marksTo find

[\text{Cu}^{2+}]

in a copper(II) sulfate solution,

25.0\ \text{mL}

is mixed with excess potassium iodide:

2\text{Cu}^{2+}(aq) + 4\text{I}^-(aq) \rightarrow 2\text{CuI}(s) + \text{I}_2(aq)

. The liberated iodine is titrated with

0.0500\ \text{mol L}^{-1}

sodium thiosulfate:

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

40.0\ \text{mL}

of thiosulfate is used. Calculate

[\text{Cu}^{2+}]

in the original solution.

Show worked answer →

Moles of thiosulfate: $n = 0.0500 \times 0.0400 = 2.00 \times 10^{-3}\ \text{mol}$ . (1 mark)

Titration ratio is $2\ \text{S}_2\text{O}_3^{2-} : 1\ \text{I}_2$ , so $n(\text{I}_2) = 2.00 \times 10^{-3} / 2 = 1.00 \times 10^{-3}\ \text{mol}$ . (1 mark)

From the first reaction ( $2\ \text{Cu}^{2+} : 1\ \text{I}_2$ ), $n(\text{Cu}^{2+}) = 2 \times 1.00 \times 10^{-3} = 2.00 \times 10^{-3}\ \text{mol}$ in the $25.0\ \text{mL}$ aliquot. (1 mark)

$c(\text{Cu}^{2+}) = n/V = 2.00 \times 10^{-3} / 0.0250 = 0.0800\ \text{mol L}^{-1}$ . (1 mark)