Volumetric analysis: acid-base titrations (SACE Stage 2 Chemistry)

Syllabus dot point

“Apply volumetric analysis using acid-base titrations to determine unknown concentrations.”

9 min readUpdated 2 Jun 2026

Quick answer

Volumetric analysis uses a titration of accurately measured volumes to find an unknown concentration. You standardise one solution against a primary standard, titrate it against the unknown to an indicator endpoint, then use $n = cV$ and the mole ratio from the balanced equation to calculate the unknown concentration. The routine is always: moles of the known, apply the mole ratio, then divide by volume (or scale through any dilution).

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What this dot point is asking
Lead worked calculation
Key terms
The central relationship
Choosing the indicator
Back-titration
Technique and reliability
Why it matters for monitoring
Exam-style questions

What this dot point is asking

SACE expects you to perform the full calculation reliably, justify your choice of indicator, and explain the role of standard solutions, primary standards and the difference between equivalence point and endpoint.

Lead worked calculation

Finding the concentration of HCl

$25.00\ \text{mL}$ of hydrochloric acid is titrated against $0.105\ \text{mol L}^{-1}$ sodium hydroxide. The average concordant titre is $22.40\ \text{mL}$ . Find the concentration of the acid.

Step 1: Write the balanced equation and identify the mole ratio

Before any numbers, you need the stoichiometry. Hydrochloric acid and sodium hydroxide react in a simple neutralisation, and writing the equation tells you exactly how many moles of one species correspond to how many moles of the other.

\text{HCl} + \text{NaOH} \rightarrow \text{NaCl} + \text{H}_2\text{O}

The coefficients are all $1$ , so the mole ratio of $\text{HCl}$ to $\text{NaOH}$ is $1:1$ . This is the link that lets you transfer information from the standard solution to the unknown.

Step 2: Calculate moles of the standard (NaOH)

The sodium hydroxide solution is the standard here: its concentration is accurately known. You use the titre, which is the volume of NaOH delivered by the burette, converted to litres, together with the known concentration to find the number of moles actually delivered at the endpoint.

n(\text{NaOH}) = cV = 0.105 \times \frac{22.40}{1000} = 0.105 \times 0.02240 = 2.352 \times 10^{-3}\ \text{mol}

Step 3: Transfer to moles of the unknown (HCl) using the ratio

Because the ratio is $1:1$ , every mole of NaOH that reacted corresponds to exactly one mole of HCl in the flask. If the ratio were different (for example $2:1$ for sulfuric acid), you would multiply or divide accordingly.

n(\text{HCl}) = n(\text{NaOH}) = 2.352 \times 10^{-3}\ \text{mol}

Step 4: Find the concentration by dividing by the volume of the unknown

The $2.352 \times 10^{-3}\ \text{mol}$ of HCl was contained in the $25.00\ \text{mL}$ pipette volume. Dividing moles by volume (in litres) gives concentration. This is simply rearranging $n = cV$ to solve for $c$ .

c(\text{HCl}) = \frac{n}{V} = \frac{2.352 \times 10^{-3}}{0.02500} = 0.0941\ \text{mol L}^{-1}

Final answer: $c(\text{HCl}) = 0.0941\ \text{mol L}^{-1}$

Key terms

Standard solution and primary standard

A standard solution has an accurately known concentration. A primary standard is a pure, stable, non-hygroscopic solid of high and known molar mass (e.g. anhydrous sodium carbonate or potassium hydrogen phthalate) that can be weighed accurately and dissolved to a known volume, giving a standard solution directly. Sodium hydroxide is not a primary standard because it absorbs water and carbon dioxide from the air, so its true mass of $\text{NaOH}$ is uncertain; it must be standardised against a primary standard before use.

The equivalence point is where stoichiometrically equivalent amounts of acid and base have reacted, as defined by the balanced equation. The endpoint is where the chosen indicator changes colour. A well-chosen indicator makes the endpoint coincide with the equivalence point as closely as possible, so the titre is a true measure of the reacting amounts.

The central relationship

Moles from concentration and volume

n = c\,V

where $n$ is amount in $\text{mol}$ , $c$ is concentration in $\text{mol L}^{-1}$ , and $V$ is volume in $\text{litres}$ . Always convert $\text{mL}$ to $\text{L}$ (divide by $1000$ ) before substituting.

The general procedure for any titration calculation:

Write the balanced equation and read off the mole ratio.
Calculate moles of the standard reactant from $n = cV$ using the titre.
Use the mole ratio to find moles of the analyte.
Account for any dilution or aliquot scaling.
Divide by the relevant volume to get the concentration (or multiply by $M$ for a mass).

Choosing the indicator

The pH at the equivalence point depends on the salt formed, so the indicator must change colour within the steep part of the titration curve at that pH.

Strong acid with strong base (e.g. $\text{HCl}$ and $\text{NaOH}$ ): equivalence pH $= 7$ ; the vertical region is wide, so methyl orange or phenolphthalein both work.
Weak acid with strong base (e.g. ethanoic acid and $\text{NaOH}$ ): the salt is basic, equivalence pH $> 7$ ; use phenolphthalein (range $8.3$ to $10.0$ ).
Strong acid with weak base (e.g. $\text{HCl}$ and $\text{NH}_3$ ): the salt is acidic, equivalence pH $< 7$ ; use methyl orange (range $3.1$ to $4.4$ ).

Back-titration

When an analyte reacts slowly, is insoluble, or is a solid such as an antacid, you add a known excess of one reagent, then titrate the leftover excess with a second standard. The amount that reacted with the analyte is the difference. For example, an antacid containing $\text{CaCO}_3$ is dissolved in a measured excess of standard $\text{HCl}$ ; the unreacted $\text{HCl}$ is then titrated with standard $\text{NaOH}$ , and $n(\text{HCl reacted with carbonate}) = n(\text{HCl added}) - n(\text{HCl titrated})$ .

Technique and reliability

Rinse the burette and pipette with the solution they will hold; rinse the conical flask with distilled water only (residual water does not change the moles delivered into it).
Perform a rough trial titre, then repeat for concordant titres within $0.10\ \text{mL}$ of each other and average only those.
Read the burette to $\pm 0.05\ \text{mL}$ at the bottom of the meniscus at eye level to avoid parallax.
The final answer can be no more precise than the least precise data; match significant figures accordingly.

Forgetting the mole ratio and the dilution scale

The two most common SACE errors are assuming a $1:1$ ratio and ignoring aliquot or dilution scaling. For $\text{H}_2\text{SO}_4 + 2\text{NaOH}$ , two moles of base react per mole of acid, so the ratio is essential. When only an aliquot of a made-up solution is titrated, you must multiply back up by $\dfrac{V_\text{total}}{V_\text{aliquot}}$ to recover the amount in the whole flask. Also always convert $\text{mL}$ to $\text{L}$ before using $n = cV$ .

Why it matters for monitoring

Acid-base titration delivers quantitative water-quality data, such as total acidity, alkalinity or the concentration of a dissolved species, using simple, low-cost equipment. It is the benchmark technique against which instrumental methods such as AAS and chromatography are calibrated, and it underpins quality control across food, environmental and industrial laboratories.

Exam-style questions

Questions in the style of SACE Board exam questions on this dot point, each with a worked answer. They are written by ExamExplained unless tagged "Past paper"; the year shows the paper a question is modelled on.

SACE 2022-style4 marks

All the sodium benzoate in a

20.0\ \text{mL}

soft-drink sample was converted to benzoic acid and made up to

100.0\ \text{mL}

. A

25.0\ \text{mL}

portion was titrated with

0.0010\ \text{mol L}^{-1}

\text{NaOH}

(benzoic acid and

\text{NaOH}

react

1:1

); the average titre was

5.05\ \text{mL}

. Calculate the concentration, in

\text{mol L}^{-1}

, of sodium benzoate in the soft drink, to the appropriate number of significant figures.

Show worked answer →

Work from the titre back to the original drink.

Step 1: $n(\text{NaOH}) = cV = 0.0010 \times \dfrac{5.05}{1000} = 5.05 \times 10^{-6}\ \text{mol}$ . (1 mark)

Step 2: the $1:1$ ratio gives $n(\text{benzoic acid in } 25.0\ \text{mL}) = 5.05 \times 10^{-6}\ \text{mol}$ . (1 mark)

Step 3: scale to the full $100.0\ \text{mL}$ : $n = 5.05 \times 10^{-6} \times \dfrac{100.0}{25.0} = 2.02 \times 10^{-5}\ \text{mol}$ . This equals the moles of sodium benzoate in the original $20.0\ \text{mL}$ sample. (1 mark)

Step 4: $c = \dfrac{n}{V} = \dfrac{2.02 \times 10^{-5}}{0.0200} = 1.0 \times 10^{-3}\ \text{mol L}^{-1}$ (2 significant figures, matching the $0.0010$ data). (1 mark)

SACE 2021-style5 marks

1.325\ \text{g}

sample of impure anhydrous sodium carbonate was dissolved and made up to

250.0\ \text{mL}

. A

25.00\ \text{mL}

aliquot required

24.10\ \text{mL}

0.1000\ \text{mol L}^{-1}

\text{HCl}

for complete reaction:

\text{Na}_2\text{CO}_3 + 2\text{HCl} \rightarrow 2\text{NaCl} + \text{H}_2\text{O} + \text{CO}_2

. Calculate the percentage by mass of

\text{Na}_2\text{CO}_3

in the sample. (

M(\text{Na}_2\text{CO}_3) = 105.99\ \text{g mol}^{-1}

Show worked answer →

Step 1: $n(\text{HCl}) = cV = 0.1000 \times 0.02410 = 2.410 \times 10^{-3}\ \text{mol}$ . (1 mark)

Step 2: the $2:1$ ratio gives $n(\text{Na}_2\text{CO}_3 \text{ in aliquot}) = \tfrac{1}{2}(2.410 \times 10^{-3}) = 1.205 \times 10^{-3}\ \text{mol}$ . (1 mark)

Step 3: scale to the full $250.0\ \text{mL}$ : $n = 1.205 \times 10^{-3} \times \dfrac{250.0}{25.00} = 1.205 \times 10^{-2}\ \text{mol}$ . (1 mark)

Step 4: $m(\text{Na}_2\text{CO}_3) = nM = 1.205 \times 10^{-2} \times 105.99 = 1.277\ \text{g}$ . (1 mark)

Step 5: $\%\text{ by mass} = \dfrac{1.277}{1.325} \times 100 = 96.4\%$ . (1 mark)

Practise this

Sources & how we know this

Stage 2 Chemistry subject outline (2025 onwards) — SACE Board
Stage 2 Chemistry: external assessment — SACE Board