SCSA WACE 2021-style practice: A 1.325\ g sample of impure anhydrous sodium carbonate is dissolved and made up to 250.0\ mL. A 25.00\ mL aliquot requires 23.10\ mL of 0.1010\ mol L^-1 hydrochloric acid for complete reaction. (M(Na_2CO_3) = 106.0\ g mol^-1.) (a) Write the balanced equation. (b) Calculate the concentration of Na_2CO_3 in the flask. (c) Calculate the percentage purity of the sample.

A 7 mark question rewards the equation, the concentration via the 1:2 ratio, and the purity. **(a)** Na_2CO_3 + 2HCl → 2NaCl + H_2O + CO_2. **(b)** Moles HCl = 0.1010 × 0.02310 = 2.333 × 10^-3\ mol. The ratio is 2 HCl : 1 carbonate, so moles Na_2CO_3 = 2.333 × 10^-32 = 1.166 × 10^-3\ mol in the 25.00 mL aliquot. $c = 1.166 × 10^-30.02500 = 0.04666\ mol L^-1.$ **(c)** Total moles in 250.0 mL = 0.04666 × 0.2500 = 1.166 × 10^-2\ mol. Mass = 1.166 × 10^-2 × 106.0 = 1.236\ g. $purity = (1.236)/(1.325) × 100 = 93.3\%.$ Markers reward the balanced equation, the 1:2 ratio giving c = 0.0467\ mol L^-1, and the 93.3\% purity (scaling the aliquot to the full flask).

§-Syllabus dot point

WAChemistrySyllabus dot point

How do we use titration to find an unknown concentration, and what shape does a titration curve take?

Perform and interpret acid-base titrations, sketch titration curves, and calculate unknown concentrations from volumetric data

A focused answer to the WACE Year 12 Chemistry dot point on volumetric analysis, the technique and calculations of acid-base titration, the shape of titration curves for different acid-base combinations, and the equivalence point, with a worked example and common exam mistakes.

Generated by Claude Opus 4.88 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

Volumetric analysis is a quantitative technique that uses an accurately measured volume of a solution of known concentration (a standard solution) to determine the concentration of another solution.

The technique

A pipette delivers a precise volume (an aliquot) of one solution into a conical flask. The other solution is added from a burette until the reaction is just complete, signalled by an indicator colour change (the end point). With careful technique the end point coincides with the equivalence point. Apparatus is rinsed appropriately: the burette and pipette with the solution they will deliver, the conical flask with distilled water only, to avoid changing the amount of substance being measured.

The titration calculation

The calculation uses three steps. First find the moles of the standard from $n = c \times V$ . Second use the mole ratio from the balanced equation to find moles of the unknown. Third divide by the unknown's volume to get its concentration.

The shape of titration curves

A titration curve plots the pH of the flask against the volume of titrant added. Its key feature is a steep, near-vertical rise through the equivalence point. The shape depends on the strengths involved:

Strong acid with strong base: curve starts low, rises gently, then jumps steeply through pH 7 (the equivalence point), then levels off high. The steep region is large (about pH 3 to 11).
Weak acid with strong base: starts at a higher pH, has a buffer region where pH changes slowly, and the equivalence point lies above pH 7 (the salt is basic). The vertical region is shorter.
Strong acid with weak base: the equivalence point lies below pH 7 (the salt is acidic).

The indicator must change colour within the steep region so the end point matches the equivalence point.

Standard solutions

A primary standard is a pure, stable substance of accurately known formula (such as anhydrous sodium carbonate) that can be weighed to make a solution of precisely known concentration. Solutions of substances that are not stable enough to be primary standards (such as sodium hydroxide, which absorbs moisture and carbon dioxide) must be standardised against a primary standard before use.

Finding an unknown concentration

A $25.0$ mL aliquot of hydrochloric acid is titrated against $0.100$ mol L $^{-1}$ sodium hydroxide. The average titre is $22.4$ mL. Find the concentration of the acid. Equation: $\text{HCl} + \text{NaOH} \rightarrow \text{NaCl} + \text{H}_2\text{O}$ (1:1 ratio).

Step 1: Find moles of sodium hydroxide

Use $n = c \times V$ , converting the titre to litres. The standard sodium hydroxide solution provides a known number of moles:

n(\text{NaOH}) = 0.100 \times 0.0224 = 2.24 \times 10^{-3} \text{ mol}

Step 2: Use the mole ratio to find moles of HCl

The balanced equation shows a 1:1 mole ratio between HCl and NaOH, so the moles of HCl that reacted equal the moles of NaOH used:

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

Step 3: Calculate the concentration of HCl

Divide the moles of HCl by the volume of the aliquot (in litres) to find the concentration:

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

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

Why this matters

Volumetric analysis is the foundation of quantitative chemistry in the laboratory, from quality control to environmental monitoring. The titration curve links the calculation back to the equilibrium ideas of pH, $K_a$ and indicator choice, tying the acid-base section of Unit 3 together.

Exam-style practice questions

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

WACE 20217 marksA

1.325\ \text{g}

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

250.0\ \text{mL}

. A

25.00\ \text{mL}

aliquot requires

23.10\ \text{mL}

0.1010\ \text{mol L}^{-1}

hydrochloric acid for complete reaction. (

M(\text{Na}_2\text{CO}_3) = 106.0\ \text{g mol}^{-1}

.) (a) Write the balanced equation. (b) Calculate the concentration of

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

in the flask. (c) Calculate the percentage purity of the sample.

Show worked answer →

A 7 mark question rewards the equation, the concentration via the 1:2 ratio, and the purity.

(a) $\text{Na}_2\text{CO}_3 + 2\text{HCl} \rightarrow 2\text{NaCl} + \text{H}_2\text{O} + \text{CO}_2$ .

(b) Moles $\text{HCl} = 0.1010 \times 0.02310 = 2.333 \times 10^{-3}\ \text{mol}$ . The ratio is 2 HCl : 1 carbonate, so moles $\text{Na}_2\text{CO}_3 = \dfrac{2.333 \times 10^{-3}}{2} = 1.166 \times 10^{-3}\ \text{mol}$ in the 25.00 mL aliquot.

c = \frac{1.166 \times 10^{-3}}{0.02500} = 0.04666\ \text{mol L}^{-1}.

(c) Total moles in 250.0 mL $= 0.04666 \times 0.2500 = 1.166 \times 10^{-2}\ \text{mol}$ . Mass $= 1.166 \times 10^{-2} \times 106.0 = 1.236\ \text{g}$ .

\text{purity} = \frac{1.236}{1.325} \times 100 = 93.3\%.

Markers reward the balanced equation, the 1:2 ratio giving $c = 0.0467\ \text{mol L}^{-1}$ , and the $93.3\%$ purity (scaling the aliquot to the full flask).

WACE 20235 marksSketch and explain the shape of the titration curve obtained when

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

ethanoic acid in the flask is titrated with

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

sodium hydroxide from the burette. Refer to the starting pH, the buffer region, the equivalence point pH, and a suitable indicator.

Show worked answer →

A 5 mark answer needs the four labelled features with reasoning.

The curve starts at a moderate pH (about 2.9), higher than for a strong acid, because ethanoic acid is weak and only partly ionised. As base is added a buffer region forms where the flask contains both ethanoic acid and ethanoate; here the pH rises only slowly because the mixture resists change (the half-equivalence point gives $\text{pH} = \text{p}K_a \approx 4.7$ ).

Near the equivalence point the pH rises steeply. The equivalence point is basic (about pH 8.7) because the product, sodium ethanoate, is the salt of a weak acid and its anion hydrolyses to give $\text{OH}^-$ . After the equivalence point the curve levels off at high pH as excess $\text{NaOH}$ dominates.

A suitable indicator is phenolphthalein (range 8.3 to 10.0), because its colour change lies on the steep region around the basic equivalence point; methyl orange would change far too early.

Markers reward the elevated start, the buffer region (with $\text{pH} = \text{p}K_a$ at half-equivalence), the basic equivalence point with hydrolysis, and phenolphthalein.