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Volumetric analysis and titration curves: WACE Year 12 Chemistry

Syllabus dot point

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

8 min read

Quick answer

Volumetric analysis (titration) finds an unknown concentration by reacting it with a solution of known concentration (a standard) until the reaction is exactly complete at the equivalence point. You calculate the unknown from the balanced equation, the standard's moles, and the volumes. A titration curve plots pH against added volume and rises steeply through the equivalence point.

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.

Equivalence point

The point in a titration at which the amount of titrant added is exactly enough to react completely with the substance being analysed, according to the stoichiometry of the balanced equation.

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×Vn = 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.

Use the mole ratio, not a 1:1 assumption

The amounts of acid and base that react are linked by the coefficients in the balanced equation. For H2SO4+2NaOH→Na2SO4+2H2O\text{H}_2\text{SO}_4 + 2\text{NaOH} \rightarrow \text{Na}_2\text{SO}_4 + 2\text{H}_2\text{O} the ratio is 1:2, not 1:1. Always balance the equation first.

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 25.025.0 mL aliquot of hydrochloric acid is titrated against 0.1000.100 mol L−1^{-1} sodium hydroxide. The average titre is 22.422.4 mL. Find the concentration of the acid. Equation: HCl+NaOH→NaCl+H2O\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×Vn = c \times V, converting the titre to litres. The standard sodium hydroxide solution provides a known number of moles:

n(NaOH)=0.100×0.0224=2.24×10−3 moln(\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(HCl)=2.24×10−3 moln(\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=nV=2.24×10−30.0250=0.0896 mol L−1c = \frac{n}{V} = \frac{2.24 \times 10^{-3}}{0.0250} = 0.0896 \text{ mol L}^{-1}

Final answer: c(HCl)=0.0896c(\text{HCl}) = 0.0896 mol L−1^{-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, KaK_a and indicator choice, tying the acid-base section of Unit 3 together.

Wrong rinsing or wrong mole ratio

Two errors recur. First, rinsing the conical flask with the analyte solution (it should be rinsed with distilled water only) adds extra moles and ruins the result. Second, assuming a 1:1 mole ratio when the acid or base is polyprotic; always balance the equation and use the correct ratio. A third is reading the burette to the wrong precision or from the top of the meniscus.

Exam-style questions

Questions in the style of SCSA 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.

WACE 2021-style7 marks
A 1.325 g1.325\ \text{g} sample of impure anhydrous sodium carbonate is dissolved and made up to 250.0 mL250.0\ \text{mL}. A 25.00 mL25.00\ \text{mL} aliquot requires 23.10 mL23.10\ \text{mL} of 0.1010 mol L−10.1010\ \text{mol L}^{-1} hydrochloric acid for complete reaction. (M(Na2CO3)=106.0 g mol−1M(\text{Na}_2\text{CO}_3) = 106.0\ \text{g mol}^{-1}.) (a) Write the balanced equation. (b) Calculate the concentration of Na2CO3\text{Na}_2\text{CO}_3 in the flask. (c) Calculate the percentage purity of the sample.
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A 7 mark question rewards the equation, the concentration via the 1:2 ratio, and the purity.

(a) Na2CO3+2HCl→2NaCl+H2O+CO2\text{Na}_2\text{CO}_3 + 2\text{HCl} \rightarrow 2\text{NaCl} + \text{H}_2\text{O} + \text{CO}_2.

(b) Moles HCl=0.1010×0.02310=2.333×10−3 mol\text{HCl} = 0.1010 \times 0.02310 = 2.333 \times 10^{-3}\ \text{mol}. The ratio is 2 HCl : 1 carbonate, so moles Na2CO3=2.333×10−32=1.166×10−3 mol\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=1.166×10−30.02500=0.04666 mol L−1.c = \frac{1.166 \times 10^{-3}}{0.02500} = 0.04666\ \text{mol L}^{-1}.

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

purity=1.2361.325×100=93.3%.\text{purity} = \frac{1.236}{1.325} \times 100 = 93.3\%.

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

WACE 2023-style5 marks
Sketch and explain the shape of the titration curve obtained when 0.10 mol L−10.10\ \text{mol L}^{-1} ethanoic acid in the flask is titrated with 0.10 mol L−10.10\ \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.
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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 pH=pKa≈4.7\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 OH−\text{OH}^-. After the equivalence point the curve levels off at high pH as excess NaOH\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 pH=pKa\text{pH} = \text{p}K_a at half-equivalence), the basic equivalence point with hydrolysis, and phenolphthalein.

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Sources & how we know this

Based on the official SCSA syllabus. Not yet individually reviewed by a person; always check SCSA and your school's materials for current rules.

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