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SAChemistrySyllabus dot point

How can we accurately measure the concentration of a solution?

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

Standard solutions, primary standards, titration technique and stoichiometric calculations to find an unknown concentration from acid-base titration data.

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  1. What this dot point is asking
  2. Key terms
  3. The titration formula
  4. Worked example
  5. Technique and reliability
  6. Why it matters for monitoring

What this dot point is asking

You must carry out the calculation reliably and explain the role of standard solutions, indicators and equivalence/endpoint.

Key terms

The equivalence point is where stoichiometrically equivalent amounts of acid and base have reacted. The endpoint is where the indicator changes colour, chosen to be as close as possible to the equivalence point.

The titration formula

The central relationship is:

The procedure:

  1. Use a balanced equation to find the mole ratio.
  2. Calculate moles of the standard reactant from n=cVn = cV using the titre.
  3. Use the mole ratio to find moles of the unknown.
  4. Divide by the volume of unknown to get its concentration.

Worked example

Technique and reliability

  • Rinse the burette and pipette with the solution they will hold; rinse the conical flask only with distilled water.
  • Titrate to a concordant set of titres (within 0.10 mL0.10\,\text{mL}) and average those.
  • Choose an indicator whose colour-change range spans the pH at the equivalence point (e.g. phenolphthalein for a strong acid-strong base or weak acid-strong base titration).
  • Read the burette to ±0.05 mL\pm 0.05\,\text{mL} at the bottom of the meniscus at eye level.

Why it matters for monitoring

Acid-base titration is used in environmental monitoring to measure quantities such as the acidity of water samples or the concentration of dissolved species, giving quantitative water-quality data with simple equipment.

Exam-style practice questions

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

2022 SACE Stage 24 marksAll the sodium benzoate in a 20.0 mL soft-drink sample was converted to benzoic acid and made up to 100.0 mL. A 25.0 mL sample was titrated with 0.0010 mol L-1 NaOH (benzoic acid and NaOH react 1:1); the average titre was 5.05 mL. Calculate the concentration, in mol L-1, of sodium benzoate in this soft drink. Give your answer to the appropriate number of significant figures.
Show worked answer →

Work from the titre back to the original drink.

  1. n(NaOH) = cV = 0.0010 x (5.05 / 1000) = 5.05 x 10^-6 mol.

  2. 1:1 ratio, so n(benzoic acid in 25.0 mL) = 5.05 x 10^-6 mol.

  3. Scale to the full 100.0 mL solution: n(benzoic acid) = 5.05 x 10^-6 x (100.0 / 25.0) = 2.02 x 10^-5 mol. This equals the moles of sodium benzoate originally in the 20.0 mL drink sample.

  4. c(sodium benzoate) = n / V = 2.02 x 10^-5 / 0.0200 = 1.0 x 10^-3 mol L-1 (2 significant figures, matching the 0.0010 data). Marks for moles of NaOH, the ratio, scaling the dilution, and the final concentration.

2022 SACE Stage 22 marksThe molar mass of sodium benzoate is 144.11 g. Calculate the concentration in %w/v of sodium benzoate in this soft drink.
Show worked answer →

%w/v is grams of solute per 100 mL of solution.

  1. Using c = 1.0 x 10^-3 mol L-1, mass per litre = c x M = 1.0 x 10^-3 x 144.11 = 0.144 g L-1.

  2. Convert to grams per 100 mL: 0.144 g L-1 = 0.0144 g per 100 mL.

So the concentration is approximately 0.014 %w/v. One mark for converting moles to mass, one for expressing it as g per 100 mL (%w/v).

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