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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, indicator choice and back-titration, with fully worked SACE-style stoichiometric calculations that take a titre back to an unknown concentration.

Reviewed by: AI editorial process; not yet individually human-reviewed

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  1. What this dot point is asking
  2. Lead worked calculation
  3. Key terms
  4. The central relationship
  5. Choosing the indicator
  6. Back-titration
  7. Technique and reliability
  8. Why it matters for monitoring

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

Key terms

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

The general procedure for any titration calculation:

  1. Write the balanced equation and read off 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 analyte.
  4. Account for any dilution or aliquot scaling.
  5. Divide by the relevant volume to get the concentration (or multiply by MM 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. HCl\text{HCl} and NaOH\text{NaOH}): equivalence pH =7= 7; the vertical region is wide, so methyl orange or phenolphthalein both work.
  • Weak acid with strong base (e.g. ethanoic acid and NaOH\text{NaOH}): the salt is basic, equivalence pH >7> 7; use phenolphthalein (range 8.38.3 to 10.010.0).
  • Strong acid with weak base (e.g. HCl\text{HCl} and NH3\text{NH}_3): the salt is acidic, equivalence pH <7< 7; use methyl orange (range 3.13.1 to 4.44.4).

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 CaCO3\text{CaCO}_3 is dissolved in a measured excess of standard HCl\text{HCl}; the unreacted HCl\text{HCl} is then titrated with standard NaOH\text{NaOH}, and n(HCl reacted with carbonate)=n(HCl added)n(HCl titrated)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 mL0.10\ \text{mL} of each other and average only those.
  • Read the burette to ±0.05 mL\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.

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 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.

SACE 20224 marksAll the sodium benzoate in a 20.0 mL20.0\ \text{mL} soft-drink sample was converted to benzoic acid and made up to 100.0 mL100.0\ \text{mL}. A 25.0 mL25.0\ \text{mL} portion was titrated with 0.0010 mol L10.0010\ \text{mol L}^{-1} NaOH\text{NaOH} (benzoic acid and NaOH\text{NaOH} react 1:11:1); the average titre was 5.05 mL5.05\ \text{mL}. Calculate the concentration, in mol L1\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(NaOH)=cV=0.0010×5.051000=5.05×106 moln(\text{NaOH}) = cV = 0.0010 \times \dfrac{5.05}{1000} = 5.05 \times 10^{-6}\ \text{mol}. (1 mark)

Step 2: the 1:11:1 ratio gives n(benzoic acid in 25.0 mL)=5.05×106 moln(\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 mL100.0\ \text{mL}: n=5.05×106×100.025.0=2.02×105 moln = 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 mL20.0\ \text{mL} sample. (1 mark)

Step 4: c=nV=2.02×1050.0200=1.0×103 mol L1c = \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 0.00100.0010 data). (1 mark)

SACE 20215 marksA 1.325 g1.325\ \text{g} sample of impure anhydrous sodium carbonate was dissolved and made up to 250.0 mL250.0\ \text{mL}. A 25.00 mL25.00\ \text{mL} aliquot required 24.10 mL24.10\ \text{mL} of 0.1000 mol L10.1000\ \text{mol L}^{-1} HCl\text{HCl} for complete reaction: Na2CO3+2HCl2NaCl+H2O+CO2\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 Na2CO3\text{Na}_2\text{CO}_3 in the sample. (M(Na2CO3)=105.99 g mol1M(\text{Na}_2\text{CO}_3) = 105.99\ \text{g mol}^{-1}.)
Show worked answer →

Step 1: n(HCl)=cV=0.1000×0.02410=2.410×103 moln(\text{HCl}) = cV = 0.1000 \times 0.02410 = 2.410 \times 10^{-3}\ \text{mol}. (1 mark)

Step 2: the 2:12:1 ratio gives n(Na2CO3 in aliquot)=12(2.410×103)=1.205×103 moln(\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 mL250.0\ \text{mL}: n=1.205×103×250.025.00=1.205×102 moln = 1.205 \times 10^{-3} \times \dfrac{250.0}{25.00} = 1.205 \times 10^{-2}\ \text{mol}. (1 mark)

Step 4: m(Na2CO3)=nM=1.205×102×105.99=1.277 gm(\text{Na}_2\text{CO}_3) = nM = 1.205 \times 10^{-2} \times 105.99 = 1.277\ \text{g}. (1 mark)

Step 5: % by mass=1.2771.325×100=96.4%\%\text{ by mass} = \dfrac{1.277}{1.325} \times 100 = 96.4\%. (1 mark)

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