Soluble salts from acid–alkali titration
Find the reacting volumes, then repeat without indicator so a dissolved salt can be crystallised cleanly.
Exam scope and assessment notes
Core and Extended · Chemistry 0971, 2026–2028. Cambridge states that IGCSE 0971 has the same subject content as IGCSE 0620, with 9–1 grading. This StudyForA pathway maps the 2026–2028 syllabus only. This 9–1 graded qualification is available only in a limited number of administrative zones. Confirm that your centre can enter you for this code and exam series with Cambridge International. Cambridge has published a separate syllabus for 2029. StudyForA has not reconciled that version yet; this page supports exams in 2026, 2027 and 2028 only.
Before you begin
- Read the acid–alkali neutralisation equation.
- Distinguish an indicator endpoint from a product and recall filtration limitations.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
Which reactant is easily removable by filtration after a suitable acid reaction?
What you will learn
- Explain why two soluble reactants need measured reacting volumes rather than an easily filtered excess.
- Describe titration followed by an indicator-free repeat and crystallisation.
The excess cannot be filtered off
For sodium chloride from hydrochloric acid and sodium hydroxide, both reactants and product are soluble. An excess of either reactant stays in solution, so ordinary filtration cannot remove it. Titration finds the appropriate reacting volumes by using a measured volume of one solution and adding the other from a burette to a suitable indicator endpoint, under teacher supervision.
Choose a suitable indicator for the acid–alkali system rather than assuming every indicator changes at exactly pH 7. The observed endpoint helps locate the required volume; it is not a mass of product. Use careful burette readings and repeat with consistent endpoints to obtain reliable reacting volumes. Mole-based concentration calculations are taught in the Extended stoichiometry lesson, while the Core focus here is preparation and purification.
Repeat without the colour contaminant
Once the appropriate volumes are established, use those reacting volumes with fresh solutions but no indicator, so indicator dye does not contaminate the intended crystals. For the stated HCl/NaOH example, HCl+NaOH→NaCl+H₂O produces the salt solution. Omitting indicator on the initial unknown-volume trial would remove the stated endpoint evidence; omitting it on the measured repeat is purposeful.
Concentrate the indicator-free salt solution appropriately and allow it to cool for crystallisation, then separate and suitably dry the product. Do not try to filter indicator molecules or dissolved excess alkali from the solution as a substitute for the controlled repeat. The measured-volume method and the excess-insoluble-solid method solve different separation problems.
Pause and explain
When should indicator be omitted in this preparation?
Worked example
Why is an indicator-free repeat needed after finding the reacting volumes of HCl and NaOH by titration?
Show the worked solution
- An indicator provides endpoint evidence during the measured trial.
- The resulting dye remains dissolved with the salt and would be a contamination source.
- Repeat with the established reacting volumes and fresh solutions without indicator, then crystallise the salt.
Answer Use indicator to find volumes; use an indicator-free measured repeat to prepare the salt solution for crystallisation.
Try a topic test
Choose Easy, Normal or Difficult. Save your answers and open worked review after finishing.
Test this topic →From guided practice to a new situation
Write your answer and reasoning first. Use a hint only if you are stuck. The model solution is for self-checking; your written work is not automatically marked.
Outline a sodium chloride preparation from HCl and NaOH using titration.
Give me a hint
Find the volume first, then prepare the clean product solution.
Compare my reasoning
- Use a measured reactant volume, burette addition and suitable indicator to establish reacting volumes.
- Repeat with the measured volumes and fresh solutions without indicator.
- Concentrate appropriately, cool to crystallise, separate and suitably dry the NaCl.
Titrate to find volumes → indicator-free repeat → crystallise and recover salt.
Look for these in your work
- Volume-finding and product-making stages differ.
- The dye contamination is avoided.
A reliable trial requires 18.6 cm³ HCl for 25.0 cm³ NaOH. State the volumes and indicator choice for a fresh product-making repeat, without calculating concentrations.
Give me a hint
Use the experimentally established amounts rather than adding an arbitrary excess.
Compare my reasoning
- Measure 25.0 cm³ fresh NaOH and 18.6 cm³ fresh HCl under the appropriate supervised method.
- Omit the indicator because the reacting volumes were established in the trial.
- Use the resulting salt solution for suitable crystallisation rather than filtering out dissolved excess.
25.0 cm³ NaOH and 18.6 cm³ HCl, without indicator in the measured repeat.
Look for these in your work
- Both established volumes are retained.
- No invented concentration calculation is required.
Compare NaCl preparation from HCl/NaOH with CuSO₄ preparation from H₂SO₄/excess CuO. Explain why their excess-removal strategies differ.
Give me a hint
One excess is dissolved; the other is an insoluble solid.
Compare my reasoning
- Excess NaOH stays dissolved, so use titration and an indicator-free measured repeat.
- Excess CuO remains solid and can be filtered away after consuming the acid.
- Both retain a soluble salt solution for crystallisation, but their reactant-removal steps differ.
HCl/NaOH needs measured reacting volumes; excess CuO can be filtered. Both then crystallise the soluble salt.
Look for these in your work
- Solubility determines the method choice.
- The shared final recovery and different earlier stages are explained.
Common mistakes
- Adding excess soluble alkali and expecting filtration to remove it.
- Omitting endpoint evidence before reacting volumes are known.
- Leaving indicator in the batch intended for clean crystals.
- Calling every indicator endpoint exactly neutral pH 7.
What can you explain now?
Why can excess NaOH not be removed by ordinary filtration from a NaCl solution?
Compare with the explanation
It is dissolved and passes through the filter with the solution.
Titration controls the reacting amount because an excess soluble reactant is not a removable solid.
After trying it yourself, choose your next review. This is your self-assessment.
Your review choice appears on Today. Sign in to sync it across devices.
Make it stick
Explain titration, the indicator-free repeat and crystallisation as three distinct stages. Compare the method with an excess-insoluble-base preparation.
If you needed the solution, close it and solve the problem again from a blank page. A correct answer is stronger when you can explain the reason for each step.