Solution concentrations and titration calculations
Combine concentration, volume and the balanced reaction ratio without losing the units.
Mapped to the named 0654 section using teaching from the corresponding single-science course. A 0654-specific review of every Core and Supplement objective is still in progress. Written practice is self-checked; it does not demonstrate supervised laboratory competence or award an official Cambridge grade.
Before you begin
- Convert cm³ to dm³ and use n=m/M.
- Balance neutralisation equations and read reaction coefficients.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
Which volume should be used in n=cV with c in mol/dm³?
What you will learn
- Calculate concentration, solute mass and amount with consistent solution-volume units.
- Use titration data and balanced coefficients to find an unknown concentration.
Convert the volume before finding the amount
For amount concentration c in mol/dm³ and final solution volume V in dm³, n=cV. Rearranging gives c=n/V and V=n/c. Convert cm³ to dm³ by dividing by 1000 before substituting. Mass concentration in g/dm³ is mass of dissolved solute divided by final solution volume in dm³; multiply a mol/dm³ concentration by molar mass in g/mol to obtain g/dm³.
For a solution prepared by dissolving a solid, the relevant volume is the final solution volume, not just the amount of solvent poured in. If a sample is diluted without losing solute, its amount is unchanged, so c₁V₁=c₂V₂. This equality relates the same solute before and after dilution; it is not a replacement for reaction coefficients when two substances react.
A titration gives one measured reacting volume
A titre is the volume delivered from a burette, found by subtracting the initial reading from the final reading. Use the instructed selection of reliable titres; if concordant results and an anomalous rough trial are identified, do not average all values blindly. The actual titration procedure and indicator choice belong to the later experimental-techniques and acids lessons; here the focus is interpreting provided data and calculating amounts.
Find the known solution's amount from concentration and reacting volume, apply the balanced mole ratio to obtain the unknown solution's amount, then divide by the unknown solution's measured volume. In H₂SO₄+2NaOH→Na₂SO₄+2H₂O, one mole of sulfuric acid reacts with two moles of sodium hydroxide. Equal volumes or equal printed concentrations do not remove that coefficient ratio. Assume the stated endpoint represents the required reaction stoichiometry.
Sources for this lesson
Pause and explain
0.0030 mol H₂SO₄ fully neutralises how much NaOH in H₂SO₄+2NaOH→Na₂SO₄+2H₂O?
Worked example
25.0 cm³ NaOH is neutralised by 20.0 cm³ of 0.100 mol/dm³ H₂SO₄. Use H₂SO₄+2NaOH→Na₂SO₄+2H₂O to find the NaOH concentration.
Show the worked solution
- Acid volume is 0.0200 dm³, so acid amount is 0.100×0.0200=0.00200 mol.
- NaOH amount is twice this: 0.00400 mol.
- NaOH volume is 0.0250 dm³, so concentration is 0.00400/0.0250=0.160 mol/dm³.
Answer 0.160 mol/dm³ NaOH.
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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.
5.85 g NaCl is dissolved to make a final volume of 500 cm³ solution. Find concentration in g/dm³ and mol/dm³, using Na=23 and Cl=35.5.
Give me a hint
Use final volume 0.500 dm³ and molar mass 58.5 g/mol.
Compare my reasoning
- Mass concentration is 5.85/0.500=11.7 g/dm³.
- Amount is 5.85/58.5=0.100 mol.
- Amount concentration is 0.100/0.500=0.200 mol/dm³; multiplying 0.200 by 58.5 reproduces 11.7 g/dm³.
11.7 g/dm³ and 0.200 mol/dm³.
Look for these in your work
- The final volume is converted correctly.
- Both concentration measures describe the same solution.
25.0 cm³ NaOH reacts with 0.100 mol/dm³ HCl in a 1:1 mole ratio. Recorded titres are rough 23.80 cm³, then 20.00, 20.10 and 20.00 cm³. Use the final three results, as instructed, to find the NaOH concentration to three significant figures.
Give me a hint
Average the selected titres, then calculate acid amount and divide the equal NaOH amount by 0.0250 dm³.
Compare my reasoning
- Mean selected titre is (20.00+20.10+20.00)/3=20.0333… cm³.
- HCl amount is 0.100×20.0333…/1000=0.00200333… mol; the 1:1 reaction gives the same NaOH amount.
- Concentration is 0.00200333…/0.0250=0.0801333… mol/dm³, rounded to 0.0801 mol/dm³.
0.0801 mol/dm³ NaOH.
Look for these in your work
- Only the specified titres enter the mean.
- Unrounded intermediates are retained until the final concentration.
10.0 cm³ of 0.500 mol/dm³ NaOH is diluted to 100 cm³. Then 25.0 cm³ of the diluted solution neutralises 20.0 cm³ H₂SO₄ by the 1:2 acid:NaOH mole ratio. Find both diluted NaOH and acid concentrations.
Give me a hint
Dilution conserves solute amount; neutralisation then uses a different substance's reaction ratio.
Compare my reasoning
- Initial NaOH amount is 0.500×0.0100=0.00500 mol; diluted concentration is 0.00500/0.100=0.0500 mol/dm³.
- The 25.0 cm³ portion contains 0.0500×0.0250=0.00125 mol NaOH, requiring 0.000625 mol H₂SO₄.
- Acid concentration is 0.000625/0.0200=0.03125 mol/dm³, or 0.0313 mol/dm³ to three significant figures.
Diluted NaOH: 0.0500 mol/dm³; H₂SO₄: 0.0313 mol/dm³ (3 significant figures).
Look for these in your work
- Dilution and reaction are treated as separate stages.
- The acid amount is half the reacting NaOH amount.
Common mistakes
- Putting cm³ directly into n=cV when c is in mol/dm³.
- Using a 1:1 mole ratio for every acid–alkali reaction.
- Averaging an explicitly excluded rough or anomalous titre.
- Treating the volume of solvent as the final solution volume.
What can you explain now?
Find the amount in 50.0 cm³ of a 0.200 mol/dm³ solution.
Compare with the explanation
0.0100 mol
50.0 cm³=0.0500 dm³; n=0.200×0.0500=0.0100 mol.
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Make it stick
Write the concentration-to-amount route with its units. Solve a new titration with a 1:2 ratio and explain why a dilution equality cannot replace the balanced equation.
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.