Titration calculations become easier when you preserve the sequence: calculate the delivered volume, choose a justified mean, find the known amount in moles, apply the balanced equation and divide by the other solution’s volume. Writing the substance beside every mole value prevents an otherwise correct calculation from changing the wrong quantity.
All numerical readings and scenarios below are invented classroom calculation data. They are not observed practical results and must not be submitted as a record of an experiment. This is a theoretical revision lesson, not a procedure for handling acids, alkalis or laboratory equipment.
The quantities and assumptions behind the calculation
A titre is the volume delivered during the titration. For the supplied burette tables here, titre = final reading − initial reading. A final reading alone is not necessarily the delivered volume. Concentration in mol dm−3 means the amount of solute in each cubic decimetre of solution.
The basic relationship is n = cV, where n is in mol, c in mol dm−3 and V in dm³. Since 1 dm³ = 1,000 cm³, divide a volume in cm³ by 1,000 before using this version of the equation. For example, 25.00 cm³ = 0.02500 dm³.
These exercises assume complete reaction in the stated mole ratio and an endpoint that adequately represents that reaction’s equivalence point. The distinction between observed endpoint and stoichiometric equivalence is described in OpenStax’s quantitative chemical analysis chapter. Real measurement quality cannot be established just by obtaining a plausible answer.
Worked example 1: readings, a mean and a 1:1 reaction
A theoretical question gives the following acid-volume readings. It explicitly labels the rough result and asks for the mean of the remaining titres if their maximum difference is no more than 0.10 cm³. This tolerance is an assumption of this exercise, not a universal examination rule.
| Run | Initial reading | Final reading | Calculated titre |
|---|---|---|---|
| Rough | 0.00 | 24.90 | 24.90 |
| 1 | 0.20 | 24.50 | 24.30 |
| 2 | 0.10 | 24.50 | 24.40 |
| 3 | 1.10 | 25.45 | 24.35 |
Question: Find the mean titre. If the acid is 0.100 mol dm−3 HCl and reacts with 25.00 cm³ of NaOH solution, find the NaOH concentration.
Solution: Run 3 illustrates the subtraction: 25.45 − 1.10 = 24.35 cm³. The three non-rough titres span 24.40 − 24.30 = 0.10 cm³, satisfying the stated inclusive limit. Their mean is (24.30 + 24.40 + 24.35) ÷ 3 = 24.35 cm³. Exclude the labelled rough value as instructed.
The balanced equation is HCl + NaOH → NaCl + H2O. Acid volume = 24.35/1,000 = 0.02435 dm³. Amount of HCl = 0.100 × 0.02435 = 0.002435 mol. The 1:1 ratio means the NaOH amount is also 0.002435 mol.
Divide by the NaOH volume, not the acid titre: c(NaOH) = 0.002435 ÷ 0.02500 = 0.0974 mol dm−3, to three significant figures. Keep the unrounded mole value until this final step.
The answer is slightly below 0.100 mol dm−3. That is sensible: the less concentrated base occupies a slightly larger volume than the acid in a 1:1 reaction. Using 24.35 dm³ would introduce a thousand-fold error, even though the subtraction and mole ratio were correct.
Worked example 2: the equation changes the mole ratio
A separate theoretical question states that 18.60 cm³ of 0.125 mol dm−3 NaOH completely neutralises 20.00 cm³ of sulfuric acid solution. Calculate the sulfuric acid concentration.
The complete-neutralisation equation is H2SO4 + 2NaOH → Na2SO4 + 2H2O. The ratio H2SO4:NaOH is 1:2. Check the coefficients before touching the calculator.
NaOH amount = 0.125 × (18.60/1,000) = 0.002325 mol. Therefore sulfuric acid amount = 0.002325 ÷ 2 = 0.0011625 mol. Acid volume = 20.00/1,000 = 0.02000 dm³. Acid concentration = 0.0011625 ÷ 0.02000 = 0.058125 mol dm−3, or 0.0581 mol dm−3 to three significant figures.
Check backwards: 0.058125 × 0.02000 gives 0.0011625 mol acid, which needs twice that amount of NaOH, 0.002325 mol. Dividing by 0.125 gives 0.01860 dm³, the stated base volume. The unrounded value is used for this check so rounding does not obscure the agreement.
Simply setting acid moles equal to base moles would produce twice the correct acid concentration. Equal reacting volumes are not required either. The equation controls the amount ratio; concentrations determine what volumes supply those amounts.
Worked example 3: identify which solution was diluted
A calculation scenario states that 25.00 cm³ of an original HCl solution has a final diluted volume of 250.0 cm³. A 20.00 cm³ portion of that diluted solution requires 16.00 cm³ of 0.0500 mol dm−3 NaOH. Find the original HCl concentration. The stated volumes are theoretical givens, not instructions to perform a dilution.
NaOH amount = 0.0500 × 0.01600 = 0.000800 mol. The 1:1 equation gives the same HCl amount in the 20.00 cm³ portion. Thus diluted HCl concentration = 0.000800 ÷ 0.02000 = 0.0400 mol dm−3.
The dilution factor is final solution volume ÷ original portion volume = 250.0 ÷ 25.00 = 10. Original concentration = 0.0400 × 10 = 0.400 mol dm−3.
The denominator is 25.00 cm³ because that is the original portion in the dilution statement. The 20.00 cm³ value belongs to the later titrated portion and must not replace it. Also, 250.0 cm³ is the final solution volume, not a volume of added water to which 25.00 must be added.
A short audit before accepting an answer
- Check each titre by subtraction, including non-zero initial readings.
- State which results enter the mean and why. Do not discard an inconvenient result without a supplied criterion or a documented reason.
- Balance atoms on both sides before choosing a mole ratio.
- Use the volume of the solution whose concentration you are calculating.
- Label whether a result describes the original solution or a diluted one.
- Retain intermediate digits, then round consistently with the question’s data and instructions.
Concordant titres indicate agreement among repeated results under a stated criterion. Agreement alone does not prove accuracy: a shared systematic error could affect all readings similarly. If no suitable set meets the supplied criterion, say so rather than manufacturing agreement through selective averaging.
Three solved practice tasks
Practice 1: find the delivered volume
Invented initial and final burette readings are 1.35 cm³ and 24.80 cm³. What volume was delivered?
Solution: Titre = 24.80 − 1.35 = 23.45 cm³. Using 24.80 alone overstates the delivered volume by the initial reading.
Practice 2: calculate the justified mean
A theoretical task supplies titres of 22.10, 22.15 and 22.90 cm³ and instructs you to average the pair agreeing within 0.10 cm³.
Solution: The first two differ by 0.05 cm³; the third differs from them by at least 0.75 cm³. Mean = (22.10 + 22.15)/2 = 22.125 cm³, reported as 22.13 cm³ to two decimal places. Keep 22.125 internally if a later calculation needs it.
Practice 3: calculate an acid volume from a 2:1 ratio
For the theoretical complete reaction Na2CO3 + 2HCl → 2NaCl + H2O + CO2, find the volume of 0.100 mol dm−3 HCl required by 25.00 cm³ of 0.0500 mol dm−3 sodium carbonate.
Solution: Carbonate amount = 0.0500 × 0.02500 = 0.00125 mol. HCl amount = 2 × 0.00125 = 0.00250 mol. Acid volume = 0.00250/0.100 = 0.0250 dm³ = 25.0 cm³.
Frequently asked questions
Can I always use c₁V₁ = c₂V₂?
Only when the relevant mole ratio is 1:1. The safer general route is to calculate moles and apply the balanced equation explicitly.
Why exclude a rough titre?
Follow the supplied instructions. A labelled rough run is usually for locating an approximate endpoint, but you should not invent a rough label for an inconvenient value.
Is the concordance tolerance always 0.10 cm³?
No. Use the criterion given by the task or applicable practical instructions. Here it is explicitly defined for the fictional exercise.
Are endpoint and equivalence point identical?
They describe different things: the detected endpoint and the stoichiometrically equivalent reacting amounts. These calculations assume their difference is negligible for the stated purpose.
Does concentration in g dm⁻³ mean molarity?
No. It is mass concentration. Divide it by molar mass in g mol⁻¹ to obtain molar concentration in mol dm⁻³.
Which volume belongs in c = n/V?
Use the volume containing the calculated moles of that same substance. For an unknown solution in the flask, this is its stated portion volume.
Why keep extra digits during working?
Repeated rounding can shift the final result. Keep calculator precision through the mole-ratio step and round the requested concentration at the end.
Practise the sequence, not a shortcut
For another question, write the substances and their mole ratio before inserting numbers. The site’s existing KCSE Chemistry topic-tests page currently requires login. This lesson’s supplied-data examples remain complete without access to those resources.
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