Search

Chemistry Clarity · chemistryclarity.com

Titration Calculations

How do you turn titration results into a concentration, a mass or a purity?

IntermediateAcids & BasesLast reviewed 6 October 2026
Download lesson (PDF)

What is it?

In a titration, a solution of known concentration (the standard solution or titrant, in the burette) is added to a measured volume of an unknown solution (the analyte, delivered by pipette as an aliquot) until the reaction is exactly complete: the equivalence point, shown by an indicator colour change (the end point).

The volume delivered from the burette is the titre. A rough titre is followed by accurate repeats until at least two or three concordant titres agree closely (usually within 0.10 mL). Their mean is used in the calculation, and the standard deviation shows how precise the titrations were.

Key idea

Every titration calculation follows the same path:

known solution (volume × concentration) → moles → mole ratio from the balanced equation → moles of unknown → concentration, mass or purity.

Why does it matter?

  • Quality control. Titrations check the acidity of vinegar and wine, the vitamin C in juice and the purity of medicines.
  • Precision chemistry. With a burette and pipette, results to 3 or 4 significant figures are routine.
  • Exams and lab work. Titration calculations appear in almost every chemistry course and practical assessment.

How does it work?

1. Choosing the titres

Leave out the rough titre and any titre that does not agree with the others. Average the concordant ones, and calculate the standard deviation to report precision.

2. The basic calculation

n=c×Vc=nVn = c \times V \qquad c = \frac{n}{V}

with VV in litres. Always write the balanced equation: the mole ratio is often not 1 : 1. For example:

HX2SOX4+2 NaOH→NaX2SOX4+2 HX2O\ce{H2SO4 + 2NaOH -> Na2SO4 + 2H2O}

3. Dilutions and aliquots

If the original sample was diluted before titrating, scale the moles back up: a 25.00 mL aliquot from a 100.0 mL flask contains one quarter of the moles in the flask.

4. Back titration

When a substance doesn’t react quickly or doesn’t dissolve (such as limestone, CaCOX3\ce{CaCO3}), add a known excess of acid, let it react completely, then titrate the leftover acid with a base. The acid that reacted = acid added − acid left over.

Think of it like this

A back titration is like finding out how much a friend ate from a full bag of sweets. You can’t count what they ate directly, but you know how many were in the bag and you can count what is left. Eaten = started with − left over.

More precisely

The end point (indicator change) and the equivalence point (exact stoichiometric amount) are not quite the same; a well-chosen indicator makes the difference smaller than the burette’s reading uncertainty. A titre is the difference of two burette readings, each uncertain by about ±0.05 mL, so the titre is uncertain by about ±0.10 mL; larger titres give a smaller percent uncertainty. Standard solutions are made from primary standards, very pure, stable solids such as sodium carbonate or potassium hydrogen phthalate.

Visualise it

Flowchart. Volume and concentration of the standard solution give moles of titrant. The mole ratio from the balanced equation gives moles of analyte. Dividing by the aliquot volume gives the concentration; multiplying by molar mass gives the mass; comparing with the sample mass gives the purity.
The calculation path for any titration.

Worked example

Worked example: Concentration from concordant titres

Question: 25.00 mL of NaOH is titrated with 0.1000 mol/L HCl. Titres: rough 24.10 mL, then 23.45 mL, 23.50 mL and 23.40 mL. Find the concentration of the NaOH.

  1. Mean of the three accurate titres:

    Vˉ=13(23.45+23.50+23.40) mL=23.45 mL\small\begin{aligned} &\bar{V} = \tfrac{1}{3}(23.45 + 23.50 \\[4pt] &\qquad + 23.40)\ \text{mL} \\[4pt] &= 23.45\ \text{mL} \end{aligned}

    The standard deviation is 0.05 mL: the titres are precise.

  2. Moles of HCl:

    n=0.1000 molL×0.02345 L=2.345×10−3 mol\small\begin{aligned} &n = 0.1000\ \tfrac{\text{mol}}{\text{L}} \\[4pt] &\quad \times 0.02345\ \text{L} \\[4pt] &= 2.345 \times 10^{-3}\ \text{mol} \end{aligned}

  3. The mole ratio is 1 : 1:

    HCl+NaOH→NaCl+HX2O\ce{HCl + NaOH -> NaCl + H2O}

    so n(NaOH)=2.345×10−3n(\ce{NaOH}) = 2.345 \times 10^{-3} mol.

  4. Concentration:

    c=2.345×10−3 mol0.02500 L=0.09380 mol/L\small\begin{aligned} &c \\[4pt] &= \frac{2.345 \times 10^{-3}\ \text{mol}}{0.02500\ \text{L}} \\[4pt] &= 0.09380\ \text{mol/L} \end{aligned}

Worked example: A diprotic acid

Question: 20.00 mL of sulfuric acid needs 24.80 mL of 0.1500 mol/L NaOH. Find the concentration of the acid.

  1. n(NaOH)=0.1500 mol/L×0.02480 L=3.720×10−3 moln(\ce{NaOH}) = 0.1500\ \text{mol/L} \times 0.02480\ \text{L} = 3.720 \times 10^{-3}\ \text{mol}

  2. The mole ratio is 1 : 2:

    HX2SOX4+2 NaOH→NaX2SOX4+2 HX2O\ce{H2SO4 + 2NaOH -> Na2SO4 + 2H2O}

    so n(HX2SOX4)=12×3.720×10−3 mol=1.860×10−3 moln(\ce{H2SO4}) = \tfrac{1}{2} \times 3.720 \times 10^{-3}\ \text{mol} = 1.860 \times 10^{-3}\ \text{mol}

  3. Concentration:

    c=1.860×10−3 mol0.02000 L=0.09300 mol/L\small\begin{aligned} &c \\[4pt] &= \frac{1.860 \times 10^{-3}\ \text{mol}}{0.02000\ \text{L}} \\[4pt] &= 0.09300\ \text{mol/L} \end{aligned}

Worked example: Dilution and an aliquot: vinegar

Question: 10.00 mL of vinegar is diluted to 100.0 mL. A 25.00 mL aliquot needs 17.60 mL of 0.1000 mol/L NaOH. Find the concentration of ethanoic acid in the original vinegar, and the mass of acid in 100 mL of vinegar.

  1. Moles in the aliquot (1 : 1 reaction):

    n=0.1000 molL×0.01760 L=1.760×10−3 mol\small\begin{aligned} &n = 0.1000\ \tfrac{\text{mol}}{\text{L}} \\[4pt] &\quad \times 0.01760\ \text{L} \\[4pt] &= 1.760 \times 10^{-3}\ \text{mol} \end{aligned}

  2. The flask holds 100.025.00=4\frac{100.0}{25.00} = 4 aliquots: 4×1.760×10−3 mol=7.040×10−3 mol4 \times 1.760 \times 10^{-3}\ \text{mol} = 7.040 \times 10^{-3}\ \text{mol}, all from the 10.00 mL of vinegar.

  3. Concentration:

    c=7.040×10−3 mol0.01000 L=0.7040 mol/L\small\begin{aligned} &c \\[4pt] &= \frac{7.040 \times 10^{-3}\ \text{mol}}{0.01000\ \text{L}} \\[4pt] &= 0.7040\ \text{mol/L} \end{aligned}

  4. Mass in 100 mL (0.1000 L): 0.7040 mol/L×0.1000 L×60.05 g/mol=4.23 g0.7040\ \text{mol/L} \times 0.1000\ \text{L} \times 60.05\ \text{g/mol} = 4.23\ \text{g}, so the vinegar is about 4.2 % acid.

Worked example: A back titration: purity of limestone

Question: 0.500 g of limestone is added to 50.00 mL of 0.2000 mol/L HCl (an excess). The leftover HCl needs 12.40 mL of 0.1000 mol/L NaOH. Find the percent of CaCOX3\ce{CaCO3} in the limestone.

  1. HCl added: 0.2000 mol/L×0.05000 L=1.000×10−2 mol0.2000\ \text{mol/L} \times 0.05000\ \text{L} = 1.000 \times 10^{-2}\ \text{mol}

  2. HCl left over (= NaOH used): 0.1000 mol/L×0.01240 L=1.240×10−3 mol0.1000\ \text{mol/L} \times 0.01240\ \text{L} = 1.240 \times 10^{-3}\ \text{mol}

  3. HCl that reacted with the limestone:

    1.000×10−2 mol−1.240×10−3 mol=8.760×10−3 mol\small\begin{aligned} &1.000 \times 10^{-2}\ \text{mol} - 1.240 \\[4pt] &\quad \times 10^{-3}\ \text{mol} \\[4pt] &= 8.760 \times 10^{-3}\ \text{mol} \end{aligned}

  4. The mole ratio is 1 : 2:

    CaCOX3+2 HCl→CaClX2+HX2O+COX2\ce{CaCO3 + 2HCl -> CaCl2 + H2O + CO2}

    so n(CaCOX3)=12×8.760×10−3 mol=4.380×10−3 moln(\ce{CaCO3}) = \tfrac{1}{2} \times 8.760 \times 10^{-3}\ \text{mol} = 4.380 \times 10^{-3}\ \text{mol}

  5. Mass and purity:

    m=4.380×10−3 mol×100.09 gmol=0.4384 gpurity=0.4384 g0.500 g×100%=87.7%\small\begin{aligned} &m = 4.380 \times 10^{-3}\ \text{mol} \\[4pt] &\quad \times 100.09\ \tfrac{\text{g}}{\text{mol}} \\[4pt] &= 0.4384\ \text{g} \\[4pt] &\text{purity} = \frac{0.4384\ \text{g}}{0.500\ \text{g}} \\[4pt] &\quad \times 100\% = 87.7\% \end{aligned}

Common mistake

Common mistake: Including the rough titre

The rough titre is only a guide to where the end point is. Average only the accurate, concordant titres.

Common mistake: Forgetting the mole ratio

Sulfuric acid is diprotic: 1 mol HX2SOX4\ce{H2SO4} reacts with 2 mol NaOH. Using 1 : 1 doubles the answer.

Common mistake: Leaving volumes in mL

n=c×Vn = c \times V needs VV in litres when cc is in mol/L: 23.45 mL = 0.02345 L.

Notation note

  • Record burette readings to two decimal places (e.g. 23.45 mL), and report the mean titre the same way.
  • Concentrations from careful titrations are usually given to 4 significant figures.

Remember this

Remember this

  • Use the mean of concordant titres (within about 0.10 mL); leave out the rough titre.
  • Path: n=cVn = cV (known) → mole ratio → moles of unknown → c=n/Vc = n/V, mass or purity.
  • Scale up for dilutions: moles in flask = moles in aliquot × (flask volume ÷ aliquot volume).
  • Back titration: reacted = added − left over.

Test yourself

Check your understanding before moving on.

Flashcards

10 cards

  1. Question
    What is a titre?
    Answer

    The volume of solution delivered from the burette to reach the end point.

  2. Question
    What are concordant titres?
    Answer

    Accurate titres that agree closely, usually within 0.10 mL.

  3. Question
    Why is the rough titre left out of the mean?
    Answer

    It is only a quick guide to the end point and is usually overshot.

  4. Question
    Give the equation linking moles, concentration and volume.
    Answer

    n = c × V, with V in litres.

  5. Question
    What is the mole ratio of H₂SO₄ to NaOH?
    Answer

    1 : 2 (H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O)

  6. Question
    What is the difference between the end point and the equivalence point?
    Answer

    Equivalence point: exact stoichiometric amount added. End point: where the indicator changes colour.

  7. Question
    A 25.00 mL aliquot is taken from a 250.0 mL flask. How do you find the moles in the flask?
    Answer

    Multiply the moles in the aliquot by 250.0 ÷ 25.00 = 10.

  8. Question
    What is a back titration?
    Answer

    A known excess of reagent is added; the leftover is titrated to find how much reacted.

  9. Question
    In a back titration, how do you find the moles that reacted?
    Answer

    Moles added − moles left over (found by titration).

  10. Question
    What is a primary standard?
    Answer

    A very pure, stable solid used to make a solution of accurately known concentration (e.g. Na₂CO₃).

Quiz

7 questions

  1. Question 1EasyTitres: rough 24.10 mL, then 23.45, 23.50 and 23.40 mL. What mean titre should be used?
    Show answer

    Answer: 23.45 mL

    Leave out the rough titre and average the three concordant ones: (23.45 + 23.50 + 23.40) mL ÷ 3 = 23.45 mL. 23.61 mL includes the rough titre.

  2. Question 2EasyHow many moles are in 23.45 mL of 0.1000 mol/L HCl?
    Show answer

    Answer: 2.345 × 10⁻³ mol

    n = c × V = 0.1000 mol/L × 0.02345 L = 2.345 × 10⁻³ mol. The volume must be in litres.

  3. Question 3Medium25.00 mL of NaOH needs 23.45 mL of 0.1000 mol/L HCl. What is the NaOH concentration?
    Show answer

    Answer: 0.09380 mol/L

    n(NaOH) = n(HCl) = 2.345 × 10⁻³ mol (1 : 1); c = 2.345 × 10⁻³ mol ÷ 0.02500 L = 0.09380 mol/L.

  4. Question 4Medium20.00 mL of H₂SO₄ needs 24.80 mL of 0.1500 mol/L NaOH. What is the acid concentration?
    Show answer

    Answer: 0.09300 mol/L

    n(NaOH) = 3.720 × 10⁻³ mol; H₂SO₄ : NaOH = 1 : 2, so n(H₂SO₄) = 1.860 × 10⁻³ mol; c = 1.860 × 10⁻³ mol ÷ 0.02000 L = 0.09300 mol/L. 0.1860 forgets the ratio.

  5. Question 5MediumA 25.00 mL aliquot taken from a 100.0 mL flask contains 1.760 × 10⁻³ mol of acid. How many moles were in the flask?
    Show answer

    Answer: 7.040 × 10⁻³ mol

    The flask holds 100.0 ÷ 25.00 = 4 aliquots: 4 × 1.760 × 10⁻³ mol = 7.040 × 10⁻³ mol.

  6. Question 6HardIn a back titration, 1.000 × 10⁻² mol HCl is added and 1.240 × 10⁻³ mol is left over. How much HCl reacted with the sample?
    Show answer

    Answer: 8.760 × 10⁻³ mol

    Reacted = added − left over = 1.000 × 10⁻² mol − 1.240 × 10⁻³ mol = 8.760 × 10⁻³ mol. 4.380 × 10⁻³ mol is the CaCO₃ it reacts with (1 : 2).

  7. Question 7MediumWhy is a back titration used for limestone (CaCO₃)?
    Show answer

    Answer: CaCO₃ is insoluble and reacts slowly, so it can't be titrated directly

    An excess of acid is added so the solid reacts completely; the leftover acid is then titrated.

Notes and downloads

  • Worksheet

    Titration Calculations Worksheet

    8 questions on concordant titres, standardisation, diprotic acids, dilutions and aliquots, back titrations and uncertainty. Answer key included.

    IntermediateFree

References

  1. Brown, T. L.; LeMay, H. E., Jr.; Bursten, B. E.; Murphy, C. J.; Woodward, P. M.; Stoltzfus, M. W. Chemistry: The Central Science, 15th ed.; Pearson, 2022.

Practise this topic with flashcards and a quiz at chemistryclarity.com/chemistry/titration-calculations/

Spotted a mistake? Let us know and we'll fix it.