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Titration Calculations Step by Step

Moles & Chemical CalculationsIntermediate6 min read
On this page
  1. The core method
  2. Example 1: 1 : 1 ratio
  3. Example 2: 1 : 2 ratio
  4. Example 3: a diluted sample
  5. Example 4: finding a molar mass
  6. Example 5: percentage purity
  7. Example 6: water of crystallisation
  8. A checklist for every titration question
  9. Practice questions
  10. Common mistakes
  11. Key takeaways

Titration questions have a reputation for being long and fiddly. But underneath the wordy setups, every one of them uses the same three-step core. Learn that core, add two or three extra steps for the harder variants, and you can handle any titration question an exam can throw at you.

If you haven’t done a titration practical yet, acid–base titration explains the method and the equipment.

The core method

Step 1. Moles of the solution you know everything about. You’ll know its concentration and the volume used (usually the titre from the burette).

moles = concentration (mol/dm³) × volume (dm³)

Remember: volume in cm³ ÷ 1000 = volume in dm³.

Step 2. Use the balanced equation to find moles of the other substance. Multiply by the mole ratio from the equation.

Step 3. Find what the question asks for. Usually concentration = moles ÷ volume (dm³). Sometimes it’s a mass, a molar mass or a percentage.

Write the equation first, every time. Most mistakes come from using the wrong ratio.

Example 1: 1 : 1 ratio

25.0 cm³ of sodium hydroxide is neutralised by 21.60 cm³ of 0.150 mol/dm³ hydrochloric acid. Calculate the concentration of the sodium hydroxide.

Equation: HCl + NaOH → NaCl + H₂O

  1. Moles HCl = 0.150 × (21.60 ÷ 1000) = 3.24 × 10⁻³ mol
  2. Ratio 1 : 1, so moles NaOH = 3.24 × 10⁻³ mol
  3. [NaOH] = 3.24 × 10⁻³ ÷ (25.0 ÷ 1000) = 0.130 mol/dm³

Example 2: 1 : 2 ratio

A 25.0 cm³ sample of sulfuric acid needs 28.40 cm³ of 0.200 mol/dm³ sodium hydroxide for neutralisation. Find the concentration of the acid.

Equation: H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O

  1. Moles NaOH = 0.200 × 0.02840 = 5.68 × 10⁻³ mol
  2. Ratio H₂SO₄ : NaOH = 1 : 2, so moles H₂SO₄ = 5.68 × 10⁻³ ÷ 2 = 2.84 × 10⁻³ mol
  3. [H₂SO₄] = 2.84 × 10⁻³ ÷ 0.0250 = 0.114 mol/dm³

The trap: dividing instead of multiplying, or ignoring the ratio altogether. Say it in words: “each sulfuric acid needs two hydroxides, so there are half as many sulfuric acid molecules”.

Example 3: a diluted sample

10.0 cm³ of concentrated cleaning solution containing sodium hydroxide was diluted to 250.0 cm³ in a volumetric flask. A 25.0 cm³ portion of the diluted solution needed 18.75 cm³ of 0.100 mol/dm³ HCl. Find the concentration of NaOH in the original cleaning solution.

  1. Moles HCl = 0.100 × 0.01875 = 1.875 × 10⁻³ mol
  2. Moles NaOH in 25.0 cm³ portion = 1.875 × 10⁻³ mol
  3. Scale up to the whole 250.0 cm³ flask: 1.875 × 10⁻³ × (250.0 ÷ 25.0) = 1.875 × 10⁻² mol
  4. All of that came from the original 10.0 cm³: [NaOH] = 1.875 × 10⁻² ÷ 0.0100 = 1.88 mol/dm³

The scaling step (×10 here) is where students most often go wrong. Keep asking: “which volume did these moles actually come from?”

The shortcut: the dilution factor is 250.0 ÷ 10.0 = 25. The diluted solution’s concentration is 1.875 × 10⁻³ ÷ 0.0250 = 0.0750 mol/dm³, so the original was 0.0750 × 25 = 1.88 mol/dm³. Same answer.

Example 4: finding a molar mass

1.260 g of an unknown solid diprotic acid, H₂X, was dissolved and made up to 250.0 cm³. 25.0 cm³ portions needed 20.00 cm³ of 0.100 mol/dm³ NaOH. Find the molar mass of the acid.

Equation: H₂X + 2NaOH → Na₂X + 2H₂O

  1. Moles NaOH = 0.100 × 0.02000 = 2.000 × 10⁻³ mol
  2. Moles H₂X in 25.0 cm³ = 2.000 × 10⁻³ ÷ 2 = 1.000 × 10⁻³ mol
  3. Moles H₂X in 250.0 cm³ = 1.000 × 10⁻² mol
  4. Molar mass = mass ÷ moles = 1.260 ÷ 1.000 × 10⁻² = 126 g/mol

That matches hydrated ethanedioic (oxalic) acid, H₂C₂O₄·2H₂O, whose molar mass is 126.07 g/mol. You can check any formula’s molar mass with the molar mass calculator.

Example 5: percentage purity

A 0.500 g sample of impure calcium carbonate was dissolved in 50.0 cm³ of 0.250 mol/dm³ HCl (an excess). The leftover acid needed 25.40 cm³ of 0.100 mol/dm³ NaOH. Find the percentage purity of the calcium carbonate.

This is a back titration: you add a known excess of acid, then titrate what’s left. It’s used when the solid doesn’t dissolve easily in water or reacts too slowly for a direct titration.

Equations: CaCO₃ + 2HCl → CaCl₂ + H₂O + CO₂ HCl + NaOH → NaCl + H₂O

  1. Moles HCl added at the start = 0.250 × 0.0500 = 0.0125 mol
  2. Moles HCl left over = moles NaOH = 0.100 × 0.02540 = 2.54 × 10⁻³ mol
  3. Moles HCl that reacted with CaCO₃ = 0.0125 − 0.00254 = 9.96 × 10⁻³ mol
  4. Moles CaCO₃ = 9.96 × 10⁻³ ÷ 2 = 4.98 × 10⁻³ mol
  5. Mass CaCO₃ = 4.98 × 10⁻³ × 100.09 = 0.498 g
  6. Percentage purity = 0.498 ÷ 0.500 × 100 = 99.7%

Always sense-check a purity answer. If you ever get more than 100%, there’s a mistake in the working (often a missed mole ratio) or the data. A result like 99.7% is plausible for a good-quality sample of chalk powder; natural limestone is often 90–98% calcium carbonate.

Example 6: water of crystallisation

2.86 g of hydrated sodium carbonate, Na₂CO₃·xH₂O, was dissolved and made up to 250.0 cm³. 25.0 cm³ portions needed 20.00 cm³ of 0.100 mol/dm³ HCl to reach the second end point (methyl orange). Find x.

Equation: Na₂CO₃ + 2HCl → 2NaCl + H₂O + CO₂

  1. Moles HCl = 0.100 × 0.02000 = 2.000 × 10⁻³ mol
  2. Moles Na₂CO₃ in 25.0 cm³ = 1.000 × 10⁻³ mol
  3. Moles in 250.0 cm³ = 1.000 × 10⁻² mol
  4. Molar mass of hydrate = 2.86 ÷ 0.01000 = 286 g/mol
  5. Molar mass Na₂CO₃ = 105.99 g/mol, so mass of water per mole = 286 − 106 = 180 g
  6. x = 180 ÷ 18.02 = 10

The formula is Na₂CO₃·10H₂O, washing soda. See hydrates and water of crystallisation for more.

A checklist for every titration question

  1. Write the balanced equation.
  2. Convert all volumes from cm³ to dm³.
  3. Calculate moles of the substance with known concentration.
  4. Apply the mole ratio.
  5. Scale for any dilution or aliquot (portion) step.
  6. Answer the actual question: concentration, mass, molar mass, purity or formula.
  7. Give an appropriate number of significant figures (usually 3) and units.
  8. Sense-check: is the answer physically possible?

Practice questions

  1. 25.0 cm³ of potassium hydroxide needs 19.80 cm³ of 0.125 mol/dm³ nitric acid. Find [KOH].
  2. 20.0 cm³ of 0.0500 mol/dm³ Ba(OH)₂ is neutralised by 16.00 cm³ of hydrochloric acid. Find [HCl].
  3. 25.0 cm³ of 0.100 mol/dm³ Na₂CO₃ reacts completely with 0.200 mol/dm³ HCl (to CO₂). What volume of acid is needed?

Answers: (1) 2.475 × 10⁻³ mol ÷ 0.0250 dm³ = 0.0990 mol/dm³. (2) Moles Ba(OH)₂ = 1.00 × 10⁻³, so moles HCl = 2.00 × 10⁻³; [HCl] = 0.125 mol/dm³. (3) Moles Na₂CO₃ = 2.50 × 10⁻³, moles HCl = 5.00 × 10⁻³, volume = 25.0 cm³.

Common mistakes

  • Using the rough titre in the mean.
  • Forgetting to divide by 1000.
  • Ignoring the mole ratio for H₂SO₄, Na₂CO₃ or Ba(OH)₂.
  • Forgetting to scale up from the pipetted portion to the whole flask.
  • Rounding too early. Keep extra figures until the final answer.

Key takeaways

  • Every titration calculation starts with moles = concentration × volume for the fully known solution.
  • Apply the mole ratio from the balanced equation, then find the requested quantity.
  • For diluted samples and portions, track which volume the moles came from.
  • Back titrations use a known excess and subtract what’s left.
  • Check your concentrations with the molarity calculator.

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