O LevelChemistryCambridge (CIE)Revision Notes3. Stoichiometry3.3 The Mole & the Avogadro ConstantTitration Calculations

Titration Calculations (Cambridge (CIE) O Level Chemistry): Revision Note

Exam code: 5070

Caroline Carroll

Written by: Caroline Carroll

Reviewed by: Lucy Kirkham

Updated on

Titration Calculations

  • Titrations are a method of analysing the concentration of solutions

  • Acid-base titrations are one of the most important kinds of titrations

  • They can determine exactly how much alkali is needed to neutralise a quantity of acid – and vice versa

  • You may be asked to calculate the moles present in a given amount, the concentration or volume required to neutralise an acid or a base

  • Once a titration is completed and the average titre has been calculated, you can now proceed to calculate the unknown variable using the formula triangle as shown below

Formula triangle showing moles above concentration and volume in dm³: moles = concentration × volume, concentration = moles ÷ volume, and volume = moles ÷ concentration.

Formula triangle showing the relationship between concentration, number of moles and volume of liquid

Worked Example

A solution of 25.0 cm3 of hydrochloric acid was titrated against a solution of 0.100 mol/dm3 NaOH and 12.1 cm3 were required for complete reaction. Determine the concentration of the acid.

Answer

Step 1: Write the equation for the reaction:

HCl (aq) + NaOH (aq) → NaCl (aq) + H2O (l)

Step 2: Calculate the number of moles of the NaOH

Moles = (volume ÷ 1000) x concentration

Moles of NaOH = 0.0121 dm3 x 0.100 mol/dm3 = 1.21 x 10-3 mol

Step 3: Deduce the number of moles of the acid

  • Since the acid reacts in a 1:1 ratio with the alkali,  the number of moles of HCl is also 1.21 x 10-3 mol

  • This is present in 25.0 cm3 of the solution (25.0 cm3 = 0.025 dm3)

Step 4: Find the concentration of the acid

Concentration = moles ÷ volume (dm3)

Concentration of HCl =1.21 x 10-3 mol  ÷ 0.025 dm3

Concentration of HCl= 0.0484 mol/dm3

Worked Example

25.0 cm3 of 0.100 mol/dm3 sodium hydroxide was completely neutralised by 20.0 cm3 of dilute sulfuric acid. Determine the concentration of the acid.

Answer

Step 1: Write the equation for the reaction:

H2SO4 (aq) + 2NaOH (aq) → Na2SO4 (aq) + 2H2O (l)

Step 2: Calculate the number of moles of the NaOH

Moles = (volume ÷ 1000) x concentration

Moles of NaOH = 0.0250 dm3 x 0.100 mol/dm3 = 2.50 x 10-3 mol

Step 3: Deduce the number of moles of the acid

  • The equation shows that 2 moles of NaOH react with 1 mole of H2SO4

  • The number of moles of H2SO4 is therefore half the number of moles of NaOH

Moles of H2SO4 = 2.50 x 10-3 ÷ 2 = 1.25 x 10-3 mol

  • This is present in 20.0 cm3 of the solution (20.0 cm3 = 0.0200 dm3)

Step 4: Find the concentration of the acid

Concentration = moles ÷ volume (dm3)

Concentration of H2SO4 = 1.25 x 10-3 mol ÷ 0.0200 dm3

Concentration of H2SO4 = 0.0625 mol/dm3

Examiner Tips and Tricks

Always write the balanced equation first and check the ratio before you divide

A 1:1 reaction lets you carry the number of moles straight across, but with an acid like sulfuric acid that supplies two hydrogen ions, you must halve the moles of alkali to get the moles of acid

Forgetting this step is one of the most common errors in titration calculations

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Caroline Carroll

Author: Caroline Carroll

Expertise: Head of Content Delivery

Caroline graduated from the University of Nottingham with a degree in Chemistry and Molecular Physics. She spent several years working as an Industrial Chemist in the automotive industry before retraining to teach. Caroline has over 12 years of experience teaching GCSE and A-level chemistry and physics. She is passionate about delivering high-quality resources to help students achieve their full potential.

Lucy Kirkham

Reviewer: Lucy Kirkham

Expertise: Content Creator

Lucy has been a passionate Maths teacher for over 12 years, teaching maths across the UK and abroad helping to engage, interest and develop confidence in the subject at all levels.Working as a Head of Department and then Director of Maths, Lucy has advised schools and academy trusts in both Scotland and the East Midlands, where her role was to support and coach teachers to improve Maths teaching for all.