Titrations (WJEC GCSE Chemistry)

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Determining Relative Concentration

Using a titration to prepare a soluble salt

  • Titrations can be used to prepare a soluble salt e.g. sodium chloride from an acid and alkali
  • The acid and alkali are reacted together in a neutralisation reaction
  • When the acid and alkali are completely neutralised only a salt and water will be present in the solution

hydrochloric acid + sodium hydroxide → sodium chloride + water

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

Steps to prepare a soluble salt

7-2-1-titration-and-forming-salt

Diagram showing the apparatus needed to prepare a salt by titration

 Method:

  • Use a pipette to measure a fixed volume of alkali into a conical flask and add a few drops of an indicator 
  • Add the acid into the burette and note the starting volume
  • Add the acid very slowly from the burette to the conical flask until the indicator changes to the appropriate colour
  • Note and record the final volume of acid in the burette and calculate the volume of acid added (final volume of acid - initial volume of acid)
  • Add this same volume of acid into the same volume of alkali without the indicator
  • Heat the resulting solution in an evaporating basin to partially evaporate, leaving a saturated solution (crystals just forming on the sides of the basin or on a glass rod dipped in and then removed)
  • Leave to crystallise, decant excess solution and allow crystals to dry

Examiner Tip

Make sure you learn the steps for how to prepare a soluble salt using:

  • An acid and insoluble base
  • A titration 

For the second method, many students forget the step of not adding an indicator so make sure you add this in.

Determining the relative concentration of acids and alkalis

  • If an alkali of known concentration is used in a titration then the relative concentration of the acid can be determined and vice versa
  • This is done by looking at the volumes of each reactant
  • If the volume of acid used is the greatest then the acid is less concentrated
    • This is because more acid is required to neutralise the alkali
  • If the volume of alkali used is the greatest then the alkali is less concentrated for the same reasons
  • This is true of neutralisation reactions where the ratio of acid to alkali is 1:1
  • For example, if 30.00 cm3 of 0.2 mol dm3 acid is neutralised by 15.00 cm3 of an unknown alkali, the concentration of the alkali must be twice the concentration of the acid
    • This is because only half the volume of the alkali is required to neutralise 30.00 cm3 of acid 
  • The method to carry out a titration can be found here

Determining Actual Concentration

Higher Tier

  • Higher tier students need to be able to carry out titration or neutralisation calculations
  • This is used to determine the actual concentration of the acid or alkali used in the titration 
  • You should be able to work out both the concentration and volume

Worked example

Calculating concentration

25.00 cmof 0.15 mol dm–3 barium hydroxide, Ba(OH)2, was required to neutralise 12.80 cm3 of nitric acid, HNO3 , during a titration. Calculate the concentration of HNO3 that was used. Give your answer to 2 decimal places.

Ba(OH)2 (aq) + 2HNO3 (aq) → Ba(NO3)2 (aq) + 2H2O (l)

Answer

Step 1: Calculate the number of moles of barium hydroxide 

  • Moles of barium hydroxide = concentration x volume (dm3) = 0.15 x 0.025 = 3.75 x 10–3 mols

Step 2: Using the equation, calculate the number of moles of nitric acid 

  • Moles of nitric acid = 3.75 x 10–3 x 2 = 7.5 x 10-3
  • The number of moles must be multiplied by 2 due to the 1:2 ratio

Step 3: Calculate the concentration of nitric acid

  • Concentration of nitric acid = fraction numerator 7.5 space cross times 10 to the power of negative 3 end exponent over denominator 0.0128 end fraction= 0.59 mol dm–3 to 2 dp

Remember to convert cm3 to dm3 by dividing by 1000 

Worked example

Calculating the volume

Calculate the volume of 0.50 mol dm–3 nitric acid, HNO3, required to neutralise 25.00 cm3 of 0.80 mol dm–3 potassium hydroxide, KOH. Give your answer in cm3.

KOH (aq) + HNO3 (aq) → KNO3 (aq) + H2O (l)

Answer

Step 1: Calculate the number of moles of potassium hydroxide

  • Moles of potassium hydroxide = concentration x volume (dm3) = 0.80 x 0.025 = 0.02 mols

Step 2: Using the equation, calculate the number of moles of nitric acid 

  • Moles of nitric acid = 0.02 mols
  • The ratio is 1:1 so the number of moles of nitric acid is the same

Step 3: Calculate the volume of nitric acid in cm

  • Volume of nitric acid =moles over concentrationfraction numerator 0.02 over denominator 0.50 end fraction = 0.040 dm3
  • Volume of nitric acid = 40 cm3 

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Philippa

Author: Philippa

Expertise: Chemistry

Philippa has worked as a GCSE and A level chemistry teacher and tutor for over thirteen years. She studied chemistry and sport science at Loughborough University graduating in 2007 having also completed her PGCE in science. Throughout her time as a teacher she was incharge of a boarding house for five years and coached many teams in a variety of sports. When not producing resources with the chemistry team, Philippa enjoys being active outside with her young family and is a very keen gardener.