Titrations (WJEC GCSE Chemistry)
Revision Note
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
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 Tips and Tricks
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 dm–3 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 cm3 of 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 = = 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 cm3
Volume of nitric acid == = 0.040 dm3
Volume of nitric acid = 40 cm3
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