Acid-base Titrations (OCR A Level Chemistry A): Revision Note
Acid-base Titrations
Volumetric Analysis
Volumetric analysis is a process that uses the volume and concentration of one chemical reactant (standard solution) to determine the concentration of another unknown solution
The technique most commonly used is a titration
The volumes are measured using two precise pieces of equipment, a volumetric or graduated pipette and a burette
Before the titration can be done, the standard solution must be prepared
Specific apparatus must be used both when preparing the standard solution and when completing the titration, to ensure that volumes are measured precisely
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Some key pieces of apparatus used to prepare a volumetric solution and perform a simple titration
Beaker
Burette
Volumetric Pipette
Conical Flask
Volumetric Flask
Making a Standard Solution
Chemists routinely prepare solutions needed for analysis, whose concentrations are known precisely
These solutions are termed volumetric solutions or standard solutions
They are made as accurately and precisely as possible using three decimal place balances and volumetric flasks to reduce the impact of measurement uncertainties
The steps are:
![Preparing a standard solution (1), downloadable IB Chemistry revision notes](https://cdn.savemyexams.com/cdn-cgi/image/f=auto,width=3840/https://cdn.savemyexams.com/uploads/2021/04/1.2.7-Preparing-a-standard-solution-1.png)
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Volumes & concentrations of solutions
The concentration of a solution is the amount of solute dissolved in a solvent to make 1 dm3 of solution
The solute is the substance that dissolves in a solvent to form a solution
The solvent is often water
A concentrated solution is a solution that has a high concentration of solute
A dilute solution is a solution with a low concentration of solute
Concentration is usually expressed in one of three ways:
moles per unit volume
mass per unit volume
parts per million
Performing the Titration
The key piece of equipment used in the titration is the burette
Burettes are usually marked to a precision of 0.10 cm3
Since they are analogue instruments, the uncertainty is recorded to half the smallest marking, in other words to ±0.05 cm3
The end point or equivalence point occurs when the two solutions have reacted completely and is shown with the use of an indicator
![Titration, downloadable IB Chemistry revision notes](https://cdn.savemyexams.com/cdn-cgi/image/f=auto,width=3840/https://cdn.savemyexams.com/uploads/2021/05/1.2.9-Titration.png)
The steps in a titration
A white tile is placed under the conical flask while the titration is performed, to make it easier to see the colour change
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The steps in a titration
The steps in a titration are:
Measuring a known volume (usually 20 or 25 cm3) of one of the solutions with a volumetric pipette and placing it into a conical flask
The other solution is placed in the burette
To start with, the burette will usually be filled to 0.00 cm3
A few drops of the indicator are added to the solution in the conical flask
The tap on the burette is carefully opened and the solution added, portion by portion, to the conical flask until the indicator starts to change colour
As you start getting near to the end point, the flow of the burette should be slowed right down so that the solution is added dropwise
You should be able to close the tap on the burette after one drop has caused the colour change
Multiple runs are carried out until concordant results are obtained
Concordant results are within 0.1 cm3 of each other
Recording and processing titration results
Both the initial and final burette readings should be recorded and shown to a precision of ±0.05 cm3, the same as the uncertainty
![Titration results, downloadable IB Chemistry revision notes](https://cdn.savemyexams.com/cdn-cgi/image/f=auto,width=3840/https://cdn.savemyexams.com/uploads/2021/05/1.2.9-Titration-results.png)
A typical layout and set of titration results
The volume delivered (titre) is calculated and recorded to an uncertainty of ±0.10 cm3
The uncertainty is doubled, because two burette readings are made to obtain the titre (V final – V initial), following the rules for propagation of uncertainties
Concordant results are then averaged, and non-concordant results are discarded
The appropriate calculations are then done
Percentage Uncertainties
Percentage uncertainties are a way to compare the significance of an absolute uncertainty on a measurement
This is not to be confused with percentage error, which is a comparison of a result to a literature value
The formula for calculating percentage uncertainty is as follows:
![](https://cdn.savemyexams.com/cdn-cgi/image/f=auto,width=3840/https://cdn.savemyexams.com/uploads/2021/06/percentage-university.png)
Adding or subtracting measurements
When you are adding or subtracting two measurements then you add together the absolute measurement uncertainties
For example,
Using a balance to measure the initial and final mass of a container
Using a thermometer for the measurement of the temperature at the start and the end
Using a burette to find the initial reading and final reading
In all these example you have to read the instrument twice to obtain the quantity
If each you time you read the instrument the measurement is ‘out’ by the stated uncertainty, then your final quantity is potentially ‘out’ by twice the uncertainty
Acid-base Titration Calculations
Volumes & concentrations of solutions
The concentration of a solution is the amount of solute dissolved in a solvent to make 1 dm3 of solution
The solute is the substance that dissolves in a solvent to form a solution
The solvent is often water
A concentrated solution is a solution that has a high concentration of solute
A dilute solution is a solution with a low concentration of solute
When carrying out calculations involve concentrations in mol dm-3 the following points need to be considered:
Change mass in grams to moles
Change cm3 to dm3
To calculate the mass of a substance present in solution of known concentration and volume:
Rearrange the concentration equation
number of moles (mol) = concentration (mol dm-3) x volume (dm3)
Multiply the moles of solute by its molar mass
mass of solute (g) = number of moles (mol) x molar mass (g mol-1)
Worked Example
Neutralisation calculation 25.0 cm3 of 0.050 dm-3 sodium carbonate was completely neutralised by 20.00 cm3 of dilute hydrochloric acid.
Calculate the concentration in mol dm-3 of the hydrochloric acid.
Answer
Step 1: Write the balanced symbol equation
Na2CO3 + 2HCl → 2NaCl + H2O + CO2
Step 2: Calculate the amount, in moles, of sodium carbonate reacted by rearranging the equation for amount of substance (mol) and dividing the volume by 1000 to convert cm3 to dm3
amount (Na2CO3) = 0.025 dm3 x 0.050 mol dm-3 = 0.00125 mol
Step 3: Calculate the moles of hydrochloric acid required using the reaction’s stoichiometry
1 mol of Na2CO3 reacts with 2 mol of HCl, so the molar ratio is 1 : 2
Therefore 0.00125 moles of Na2CO3 react with 0.00250 moles of HCl
Step 4: Calculate the concentration, in mol dm-3, of hydrochloric acid
![Mole Calculations Worked example Neutralisation calculation equation 1](https://cdn.savemyexams.com/cdn-cgi/image/f=auto,width=3840/https://cdn.savemyexams.com/uploads/2020/11/9.-Mole-Calculations-Worked-example-Neutralisation-calculation-equation-1.png)
![Mole Calculations Worked example Neutralisation calculation equation 2](https://cdn.savemyexams.com/cdn-cgi/image/f=auto,width=3840/https://cdn.savemyexams.com/uploads/2020/11/9.-Mole-Calculations-Worked-example-Neutralisation-calculation-equation-2.png)
concentration (HCl) (mol dm-3) = 0.125 mol dm-3
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