Titrations (AQA A Level Chemistry): Revision Note
Exam code: 7405
Titrations
Volumetric analysis is a process that uses the volume and concentration of one chemical reactant (a 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
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 stoichiometric point or equivalence point occurs when the two solutions have reacted completely and is shown with the use of an indicator
The steps in a titration are:
Measuring a known volume (usually 20 or 25 cm3) of one of the solutions with a volumetric or graduated pipette and placing it into a conical flask
The other solution is placed in the burette
A few drops of the indicator are added
The tap on the burette is carefully opened and the solution added, portion by portion, to the conical flask until the indicator just changes colour(this is the end point)
Multiple trials are carried out until concordant results are obtained
Related topic links
Titration Calculations
Titration calculations are used to find the concentration of unknown solutions
They can also be used to calculate the pH after a given point during a titration
Worked Example
Example 1: Calculations from titration results
In a titration, 25.00 cm3 of 0.05 mol dm-3 hydrochloric acid was neutralised by 8.50 cm3 of sodium hydroxide solution. Calculate the concentration of the sodium hydroxide solution.
Answer:
Step 1: Find the number of moles of acid
moles of acid = concentration x volume in dm3
moles of acid = 0.05 x 25/1000 = 1.25 x 10-3 mol
Step 2: Deduce the number of moles of alkali
The equation for the reaction shows the mole ratio is 1:1
HCl (aq) + NaOH (aq) → NaCl (aq) + H2O (l)
∴ moles of alkali = 1.25 x 10-3 mol
Step 3: Work out the concentration of the alkali
concentration = moles/volume in dm3
concentration = 1.25 x 10-3/0.0085 = 0.15 mol dm-3
Worked Example
Example 2: Calculating the pH in a strong acid-strong base titration
50.0 cm3 of 0.10 mol dm-3 NaOH is gradually added to 25.0 cm3 of 0.15 mol dm-3 hydrochloric acid. Determine the pH after 45 cm3 of NaOH has been added. (Kw = 1 x 10-14 mol2 dm-6 at 298 K).
Answer:
Step 1: Find the number of moles of acid
moles of acid = concentration x volume in dm3
moles of acid = 0.15 x 25/1000 = 3.75 x 10-3 mol
Step 2: Deduce the number of moles of alkali added
The equation for the reaction shows the mole ratio is 1:1
HCl (aq) + NaOH (aq) → NaCl (aq) + H2O (l)
moles of alkali added = 0.10 x 45/1000 = 4.50 x 10-3 mol
∴ moles of alkali in excess = (4.50 x 10-3- 3.75 x 10-3) = 7.5 x 10-4 mol
Step 3: Work out the concentration of the alkali
concentration = moles/volume in dm3
concentration = 7.5 x 10-4/0.070 = 0.0107 mol dm-3
Step 4: Use Kw to find the concentration of H+
Kw = [H+][OH-]
[H+] = Kw /[OH-] = 1.00 x 10-14/0.0107 = 9.35 x 10-13
Step 5: Find the pH
-log[H+] = -log(9.35 x 10-13)
pH = 12.03
Examiner Tips and Tricks
pH-curve questions are limited to monoprotic acids (one acidic hydrogen, like HCl), but volumetric concentration calculations can involve diprotic acids such as H₂SO₄ — just remember the 1:2 acid-to-base ratio.
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