3.1.3Unit 3. Practical Skills in Chemistry I
Determining Concentrations — IAL Edexcel Linear (Chemistry) Revision Notes & Topical Questions
5 min read 6 parts Lesson 5 of 10 Unit quiz
Section 1 of 6The Apparatus
Accurate titrations depend on precise glassware.
Burette
- A burette delivers variable volumes; it is marked every 0.10 cm³ and read to ±0.05 cm³.
Pipette
- A volumetric pipette delivers a fixed volume, such as 25.0 cm³, into the conical flask with high accuracy.
Indicator
- A few drops of indicator show the end point, where the solutions have reacted exactly in the equation ratio.
White tile
- A white tile placed under the flask makes the colour change at the end point much easier to see.
Section 2 of 6Performing the Titration
A trial run first shows roughly where the end point lies.
1
Rinse the burette with the solution it will contain, fill it, and record the initial reading to two decimal places.
2
Transfer the measured volume of the other solution to the flask using the pipette, then add a few drops of indicator.
3
Add solution from the burette while swirling the flask, quickly at first during the trial run.
4
Near the end point add the solution drop by drop until one drop changes the indicator colour permanently.
5
Record the final reading, repeat until two or more titres agree to within 0.10 cm³, and average only those concordant titres.
Section 3 of 6Recording and Uncertainty
Recording and Uncertainty
Readings
- Record initial and final burette readings to two decimal places, usually ending in 0 or 5.
Titre
- The titre is the final reading minus the initial reading, the volume of solution delivered from the burette.
Uncertainty
- Each reading has ±0.05 cm³, so the titre has an uncertainty of ±0.10 cm³ because two readings are made.
Percentage
- Percentage uncertainty = (absolute uncertainty ÷ measured value) × 100, which falls when larger titres are used.
Section 4 of 6Processing Results
Average
- Discard the trial and any non-concordant results, then calculate the mean of the concordant titres.
Moles
- Work out the moles of the solution of known concentration from n = c × V, with V in dm³.
Ratio
- Use the balanced equation to convert these moles to moles of the other reactant.
Concentration
- Divide the moles of the unknown by its volume in dm³ to find the concentration in mol dm⁻³.
Section 5 of 6Worked Example
Question
- 25.0 cm³ of sodium hydroxide solution is neutralised by a mean titre of 21.35 cm³ of 0.100 mol dm⁻³ hydrochloric acid.
- NaOH + HCl → NaCl + H₂O.
- Calculate the concentration of the sodium hydroxide and the percentage uncertainty in the titre.
- n(HCl) = 0.100 × 21.35 ÷ 1000 = 2.135 × 10⁻³ mol.
- The mole ratio is 1 : 1, so n(NaOH) = 2.135 × 10⁻³ mol.
- c(NaOH) = 2.135 × 10⁻³ ÷ 0.0250 = 0.0854 mol dm⁻³.
- Percentage uncertainty in the titre = (0.10 ÷ 21.35) × 100 = 0.47 %.
Answer
- [NaOH] = 0.0854 mol dm⁻³;
- percentage uncertainty in the titre = 0.47 %
Section 6 of 6Exam Checklist & Tips
3.1.3 – Key Points for the Exam
- Titre The volume of solution delivered from the burette to reach the end point.
- Concordant Titres that agree to within 0.10 cm³ of each other.
- End point The point where the indicator changes colour, showing that the reaction is complete.
- Standard solution A solution whose concentration is known accurately.
Common Mistakes
- Averaging all titres including the trial run and non-concordant values, which spoils the mean.
- Quoting the titre uncertainty as ±0.05 cm³ when two burette readings give ±0.10 cm³.
- Using the volume in cm³ in c = n ÷ V without converting to dm³.
- Ignoring the mole ratio in the balanced equation when the acid and alkali do not react 1 : 1.
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