Section 1 of 5Units and Key Quantities
Every titration calculation uses the same three quantities, and the units must be consistent before any arithmetic starts.
Concentration
- Concentration Amount of substance dissolved per volume, measured in mol/dm³, which is the same as moles per litre.
Volume conversion
- Volume conversion Volumes are measured in cm³ in the lab but must be divided by 1000 to give dm³ in the calculation.
Moles
- Moles Found by multiplying concentration in mol/dm³ by volume in dm³, so the units cancel correctly.
Concordant results
- Concordant results Repeated titres that agree closely, usually within about 0.10 cm³, and are averaged to give the final volume.
Section 2 of 5Calculation Steps
These steps work for any strong acid and strong alkali titration where the equation is 1 : 1 for the acid and alkali.
1
Convert the volume of the solution in the flask from cm³ into dm³ by dividing by 1000.
2
Calculate the moles of the known solution using moles = concentration × volume.
3
Use the mole ratio from the balanced equation to find the moles of the unknown solution.
4
Divide the moles of the unknown by its volume in dm³ to find its concentration.
Section 3 of 5Choosing Reliable Results
The quality of the final answer depends on how carefully the titration is carried out and recorded.
Choosing Reliable Results
Rough titre
- Rough titre A quick first trial used only to find the approximate end point, and it is not included in the mean.
Indicator
- Indicator A substance such as phenolphthalein that changes colour at the end point, showing the solution is neutralised.
Mean titre
- Mean titre The average of the concordant results, which reduces random error in the measured volume.
| Quantity | Known solution | Unknown solution |
|---|---|---|
| Solution | 0.100 mol/dm³ sodium hydroxide | Hydrochloric acid of unknown concentration |
| Volume | 25.0 cm³ in the flask | 20.0 cm³ from the burette |
| Equation | HCl + NaOH → NaCl + H₂O | HCl + NaOH → NaCl + H₂O |
Section 4 of 5Worked Example
Question
- 25.0 cm³ of 0.100 mol/dm³ sodium hydroxide is neutralised by 20.0 cm³ of hydrochloric acid.
- Calculate the concentration of the hydrochloric acid in mol/dm³.
- Volume of NaOH in dm³ = 25.0 ÷ 1000 = 0.0250 dm³.
- Moles of NaOH = 0.100 × 0.0250 = 0.00250 mol.
- Mole ratio HClNaOH from the equation is 1 : 1, so moles of HCl = 0.00250 mol.
- Volume of HCl in dm³ = 20.0 ÷ 1000 = 0.0200 dm³, so concentration = 0.00250 ÷ 0.0200 = 0.125 mol/dm³.
Answer 0.125 mol/dm³ hydrochloric acid.
Section 5 of 5Exam Checklist & Tips
4.2.10 – Key Points for the Exam
- Titration A method of accurately finding the volume of one solution needed to react exactly with another.
- End point The point at which the indicator changes colour, showing the reaction is complete.
- Concordant results Titres that agree closely and can be used to calculate a mean.
- Standard solution A solution whose concentration is known accurately and used in a titration.
- Neutralisation A reaction between an acid and an alkali that produces a salt and water.
Common Mistakes
- Forgetting to convert cm³ into dm³, which makes the calculated concentration 1000 times too large or small.
- Using the rough titre in the mean, when it was only a trial to find the approximate end point.
- Using the wrong mole ratio, which happens when the equation is not balanced before it is read.
- Mixing up concentration and moles, and dividing moles by the volume before converting the volume to dm³.