Titration Calculations
This topic covers titration calculations, a method to determine an unknown solution's concentration by reacting it with a solution of known concentration. It's a fundamental quantitative skill combining solution chemistry with stoichiometry.
Part of the ESAT Chemistry syllabus — revision for the Engineering and Science Admissions Test (ESAT), the UAT-UK admissions test for Cambridge, Imperial, Oxford and UCL.
Key points
- Always start with a correctly balanced chemical equation to establish the molar ratio between reactants.
- Calculate the moles of the 'known' substance, for which you have both concentration and volume.
- Use the molar ratio from the balanced equation to determine the number of moles of the 'unknown' substance that reacted.
- Calculate the concentration of the 'unknown' substance using its calculated moles and the volume used in the titration.
- Ensure all volumes are consistently in dm³ for calculations involving concentration in mol dm⁻³.
Formulae
moles = concentration (mol dm-3) × volume (dm3) To find the amount of solute in a solution when concentration and volume are known. This is the first step in most titration calculations.
concentration (mol dm-3) = moles / volume (dm3) To find the concentration of a solution when the amount of solute (moles) and the solution volume are known. This is the final step in most titration calculations.
Definitions
- Concentration
- The amount of a solute dissolved in a given volume of solution, typically expressed in moles per decimetre cubed (mol dm⁻³) or grams per decimetre cubed (g dm⁻³).
- Titration
- A quantitative chemical analysis method where the volume of a solution of known concentration (the titrant) is measured, which is required to react completely with a measured volume of a solution of unknown concentration (the analyte).
- Equivalence Point
- The point in a titration where the amount of titrant added is stoichiometrically equal to the amount of analyte in the sample, meaning the reaction is exactly complete.
Worked example
25.0 cm³ of a sodium hydroxide (NaOH) solution is completely neutralised by 12.5 cm³ of 0.20 mol dm⁻³ sulfuric acid (H₂SO₄). What is the concentration of the sodium hydroxide solution in mol dm⁻³?
- 1
Step 1:
Write the balanced equation.
H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O.
The molar ratio of acid to alkali is 1:2.
- 2
Step 2:
Calculate moles of the known substance, H₂SO₄.
First, convert volume to dm³:
12.5 cm³ = 0.0125 dm³Moles H₂SO₄ = 0.20 mol dm⁻³ × 0.0125 dm³ = 0.0025 mol - 3
Step 3:
Use the molar ratio to find moles of NaOH.
Moles NaOH = 2 × Moles H₂SO₄ = 2 × 0.0025 mol = 0.0050 mol - 4
Step 4:
Calculate the concentration of NaOH.
First, convert volume to dm³:
25.0 cm³ = 0.0250 dm³Concentration NaOH = moles / volume = 0.0050 mol / 0.0250 dm³.
- 5
Step 5:
Simplify the final calculation.
0.0050 / 0.0250 is the same as 50 / 250, which simplifies to 5 / 25 = 1/5 = 0.2.
So the concentration is 0.20 mol dm⁻³.
Answer: 0.20 mol dm⁻³
Common mistakes
- ×Forgetting the molar ratio: The most common error is assuming a 1:1 reaction ratio. Always write the balanced equation first to avoid this 'off by a factor' mistake (e.g., for H₂SO₄ and NaOH, the ratio is 1:2).
- ×Volume unit errors: Forgetting to convert volumes from cm³ to dm³ by dividing by 1000 is a frequent source of error. A quick check is that your final answer should be a sensible concentration, usually between 0.01 and 2 mol dm⁻³.
- ×Inverting the molar ratio: Accidentally multiplying by the ratio instead of dividing, or using the inverse ratio (e.g., 1:2 instead of 2:1). Double-check which substance you are converting 'to' and 'from'.
No-calculator tips
- ✓Work with fractions to simplify calculations. For instance, a concentration of 0.20 mol dm⁻³ is 1/5. Multiplying 12.5 by 1/5 is easier than by 0.2.
- ✓Simplify the final division by adjusting decimal places. For 0.0050 / 0.0250, multiply both top and bottom by 10000 to get 50 / 250, which is much easier to see as 1/5.
- ✓Recognise common volume relationships. Since 12.5 cm³ is half of 25.0 cm³, if the ratio were 1:1, the concentration would be halved. Since the ratio is 1:2 (requiring double the alkali), this effect is cancelled out, leading to the same concentration.