C14: Titrations
Higher tier only — the titration technique for accurately determining the concentration of an unknown solution, recording concordant results, calculating the mean titre, and using titration data to find concentration.
What is a Titration?
A titration is a quantitative analytical technique used to determine the exact concentration of an unknown solution by reacting it with a solution of known concentration.
Titrations are commonly used for acid-base neutralisation reactions. The solution of known concentration is called the standard solution or titrant. The unknown solution is the analyte.
The point at which the reaction is complete — when the acid and base have reacted in exactly the mole ratio shown by the equation — is called the end point. An indicator is used to show when this point is reached.
Titration Equipment and Technique
The key equipment for a titration includes: a burette (to deliver the titrant), a pipette (to measure a precise volume of the unknown solution), a conical flask (to hold the reaction mixture), and a white tile (to see the colour change clearly).
Standard titration procedure:
- Fill the burette with the solution of known concentration and record the initial reading
- Use a pipette to transfer a measured volume (usually 25.0 cm³) of the unknown solution into a clean conical flask
- Add a few drops of a suitable indicator to the conical flask
- Slowly add the solution from the burette, swirling the flask constantly
- Stop adding when the indicator just changes colour — this is the end point
- Record the final burette reading and calculate the volume added (the titre)
- Repeat the titration until you obtain concordant results
The volume of solution added from the burette is called the titre. It is calculated by subtracting the initial burette reading from the final reading.
Titre = final burette reading − initial burette reading
Burette readings should be recorded to 0.05 cm³ (e.g. 23.45, not 23.4 or 23.5). Always read from the bottom of the meniscus at eye level.
Concordant Results and Mean Titre
Concordant results are titre values that are within 0.10 cm³ of each other. Only concordant results should be used to calculate the mean titre.
When calculating the mean titre, discard any anomalous (non-concordant) results first, then find the mean of the concordant results. A rough initial titration can be used to get an approximate end point but should not be included in the mean.
Worked Example: Calculating the Mean Titre
A student obtains the following titre values:
Rough: 24.30 cm³
Trial 2: 23.75 cm³
Trial 3: 23.80 cm³
Trial 4: 24.50 cm³
Trial 5: 23.70 cm³
Concordant results: 23.75, 23.80 and 23.70 (all within 0.10 cm³ of each other)
Trial 4 (24.50) is not concordant — discard it. The rough titration is also discarded.
Mean titre = (23.75 + 23.80 + 23.70) / 3 = 71.25 / 3 = 23.75 cm³
Never include the rough titration in the mean. Never round a mean titre to fewer decimal places than the individual readings — always give it to 0.05 cm³ (2 decimal places).
Indicators for Titrations
Not all indicators are suitable for all titrations. The choice depends on the strength of the acid and base involved.
| Indicator | Colour in Acid | Colour in Alkali | Suitable For |
|---|---|---|---|
| Phenolphthalein | Colourless | Pink | Strong acid–strong base and weak acid–strong base |
| Methyl orange | Red | Yellow | Strong acid–strong base and strong acid–weak base |
| Universal indicator | Red (pH 1) | Purple (pH 14) | NOT suitable — gradual colour change, not sharp enough |
Phenolphthalein is the most commonly used indicator for GCSE titrations. It gives a very sharp colour change from colourless to pink at the end point, making it easy to detect.
Calculating Concentration from Titration Data
Once you have the mean titre, you can calculate the concentration of the unknown solution. Use a step-by-step method:
Step 1: Calculate moles of the known solution using: moles = concentration × volume (dm³)
Step 2: Use the balanced equation to find the mole ratio and calculate moles of the unknown solution
Step 3: Calculate the concentration of the unknown: concentration = moles / volume (dm³)
Worked Example: Finding the Concentration of an Acid
25.0 cm³ of hydrochloric acid of unknown concentration is neutralised by 23.50 cm³ of 0.100 mol/dm³ sodium hydroxide solution. Calculate the concentration of the HCl.
NaOH + HCl → NaCl + H2O (mole ratio = 1:1)
Step 1: moles of NaOH = 0.100 × (23.50 / 1000) = 0.002 350 mol
Step 2: From 1:1 ratio, moles of HCl = 0.002 350 mol
Step 3: concentration of HCl = 0.002 350 / (25.0 / 1000) = 0.002 350 / 0.0250
concentration = 0.0940 mol/dm³
Worked Example: Finding the Concentration of an Alkali
25.0 cm³ of sodium carbonate solution is titrated with 0.200 mol/dm³ HCl. The mean titre is 26.40 cm³. Calculate the concentration of the Na2CO3 solution.
Na2CO3 + 2HCl → 2NaCl + H2O + CO2 (mole ratio Na2CO3:HCl = 1:2)
Step 1: moles of HCl = 0.200 × (26.40 / 1000) = 0.005 280 mol
Step 2: moles of Na2CO3 = 0.005 280 / 2 = 0.002 640 mol
Step 3: concentration of Na2CO3 = 0.002 640 / (25.0 / 1000) = 0.002 640 / 0.0250
concentration = 0.1056 mol/dm³
Worked Example: Concentration in g/dm³
Using the result from the previous example (0.1056 mol/dm³), express the concentration of Na2CO3 in g/dm³. (Mr of Na2CO3 = 106)
Concentration (g/dm³) = 0.1056 × 106 = 11.2 g/dm³
Always convert cm³ to dm³ by dividing by 1000 BEFORE using the volume in calculations. A common mistake is forgetting this conversion — it will give an answer 1000 times too large or too small.
Common Titration Errors
Understanding common errors helps you interpret titration results and answer questions about accuracy:
| Error | Effect on Titre | Effect on Calculated Concentration |
|---|---|---|
| Rinsing burette with water instead of the titrant | Titre too high (titrant is diluted) | Calculated concentration too high |
| Rinsing conical flask with the unknown solution | No effect on titre (extra moles in flask) | Calculated concentration too low |
| Reading burette from top of meniscus | Titre too low (reads less volume than used) | Calculated concentration too low |
| Overshooting the end point | Titre too high (too much titrant added) | Calculated concentration too high |
| Not swirling the flask during addition | Titre too high (localised colour change missed) | Calculated concentration too high |
| Air bubble in burette tip | First titre too high (bubble replaced by liquid) | First calculated concentration too high |
The conical flask should be rinsed with distilled water only, NOT with the solution being measured. Rinsing with the solution would add extra moles of solute, making the concentration appear lower when calculated.
Practice Questions
In a titration, a student obtains titre values of 22.30 cm³, 22.35 cm³, 22.40 cm³ and 23.10 cm³. Which values are concordant? Calculate the mean titre.
22.30, 22.35 and 22.40 are concordant (all within 0.10 cm³ of each other). 23.10 is not concordant.
Mean titre = (22.30 + 22.35 + 22.40) / 3 = 67.05 / 3 = 22.35 cm³
25.0 cm³ of sulfuric acid is neutralised by 20.00 cm³ of 0.150 mol/dm³ NaOH. H2SO4 + 2NaOH → Na2SO4 + 2H2O. Calculate the concentration of the sulfuric acid in mol/dm³.
moles of NaOH = 0.150 × (20.00 / 1000) = 0.003 000 mol
Mole ratio H2SO4:NaOH = 1:2
moles of H2SO4 = 0.003 000 / 2 = 0.001 500 mol
concentration = 0.001 500 / 0.0250 = 0.0600 mol/dm³
A student rinsed the burette with water before filling it with the standard solution. Explain how this affects the titre value and the final calculated concentration.
Rinsing with water dilutes the standard solution in the burette. This means more volume of the diluted titrant is needed to reach the end point, so the titre will be too high. Since more moles of titrant appear to have been used, the calculated concentration of the unknown solution will be too high.
25.0 cm³ of potassium hydroxide solution is titrated with 0.120 mol/dm³ nitric acid. The mean titre is 18.75 cm³. KOH + HNO3 → KNO3 + H2O. Calculate the concentration of KOH in g/dm³. (Mr of KOH = 56)
moles of HNO3 = 0.120 × (18.75 / 1000) = 0.002 250 mol
Mole ratio 1:1, so moles of KOH = 0.002 250 mol
Concentration in mol/dm³ = 0.002 250 / 0.0250 = 0.0900 mol/dm³
Concentration in g/dm³ = 0.0900 × 56 = 5.04 g/dm³
Required Practical
Carrying Out a Titration Higher
This IS the titration required practical. You must be able to carry out a titration to determine the concentration of an unknown acid or alkali. The key steps are: fill the burette with the solution of known concentration (rinsed with the same solution first, not water), use a pipette to measure a precise volume of the unknown solution into a clean conical flask, add indicator, and add the titrant dropwise near the end point while swirling constantly.
Record burette readings to 0.05 cm³ and repeat until you obtain concordant results (within 0.10 cm³). Only use concordant results to calculate the mean titre — discard the rough titration and any anomalous results. The rough titration is used only to find the approximate end point.
Safety: wear eye protection, as acids and alkalis are corrosive. If any solution gets on your skin, wash it off immediately with plenty of water. Clean up spills promptly. Work in a well-ventilated area when using volatile or fuming substances.
Variables: The independent variable is the concentration of the unknown solution. The dependent variable is the volume of standard solution needed (the titre). Control variables include: the volume of unknown solution (use the same pipette each time), the concentration of the standard solution, the indicator used, and the temperature of the solutions.
Improving accuracy: Rinse the burette with the standard solution (not water) before filling. Read the bottom of the meniscus at eye level. Add the titrant dropwise near the end point. Use a white tile to see the colour change clearly. Swirl constantly to ensure thorough mixing. Repeat to obtain concordant results.
Maths Skills
Calculating Concentration from Titration Data Higher
To calculate concentration from titration data: (1) calculate moles of the known solution using moles = concentration × volume (dm³), (2) use the balanced equation to find moles of the unknown, (3) calculate concentration = moles / volume (dm³).
Always convert volumes from cm³ to dm³ by dividing by 1000 before using them in calculations. A common error is forgetting this step, giving an answer 1000 times too large or too small.
Mean Titre and Concordant Results Higher
When calculating the mean titre, discard the rough titration and any anomalous (non-concordant) results. Only average results that are within 0.10 cm³ of each other. Give the mean to 2 decimal places (to the same precision as individual readings).
Uncertainty in Titration Higher
Each burette reading has an uncertainty of ±0.05 cm³. Since a titre involves two readings (initial and final), the total uncertainty in the titre is ±0.10 cm³. For a mean titre of 23.60 cm³, the percentage uncertainty = (0.10 / 23.60) × 100 = 0.42%. Larger titres reduce the percentage uncertainty, which is why using a larger volume of unknown solution or a more dilute standard solution can improve accuracy.
Common Misconceptions
Concordant Results Higher
Wrong: Any two results can be used to calculate the mean titre Correct: Only concordant results (within 0.10 cm³ of each other) should be used. Non-concordant results may contain errors such as overshooting the end point. The rough titration must always be discarded
Indicators and the End Point Higher
Wrong: The indicator changes colour at the exact equivalence point Correct: The indicator changes colour at the end point, which is close to but not exactly the same as the equivalence point. Different indicators change colour over different pH ranges. Phenolphthalein and methyl orange are suitable because their colour changes are sharp enough that the end point is very close to the equivalence point in strong acid–strong base titrations
Wrong: The first (rough) titration result should be included in the mean Correct: The rough titration is only used to find the approximate end point. It is not included in the mean because it is likely to be less accurate due to overshooting
Wrong: You only need to do the titration once Correct: You must repeat the titration until you obtain concordant results (at least two within 0.10 cm³). This ensures reliability and allows you to identify anomalies
6-Mark Extended Question
Describe how to carry out a titration to determine the concentration of a solution of sodium hydroxide. Explain how to obtain accurate and reproducible results.
Use a pipette to transfer 25.0 cm3 of the sodium hydroxide solution into a clean conical flask. Add a few drops of phenolphthalein indicator, which turns pink in alkali and colourless in acid. Fill a burette with hydrochloric acid of known concentration and record the initial burette reading. [2 marks]
Slowly add the acid from the burette to the flask, swirling constantly to mix. As you approach the end point, add the acid drop by drop. Stop when the indicator just changes from pink to colourless — this is the end point. Record the final burette reading and calculate the titre. Repeat the titration at least twice more to obtain concordant results. [2 marks]
To ensure accuracy: rinse the burette with the acid solution first (not water), read the burette from the bottom of the meniscus at eye level, add acid dropwise near the end point, swirl constantly, and use a white tile to see the colour change clearly. To ensure reproducibility: repeat until at least two concordant results are obtained, discard the rough titration from the mean, and record all readings to 0.05 cm3. [2 marks]
AO3: Analyse and Evaluate
A student carries out a titration and records the following titre values: Rough = 24.80 cm3, Trial 2 = 23.60 cm3, Trial 3 = 23.65 cm3, Trial 4 = 25.10 cm3, Trial 5 = 23.55 cm3. The standard solution was 0.150 mol/dm3 HCl and the unknown was 25.0 cm3 of NaOH. Calculate the mean titre and the concentration of the NaOH.
Concordant results: 23.60, 23.65 and 23.55 (all within 0.10 cm3). Trial 4 (25.10) is anomalous — discard it. Rough titration is also discarded. Mean titre = (23.60 + 23.65 + 23.55) / 3 = 23.60 cm3. Moles of HCl = 0.150 × (23.60 / 1000) = 0.003 540 mol. NaOH + HCl → NaCl + H2O (1:1 ratio), so moles of NaOH = 0.003 540 mol. Concentration of NaOH = 0.003 540 / (25.0 / 1000) = 0.1416 mol/dm3.
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