Volumetric Solutions & Analysis (Oxford AQA International A Level (IAL) Chemistry): Revision Note
Exam code: 9622
Required Practical 1: Volumetric Solutions & Acid-Base Titration
Required practical 1 is split into:
- Part A: Make up a volumetric solution 
- Part B: Carry out a simple acid-base titration 
Part A: Make up a volumetric solution
Objective
To prepare 250 cm3 of 0.100 mol dm-3 sodium hydrogensulfate solution, NaHSO4 (aq).
Apparatus
- Weighing bottle 
- 250 cm3 volumetric flask 
- Solid sodium hydrogensulfate, NaHSO4 (s) - Mr = 120.1 g mol-1 
 
- Filter funnel 
- Spatula 
- Deionised / distilled water 
- 250 cm3 beaker 
- Glass rod 
- Digital mass balance (reading to 2 or 3 decimal places) 
Method
- Calculate the mass of solid sodium hydrogensulfate required to make 250 cm3 of a 0.100 mol dm-3 solution - 1 mol dm-3 NaHSO4 = 120.1 g in 1 dm3 
- 0.100 mol dm-3 NaHSO4 = 12.01 g in 1 dm3 
- 0.100 mol dm-3 NaHSO4 = 3.0225 g in 250 cm3 
 
- Measure out 3.0225 g of solid sodium hydrogensulfate into the weighing bottle - The exact mass needs to be recorded because it is not possible to measure to this precision with the mass balances for this experiment 
 
- Transfer the solid sodium hydrogensulfate from the weighing bottle to the beaker 
- Rinse the weighing bottle with deionised water and add the washings to the beaker 
- Add 100 cm3 of deionised water and stir until all of the solid dissolves 
- Transfer the sodium hydrogen carbonate solution from the beaker to the volumetric flask 
- Rinse the beaker and glass rod with deionised water and add the washings to the volumetric flask 
- Make the volumetric flask up to the graduation mark 
- Insert the stopper and shake / invert the volumetric flask to mix the contents 
- Calculate the exact concentration of the sodium hydrogensulfate solution in mol dm-3 
Diagram

Practical Tip
- The most common source of error in the preparation of a standard solution is the loss of solid, when it is transferred from one vessel to another 
Results
- The only result for this part of the required practical is the mass of sodium hydrogensulfate - e.g. mass of sodium hydrogensulfate = 3.02 g 
 
- Remember: The exact mass of sodium hydrogensulfate is determined by the number of decimal places on the mass balance 
Evaluation
- The exact concentration of the sodium hydrogensulfate solution is: - Mass of NaHSO4 = 3.02g 
- Moles of NaHSO4 = - = 0.0251 
- Concentration = 
- Concentration = - = 0.1004 mol dm-3 
 
Worked Example
Calculate the mass of sodium hydrogensulfate monohydrate, NaHSO4.H2O, required to prepare 250 cm3 of a 0.050 mol dm-3 solution.
Give your answer to 3 significant figures.
Answer:
- Calculate the number of moles of NaHSO4.H2O needed from the concentration and volume: - n(NaHSO4.H2O) = concentration (mol dm-3) x volume (dm3) 
- n(NaHSO4.H2O) = 0.050 mol dm–3 x 0.250 dm3 
- n(NaHSO4.H2O) = 0.0125 mol 
 
- Calculate the molar mass of NaHSO4.H2O: - Mr = 23.0 + (3 x 1.0) + 32.1 + (5 x 16.0) = 138.1 g mol–1 
 
- Calculate the mass of NaHSO4.H2O required: - mass = moles x molar mass 
- mass = 0.0125 mol x 138.1 g mol–1 =1.73 g 
 
Part B: Carry Out a Simple Acid-Base Titration
Objective
To determine the concentration of a solution of sodium hydroxide by titration using a sodium hydrogensulfate solution that has a known concentration.
Apparatus
- Burette, stand and clamp 
- 25 cm3 volumetric pipette with pipette bulb / filler 
- 250 cm3 conical flasks 
- Funnel 
- Deionised / distilled water wash bottle 
- Phenolphthalein indicator 
- 150 cm3 sodium hydrogensulfate (standard) solution 
- 150 cm3 sodium hydroxide (unknown concentration) solution 
Method
- Rinse the burette with the sodium hydrogensulfate standard solution 
- Then fill the burette with the sodium hydrogensulfate standard solution 
- Rinse a 250 cm3 conical flask with deionised / distilled water 
- Rinse the 25 cm3 volumetric pipette with the sodium hydroxide solution 
- Use the volumetric pipette to transfer exactly 25.0 cm3 of sodium hydroxide solution into the rinsed 250 cm3 conical flask 
- Add two to three drops of phenolphthalein indicator to the solution in the conical flask 
- Construct a results table 
- Record the initial burette reading to the appropriate precision 
- Titrate the contents of the conical flask by adding sodium hydrogensulfate solution to it from the burette - Add the sodium hydrogensulfate solution slowly with swirling to mix the solution 
- Add the sodium hydrogensulfate solution dropwise near the end-point 
- At the end-point the indicator undergoes a definite colour change 
 
- Record the final burette reading in your results table 
- In your results table, calculate / record the volume of sodium hydrogensulfate solution used 
- Repeat the titration until you obtain two results which are concordant - Concordant results are within 0.1 cm3 
- You should normally carry out at least three titrations 
- Record all of the results that you obtain. 
 
- Calculate / record the mean (average) volume of sodium hydrogensulfate solution used in the - titration - Show your working 
 
- Use your results to calculate the concentration of the sodium hydroxide - Show your working 
 
Diagram

Practical Tip
- Completing a rough titration first helps to determine the approximate end-point of the titration 
- This also allows you to run the burette until 2 - 3 cm3 before the end-point and slowly add the sodium hydrogensulfate solution dropwise 
Results
- Record your results for each test carefully in a suitable table like the one below: 
| Rough | Run 1 | Run 2 | Run 3 | |
|---|---|---|---|---|
| Initial burette reading (cm3) | ||||
| Final burette reading (cm3) | ||||
| Titre volume (cm3) | 
Evaluation
- Identify the concordant results - Remember: Concordant results are within 0.1 cm3 
 
- Calculate the mean average titre 
- Calculate the moles of the chemical with the known concentration - In this case, this is the sodium hydrogensulfate 
 
- Use the balanced chemical equation to deduce the moles of the chemical with the unknown concentration - In this case, this is the sodium hydroxide 
 
- Calculate the concentration of the chemical with the unknown concentration - Careful: This may require the volume converting from cm3 to dm3 
 
Worked Example
A sodium hydroxide solution of unknown concentration was titrated against a 0.100 mol dm-3 solution of sodium hydrogen sulfate.
NaHSO4 (aq) + NaOH (aq) → Na2SO4 (aq) + H2O (l)
The titration results are shown in the table below.
| Rough | Run 1 | Run 2 | Run 3 | |
|---|---|---|---|---|
| Initial burette reading (cm3) | 0.00 | 0.00 | 0.10 | 0.10 | 
| Final burette reading (cm3) | 26.10 | 25.10 | 24.85 | 25.10 | 
| Titre volume (cm3) | 26.10 | 25.10 | 24.75 | 25.00 | 
Determine the concentration of the sodium hydroxide.
Answer:
Identify the concordant results:
- The concordant results are runs 1 and 3 
- 25.10 cm3 and 25.00 cm3 are within 0.1 cm3 
Calculate the average titre:
- Average titre = - = 25.05 cm3 
Calculate the moles of sodium hydrogensulfate:
- n(NaHSO4) = concentration x volume 
- n(NaHSO4) = 0.100 mol dm-3 x - dm3 
- n(NaHSO4) = 0.002505 moles 
Deduce the moles of sodium hydroxide:
- From the balanced equation, one mole of sodium hydroxide reacts with 1 mole of sodium hydrogensulfate - i.e. n(NaHSO4) = n(NaOH) 
 
- Therefore, n(NaOH) = 0.002505 moles 
Convert the volume of sodium hydroxide solution from cm3 to dm3
- = 0.025 dm3 
Calculate the concentration of the unknown sodium hydroxide solution:
- [NaOH (aq)] = 
- [NaOH (aq)] = 
- [NaOH (aq)] = 0.1002 mol dm-3 
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