Volumetric Solutions & Analysis (Oxford AQA International A Level Chemistry)

Revision Note

Richard Boole

Written by: Richard Boole

Reviewed by: Stewart Hird

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

Diagram showing the steps involved to make a standard (volumetric) solution

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 = fraction numerator 3.02 over denominator 120.1 end fraction = 0.0251

    • Concentration = fraction numerator m o l e s over denominator v o l u m e space open parentheses d m cubed close parentheses end fraction

    • Concentration = fraction numerator 0.0251 over denominator 0.25 end fraction = 0.1004 mol dm-3

Worked Example

Calculate the mass of sodium hydrogensulfate monohydrate, NaHSO4.H2O, required to prepare 250 cmof a 0.050 mol dm-3 solution.

Give your answer to 3 significant figures.

Answer: 

  1. 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

  2. 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

  3. 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

  1. Rinse the burette with the sodium hydrogensulfate standard solution

  2. Then fill the burette with the sodium hydrogensulfate standard solution

  3. Rinse a 250 cm3 conical flask with deionised / distilled water

  4. Rinse the 25 cm3 volumetric pipette with the sodium hydroxide solution

  5. Use the volumetric pipette to transfer exactly 25.0 cm3 of sodium hydroxide solution into the rinsed 250 cm3 conical flask

  6. Add two to three drops of phenolphthalein indicator to the solution in the conical flask

  7. Construct a results table

  8. Record the initial burette reading to the appropriate precision

  9. 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

  10. Record the final burette reading in your results table

  11. In your results table, calculate / record the volume of sodium hydrogensulfate solution used

  12. 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.

  13. Calculate / record the mean (average) volume of sodium hydrogensulfate solution used in the

    titration

    • Show your working

  14. Use your results to calculate the concentration of the sodium hydroxide

    • Show your working

Diagram

Indicators are added to some titrations to make the endpoint visible / more visible
Only a few drops of indicator are added, if necessary, because they are typically weak acids and can influence the results

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 = fraction numerator open parentheses 25.10 plus 25.00 close parentheses over denominator 2 end fraction = 25.05 cm3

Calculate the moles of sodium hydrogensulfate:

  • n(NaHSO4) = concentration x volume

  • n(NaHSO4) = 0.100 mol dm-3 x fraction numerator 25.05 over denominator 1000 end fraction 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

  • fraction numerator 25.0 over denominator 1000 end fraction = 0.025 dm3

Calculate the concentration of the unknown sodium hydroxide solution:

  • [NaOH (aq)] = fraction numerator moles over denominator volume space open parentheses dm cubed close parentheses end fraction

  • [NaOH (aq)] = fraction numerator 0.002505 over denominator 0.025 end fraction

  • [NaOH (aq)] = 0.1002 mol dm-3

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Richard Boole

Author: Richard Boole

Expertise: Chemistry

Richard has taught Chemistry for over 15 years as well as working as a science tutor, examiner, content creator and author. He wasn’t the greatest at exams and only discovered how to revise in his final year at university. That knowledge made him want to help students learn how to revise, challenge them to think about what they actually know and hopefully succeed; so here he is, happily, at SME.

Stewart Hird

Author: Stewart Hird

Expertise: Chemistry Lead

Stewart has been an enthusiastic GCSE, IGCSE, A Level and IB teacher for more than 30 years in the UK as well as overseas, and has also been an examiner for IB and A Level. As a long-standing Head of Science, Stewart brings a wealth of experience to creating Topic Questions and revision materials for Save My Exams. Stewart specialises in Chemistry, but has also taught Physics and Environmental Systems and Societies.