Acid-Base Titration (OCR AS Chemistry A): Revision Note

Exam code: H032

Richard Boole

Written by: Richard Boole

Reviewed by: Philippa Platt

Updated on

PAG 2: Acid-base titration

  • There are a series of suggested practicals for PAG 2 - Acid-base titration

    • PAG 2.1: Determination of the concentration of hydrochloric acid

      • Making a standard solution of sodium hydrogencarbonate

      • Titrating this against hydrochloric acid

      • Determining the concentration of the acid

    • PAG 2.2: Determination of the molar mass of an acid

      • Making a solution of an unknown hydrated acid (e.g. citric acid monohydrate)

      • Titrating this against a sodium hydroxide solution of known concentration.

      • Determining the molar mass of the acid and, with additional information, the formula of the acid

    • PAG 2.3: Identification of an unknown carbonate

      • Making a solution of an unknown Group 1 carbonate

      • Titrating this solution against hydrochloric acid of a known concentration

      • Determine the molar mass of the carbonate and the identity of the metal

  • All three suggested practicals use volumetric analysis

    • This requires making standard solutions and completing titration experiments

    • The main difference between the practicals is how the titration results are processed

Volumetric analysis

  • Volumetric analysis uses the volume and concentration of one chemical reactant (a volumetric solution) to determine the concentration of another unknown solution

  • The technique most commonly used is a titration

  • The volumes are measured using two precise pieces of equipment:

    • volumetric or graduated pipette 

    • burette

  • Before the titration can be done, the standard solution must be prepared

  • Specific apparatus must be used both when preparing the standard solution and when completing the titration, to ensure that volumes are measured precisely

Five lab equipment items labelled 1 to 5: beaker, burette, pipette, conical flask, and volumetric flask.
Some key pieces of apparatus used to prepare a volumetric solution and perform a simple titration 
  1. Beaker

  2. Burette

  3. Volumetric Pipette

  4. Conical Flask

  5. Volumetric Flask

Making a volumetric solution

  • Chemists routinely prepare solutions needed for analysis, whose concentrations are known precisely

  • These solutions are termed volumetric solutions or standard solutions

  • They are made as accurately and precisely as possible using three decimal place balances and volumetric flasks

    • This reduces the impact of measurement uncertainties

  • The steps are:

Preparing a standard solution (1)
Preparing a standard solution (2)

Volumes & concentrations of solutions

  • The concentration of a solution is the amount of solute dissolved in a solvent to make 1 dmof  solution

    • The solute is the substance that dissolves in a solvent to form a solution

    • The solvent is often water

  • concentrated solution is a solution that has a high concentration of solute

  • dilute solution is a solution with a low concentration of solute

  • Concentration is usually expressed in one of three ways:

    • moles per unit volume (mol dm-3)

    • mass per unit volume (g dm-3)

    • parts per million (ppm)

Worked Example

Calculate the mass of sodium hydrogencarbonate, NaHCO3, required to prepare 250 cmof a 0.200 mol dm-3 solution

Answer:

  1. Calculate the moles of NaHCO3 needed from the concentration and volume:

number of moles  = concentration (mol dm-3) x volume (dm3)  

n = 0.200 mol dm-3 x 0.250 dm3

n0.0500 mol

  1. Calculate the molar mass of NaHCO3 

Mr = 23.0 + 1.0 + 12.0 + (16.0 x 3) = 84.0 g mol-1

  1. Calculate the mass of NaHCO3 required

mass = moles x molar mass

mass =  0.0500 mol x 84.0 g mol-1 = 4.2 g

Performing the titration

  • The key piece of equipment used in the titration is the burette

  • Burettes are usually marked to a precision of 0.10 cm3

    • Since they are analogue instruments, the uncertainty is recorded to half the smallest marking, in other words to ±0.05 cm3

  • The end point or equivalence point occurs when the two solutions have reacted completely and is shown with the use of an indicator

Titration, downloadable IB Chemistry revision notes

The steps in a titration

  • A white tile is placed under the conical flask while the titration is performed, to make it easier to see the colour change

Titration apparatus, downloadable AS & A Level Chemistry revision notes

Titrating 

  • The steps in a titration are:

    • Measuring a known volume (usually 20 or 25 cm3) of one of the solutions with a volumetric pipette and placing it into a conical flask

    • The other solution is placed in the burette

      • To start with, the burette will usually be filled to 0.00 cm3

    • A few drops of the indicator are added to the solution in the conical flask

    • The tap on the burette is opened and the solution is slowly added to the conical flask until the indicator starts to change colour

    • As you start getting near to the end point, the flow of the burette should be slowed right down so that the solution is added dropwise

      • You should be able to close the tap on the burette after one drop has caused the colour change

    • Multiple runs are carried out until concordant results are obtained

      • Concordant results are within 0.1 cm3 of each other

Recording and processing titration results

  • Both the initial and final burette readings should be recorded and shown to a precision of  ±0.05 cm3, the same as the uncertainty

Titration results, downloadable IB Chemistry revision notes

A typical layout and set of titration results

  • The volume delivered (titre) is calculated and recorded to an uncertainty of ±0.10 cm3

    • The uncertainty is doubled

    • This is because two burette readings are made to obtain the titre (V final – V initial)

    • This follows the rules for propagation of uncertainties

  • Concordant results are then averaged, and non-concordant results are discarded

  • The appropriate calculations are then done

Worked Example

PAG 2.1: Determining hydrochloric acid concentration

25.0 cm3 of hydrochloric acid was titrated with a 0.200 mol dm-3 solution of sodium hydrogencarbonate, NaHCO3

NaHCO3 (aq) + HCl (aq) → NaCl (aq)+ H2O (l) + CO2 (g) 

Use the following results to calculate the concentration of the acid, to 3 significant figures.

 

Rough

Run 1

Run 2

Run 3

Initial burette reading / cm3 (±0.05 cm3)

0.00

23.15

0.20

23.00

Final burette reading / cm3 (±0.05 cm3)

23.75

45.95

23.00

46.10

Volume delivered / cm3 (±0.10 cm3)

23.75

22.80

22.80

23.10

Answer:

  1. Calculate the average titre from concurrent titrations only

    • Concurrent titrations are within 0.10 cm3 of each other

Average titre = fraction numerator 22.80 plus 22.80 over denominator 2 end fraction = 22.80 cm3

  1. Calculate the number of moles of sodium hydrogencarbonate

Moles of NaHCO3fraction numerator 22.80 over denominator 1000 end fraction x 0.200 = 4.56 x 10-3 moles

  1. Calculate (or deduce) the number of moles of hydrochloric acid

    • The stoichiometry of NaHCO3 : HCl is 1 : 1

Moles of HCl = 4.56 x 10-3 moles

  1. Calculate the concentration of hydrochloric acid

Concentration =moles over volume space

Concentration = fraction numerator 4.56 space cross times space 10 to the power of negative 3 end exponent over denominator left parenthesis 25.0 space divided by space 1000 right parenthesis end fraction = 0.182 mol dm-3 

Worked Example

PAG 2.2: Determining the molar mass of an acid

0.133 g of an unknown monoprotic acid is added to a 1.0 dm3 volumetric flask and made up to the mark with distilled water. This solution is titrated with 0.070 mol dm-3 sodium hydroxide solution. The average titre required for the neutralisation of the unknown monoprotic acid is 23.40 cm3.

i) Calculate the molar mass of the unknown acid.

ii) The unknown monoprotic acid is a halogenated acid. Identify the unknown monoprotic acid.

Answer:

Part i)

  1. Calculate the number of moles of sodium hydroxide

Moles of NaOHequals 0.070 cross times fraction numerator 23.40 over denominator 1000 end fraction equals1.638 x 10-3 moles

  1. State the number of moles of the unknown monoprotic acid

    1. The stoichiometry of HX : NaOH is 1 : 1

Moles of HX = 1.638 x 10-3 moles

  1. Calculate the molar mass of the unknown monoprotic acid

Mr = mass over moles

Mr = fraction numerator 0.133 over denominator 1.638 cross times 10 to the power of negative 3 end exponent end fraction = 81.20 g mol-1 

Part ii)

  1. Calculate the mass of the halogen

81.20 - 1.0 = 80.2

  1. Use the periodic table to identify the halogen

80.2 ≈ Bromine (79.9)

  1. State the final answer

    • The unknown monoprotic acid is HBr / hydrobromic acid   

Examiner Tips and Tricks

  • You can potentially be asked to calculate the molar mass of any unknown acid

    • This includes monoprotic, diprotic and even triprotic acids

  • You must account for the stoichiometry of the neutralisation reaction

  • You have to be given information to deduce the identity of the unknown acid

Worked Example

PAG 2.3: IdentifYING an unknown carbonate

1.19 g of an unidentified Group 1 metal carbonate, X2CO3, was dissolved in water to produce a 250.0 cm³ standard solution. 25.0 cm³ aliquots of this solution were titrated with 0.150 mol dm3 hydrochloric acid. The average titre for this experiment was found to be 14.95 cm3.

Identify the Group 1 metal in the unidentified metal carbonate X2CO3.

Answer

  1. Calculate the number of moles of hydrochloric acid

Moles of HClequals 0.150 cross times fraction numerator 14.95 over denominator 1000 end fraction equals2.2425 x 10-3 moles

  1. Calculate the number of moles of X2CO3

    • The stoichiometric ratio is 1X2CO3 : 2HCl

Moles of X2CO3equals fraction numerator 2.2425 cross times 10 to the power of negative 3 end exponent over denominator 2 end fraction equals1.12125 x 10-3 moles

  3. Calculate the mass of X2CO3 in the 25.0 cm3 aliquot

fraction numerator 1.19 over denominator 10 end fraction= 0.119 g

  1. Calculate the molar mass of X2CO3

Mr = mass over moles

Mr =fraction numerator 0.119 over denominator 1.12125 cross times 10 to the power of negative 3 end exponent end fraction = 106.132 g mol-1 

  1. Calculate the mass of X in X2CO3 

fraction numerator 106.132 minus 12.0 minus open parentheses 3 cross times 16.0 close parentheses over denominator 2 end fraction= 23.066

  1. Use the periodic table to identify the Group 1 element in X2CO3 

23.066 ≈ Sodium / Na (relative mass of 23.0)

Examiner Tips and Tricks

  • Careful: Examiners have been known to construct this question to get the mass of X as around 86.0 - 86.5

    • This then gets a mixture of answers of X = rubidium or francium

    • This happens because students use the wrong number to identify the element

      • Rubidium has a mass number of 85.5

      • Francium has an atomic number of 87

Practical skills reminder

  • These titration-based practicals help develop essential experimental techniques, including:

    • Accurately measuring volumes using volumetric pipettes, burettes, and volumetric flasks

    • Preparing standard solutions with appropriate equipment and minimising measurement uncertainties

    • Recording burette readings to the correct precision and accounting for instrumental uncertainty

    • Performing titrations safely and effectively, using indicators and controlling flow rates near the endpoint

    • Calculating concentrations, molar masses, and formulae using balanced equations and volume data

    • Interpreting titration data to identify unknown compounds, following correct stoichiometric reasoning

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Richard Boole

Author: Richard Boole

Expertise: Chemistry Content Creator

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.

Philippa Platt

Reviewer: Philippa Platt

Expertise: Chemistry Content Creator

Philippa has worked as a GCSE and A level chemistry teacher and tutor for over thirteen years. She studied chemistry and sport science at Loughborough University graduating in 2007 having also completed her PGCE in science. Throughout her time as a teacher she was incharge of a boarding house for five years and coached many teams in a variety of sports. When not producing resources with the chemistry team, Philippa enjoys being active outside with her young family and is a very keen gardener