OCR AS Chemistry

Revision Notes

1.1.2 Acid-Base Titration

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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
      • This involves making a standard solution of sodium hydrogencarbonate and titrating this against a solution of hydrochloric acid in order to work out the concentration of the acid.
    • PAG 2.2: Determination of the molar mass of an acid
      • This involves making a solution of an unknown hydrated acid (e.g. citric acid monohydrate), then titrating this against a sodium hydroxide solution of known concentration. This allows the molar mass of the acid to be determined, and with additional information, the formula of the acid.
    • PAG 2.3: Identification of an unknown carbonate
      • This involves making a solution of an unknown Group 1 carbonate and then titrating this solution against hydrochloric acid of a known concentration to determine the molar mass of the carbonate and hence the identity of the metal.
  • All three suggested practicals use volumetric analysis, which requires making standard solutions and completing titration experiments.
    • The main difference between the suggested practicals is how the titration results are processed

Volumetric Analysis

  • Volumetric analysis is a process that 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, a volumetric or graduated pipette and a 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

Volumetric analysis apparatus, downloadable AS & A Level Chemistry revision notes

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 to reduce the impact of measurement uncertainties
  • The steps are:

Preparing a standard solution (1), downloadable IB Chemistry revision notesPreparing a standard solution (2), downloadable IB Chemistry revision notes

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
    • mass per unit volume
    • parts per million

Worked example

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

Answer:

Step 1: Find the number of 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

Step 2: Find the molar mass of NaHCO3 

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

Step 3: 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 carefully opened and the solution added, portion by portion, 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, because two burette readings are made to obtain the titre (V final – V initial), following 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: Determination of the concentration of hydrochloric acid

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

Step 1: Calculate the average titre from concurrent titrations only (those within 0.10 cm3 of each other)

    • Average titreequals space fraction numerator 22.80 space plus space 22.80 over denominator 2 end fraction space equals22.80 cm3

Step 2: Calculate the number of moles of sodium hydrogencarbonate

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

Step 3: Calculate (or deduce) the number of moles of hydrochloric acid

    • The stoichiometry of NaHCO3 : HCl is 1 : 1
    • Therefore, the number of moles of sodium hydrogencarbonate is also 4.56 x 10-3 moles

Step 4: Calculate the concentration of hydrochloric acid

    • Concentration = equals space moles over volume space equals 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 space equals space0.182 mol dm-3 

Worked example

PAG 2.2: Determination of 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)

   Step 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

   Step 2: State the number of moles of the unknown monoprotic acid

    • Moles of acid = 1.638 x 10-3 moles since the stoichiometric ratio is 1HX : 1NaOH

   Step 3: Calculate the molar mass of the unknown monoprotic acid

    • M subscript straight r equals mass over moles equals fraction numerator 0.133 over denominator 1.638 cross times 10 to the power of negative 3 end exponent end fraction equals81.20 g mol-1 

   Part ii)

   Step 1: Calculate the mass of the halogen

    • 81.20 - 1.0 = 80.2

   Step 2: Use the periodic table to identify the halogen

    • 80.2 ≈ Bromine (79.9)

   Step 3: State the final answer

    • The unknown monoprotic acid is HBr / hydrobromic acid

Exam Tip

  • You can potentially be asked to calculate the molar mass of any unknown acid including monoprotic, diprotic and even triprotic acids
    • You must account for the stoichiometry of the neutralisation reaction
  • To identify the unknown acid, you have to be given more information in order to deduce it's identity

Worked example

PAG 2.3: Identification of 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

   

   Step 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

   Step 2: Calculate the number of moles of X2CO3

    • 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 (since the stoichiometric ratio is 1X2CO3 : 2HCl)

   Step 3: Calculate the mass of X2CO3 in the 25.0 cm3 aliquot

    • fraction numerator 1.19 over denominator 10 end fraction= 0.119 g

Step 4: Calculate the molar mass of X2CO3

    • M subscript straight r equals mass over moles equals fraction numerator 0.119 over denominator 1.12125 cross times 10 to the power of negative 3 end exponent end fraction equals106.132 g mol-1 

 Step 5: 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

   Step 6: Use the periodic table to identify the Group 1 element in X2CO3 

    • 23.066 ≈ Sodium/ Na (relative mass of 23.0)

Exam Tip

  • 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

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