AQA A Level Chemistry

Revision Notes

4.1.1 Making a Volumetric Solution

Making a Volumetric Solution


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 hydroxide, NaOH, required to prepare 250 cmof a 0.200 mol dm-3 solution


Step 1: Find the number of moles of NaOH needed from the concentration and volume:

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

n = 0.200 mol dm-3 x 0.250 dm3

n = 0.0500 mol

Step 2: Find the molar mass of NaOH

M = 22.99 + 16.00 + 1.01 = 40.00 g mol-1

Step 3: Calculate the mass required

mass = moles x molar mass

mass =  0.0500 mol x 40.00 g mol-1   = 2.00 g


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