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Concentration of solutions

Calculate concentration, volume and moles for solutions and use titration data.

Start here

The key idea

Concentration (mol/dm³) = moles ÷ volume (dm³).A more concentrated solution has more moles of solute per dm³ of solution.

Equation to know

concentration (mol/dm³) = moles ÷ volume (dm³)

Concentration Solutions
fixed volumemore solute means higher concentration

Use the labels to explain the scientific relationship shown.

Revision notes

The bit that matters

Keep the idea tight, then use the worked example to practise the exact exam wording.

1

Concentration and the mole

Concentration measures how much solute is dissolved in a given volume of solution.At GCSE, concentration is measured in mol/dm³ (moles per cubic decimetre, which is the same as moles per litre).The equation is: concentration = moles ÷ volume (dm³).You must convert cm³ to dm³ by dividing by 1000 before using this equation.

2

Titration

A titration is a practical technique for finding the exact volume of one solution that reacts completely with a known volume of another.A burette delivers one solution accurately; the other solution is placed in a conical flask with an indicator.The endpoint is reached when the indicator changes colour permanently, showing the neutralisation is complete.

3

Calculating concentration from titration

From a titration, you know the volume and concentration of one solution and the volume of the other.Use: moles = concentration × volume to find moles of the first solution; use the equation's mole ratio to find moles of the second; then use concentration = moles ÷ volume.Always read the mean titre from concordant results.

4

Preparing standard solutions

A standard solution has a precisely known concentration.It is prepared by accurately weighing the solute, dissolving it in a small volume of distilled water, transferring to a volumetric flask, and making up to the mark.Standard solutions are used in titrations as the solution of known concentration.

Key terms

Definitions to learn

Concentration

The amount of solute dissolved in a given volume of solution, in mol/dm³.

Titration

A method for finding the exact volume of one solution that reacts completely with a known volume of another.

Burette

A graduated glass tube with a tap used to deliver accurate volumes of solution in a titration.

Concordant results

Titration readings that agree to within 0.1 cm³ of each other, used to calculate a reliable mean.

Standard solution

A solution of precisely known concentration used in quantitative analysis.

Worked example

Calculate the concentration of a solution containing 0.2 mol of sodium hydroxide in 500 cm³ of solution.

1

Convert volume to dm³: 500 cm³ ÷ 1000 = 0.5 dm³.

2

Use concentration = moles ÷ volume.

3

Concentration = 0.2 ÷ 0.5 = 0.4 mol/dm³.

Final answer

Concentration = 0.4 mol/dm³.

Exam habit

Convert cm³ to dm³ as the very first step in every concentration calculation — divide by 1000.

Watch out

Always convert cm³ to dm³ by dividing by 1000 before using the concentration equation.

Examiner tips

How to score full marks

  • 1ALWAYS divide cm³ by 1000 to get dm³ before substituting into the concentration formula — this is the most common arithmetic error.
  • 2In titration calculations, state the mole ratio from the balanced equation explicitly before using it.
  • 3Use only concordant titre readings (within 0.1 cm³) when calculating the mean — reject anomalous rough or inaccurate runs.
Practice questions

Try these yourself

Open each answer only after you have explained the full chemical process.

1State the units of concentration used in chemistry.[1 mark]
Mark scheme
  1. 1.Recall the standard unit.
mol/dm³ (1).
2Calculate the number of moles in 250 cm³ of a 0.1 mol/dm³ solution.[2 marks]
Mark scheme
  1. 1.Convert cm³ to dm³ first, then use moles = concentration × volume.
Volume = 0.250 dm³; moles = 0.1 × 0.250 = 0.025 mol (1).
3A solution contains 5.85 g of NaCl dissolved in 1 dm³ of water. Calculate its concentration in mol/dm³. (Na = 23, Cl = 35.5)[2 marks]
Mark scheme
  1. 1.Find moles of NaCl, then use concentration = moles ÷ volume.
M_r of NaCl = 23 + 35.5 = 58.5; moles = 5.85 ÷ 58.5 = 0.1 mol (1); concentration = 0.1 ÷ 1 = 0.1 mol/dm³ (1).
4In a titration, 25 cm³ of 0.1 mol/dm³ NaOH exactly neutralises HCl. The equation is NaOH + HCl → NaCl + H₂O. Calculate the moles of NaOH used.[1 mark]
Mark scheme
  1. 1.Use moles = concentration × volume in dm³.
moles NaOH = 0.1 × 0.025 = 0.0025 mol (1).
5Using the titration data above, state the moles of HCl in 25 cm³ of the acid, and calculate its concentration.[3 marks]
Mark scheme
  1. 1.Use the 1:1 mole ratio.
Moles HCl = 0.0025 mol (1) (1:1 ratio); concentration = 0.0025 ÷ 0.025 = 0.1 mol/dm³ (1).
6Explain why multiple concordant titration readings should be used to calculate an accurate mean titre.[3 marks]
Mark scheme
  1. 1.Link repeats to reliability.
Multiple readings reduce the effect of random errors (1); concordant results (within 0.1 cm³ of each other) confirm the result is reliable (1), and averaging gives a more accurate value (1).
7Calculate the mass of solute needed to make 200 cm³ of a 0.5 mol/dm³ copper sulfate solution (CuSO₄). (Cu = 64, S = 32, O = 16)[2 marks]
Mark scheme
  1. 1.Find moles needed, then mass.
M_r of CuSO₄ = 64 + 32 + 64 = 160; moles = 0.5 × 0.2 = 0.1 mol (1); mass = 0.1 × 160 = 16 g (1).
8Explain what a rough titration is used for.[2 marks]
Mark scheme
  1. 1.State its purpose before the accurate runs.
A rough titration gives an approximate endpoint (1) so that accurate titrations can be carried out more quickly by adding rapidly until close to the endpoint (1).
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