Concentration and titrations

The concentration of a solution tells you how much solute is dissolved in each dm³ (litre) of it. It can be given in g/dm³ or in mol/dm³. In a titration, a burette adds one solution to a measured volume of another until they just react, shown by an indicator changing colour. The volumes and the balanced equation then give the unknown concentration.

Key facts

  • 1 dm³ = 1 litre = 1000 cm³: divide a volume in cm³ by 1000 to get dm³.
  • concentration (g/dm³) = mass (g) ÷ volume (dm³).
  • concentration (mol/dm³) = moles ÷ volume (dm³), so moles = concentration × volume.
  • To change mol/dm³ into g/dm³, multiply by the Mᵣ.
  • In a titration a pipette measures one solution (often 25.0 cm³) into a flask, and a burette adds the other until the indicator changes colour: the end point.
  • Repeat until the titres are concordant (within 0.10 cm³ of each other) and use their mean.

Worked examples

Example 1

4 g of sodium hydroxide is dissolved to make 250 cm³ of solution. Find its concentration in g/dm³ and in mol/dm³. (Mᵣ of NaOH = 40)

  1. Change the volume to dm³. 250 ÷ 1000 = 0.25 dm³
  2. Concentration in g/dm³. 4 ÷ 0.25 = 16 g/dm³
  3. Moles of NaOH. 4 ÷ 40 = 0.1 mol
  4. Concentration in mol/dm³. 0.1 ÷ 0.25 = 0.4 mol/dm³

Answer: 16 g/dm³; 0.4 mol/dm³

Example 2

25.0 cm³ of 0.100 mol/dm³ sodium hydroxide is neutralised by 20.0 cm³ of hydrochloric acid. NaOH + HCl → NaCl + H2O. Find the concentration of the acid.

  1. Moles of NaOH in the flask. 0.100 × 25.0 ÷ 1000 = 0.00250 mol
  2. The ratio NaOH : HCl is 1 : 1, so the acid contains 0.00250 mol of HCl.
  3. Divide by the acid’s volume in dm³. 0.00250 ÷ 0.0200 = 0.125 mol/dm³

Answer: 0.125 mol/dm³

Example 3

25.0 cm³ of 0.200 mol/dm³ potassium hydroxide is neutralised by 12.5 cm³ of sulfuric acid. 2KOH + H2SO4 → K2SO4 + 2H2O. Find the concentration of the sulfuric acid.

  1. Moles of KOH. 0.200 × 25.0 ÷ 1000 = 0.00500 mol
  2. 2 mol of KOH react with 1 mol of H2SO4, so halve it. 0.00500 ÷ 2 = 0.00250 mol
  3. Divide by the volume in dm³. 0.00250 ÷ 0.0125 = 0.200 mol/dm³

Answer: 0.200 mol/dm³