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You are expected to use titration data to calculate either the concentration or the volume of a solution. In these calculations, you are always told the concentration and volume of one of the solutions (so that you can work on the number of moles) as well as the balanced symbol equation for the reaction. Here are some examples.

Concentration calculations

This type of question will give you information and ask you to work out the concentration of one of the solutions. Here is an example:

50.00 cm³ of aqueous HNO₃ is neutralised by 25.00 cm³ of 0.2 mol/dm³ barium hydroxide. Calculate the concentration of the nitric acid.

Ba(OH)₂ + 2HNO₃ → Ba(NO₃)₂ + 2H₂O

1. We have the concentration and volume of the barium hydroxide solution, so we can work out the moles of barium hydroxide (note: 25.00 cm³ = 0.025 dm³).

Mol(Ba(OH)₂) = 0.025 × 0.2 = 0.005 mol

2. We now look at the balanced equation to work out the number of moles of nitric acid.

Mol(HNO₃) = 2 × Mol(Ba(OH)₂) = 2 × 0.005 = 0.01 mol

3. Now we have the number of moles of nitric acid, we use the moles and the volume to work out the concentration (note: 50.00 cm³ = 0.05 dm³).

Conc(HNO₃) = 0.01 ÷ 0.05 = 0.2 mol/dm³

Volume calculations

This type of question will give you information and ask you to work out the volume of one of the solutions. Here is an example:

Calculate the volume of 0.5 mol/dm³ H₂SO₄ required to react exactly with 20.00cm³ of 2 mol/dm³ KOH solution.

H₂SO₄ + 2KOH → K₂SO₄ + 2H₂O

1. We have the concentration and volume of the potassium hydroxide solution, so we can work out the moles of potassium hydroxide (note: 20.00 cm³ = 0.02 dm³).

Mol(KOH) = 0.02 × 2 = 0.04 mol

2. We now look at the balanced equation to work out the number of moles of sulfuric acid.

Mol(H₂SO₄) = 0.5 × Mol(KOH) = 0.5 × 0.04 = 0.02 mol

3. Now we have the number of moles of sulfuric acid, we use the moles and the concentration to work out the volume.

Vol(H₂SO₄) = 0.02 ÷ 0.5 = 0.04 dm³ (40 cm³)