Working out concentration

The concentration of a reactant can be calculated from the results of a redox titration.

In order to calculate concentration we have to work out the number of moles reacting and multiply that by the volume. The number of moles of electrons involved must also be taken into account.

This can be done using the following relationship:

C_x × V_x × n_x = C_y × V_y × n_y

So, C_x = (C_y × V_y × n_y) ÷ (V_x × n_x)

Where:

• C_x = concentration to be calculated
• V_x = volume of X
• n_x = number of moles of electrons gained or lost by X
• C_y = concentration of y
• V_y = volume of Y
• n_y = number of moles of electrons gained or lost by Y

Example:

Calculate the concentration of an iron(II) sulphate solution given that 20 cm³ of the solution reacts completely with 23.5 cm³ of 0.02 mol l^-1 potassium permanganate solution, in the presence of acid.

The oxidation equation is:

Fe²⁺(aq)→Fe³⁺ + e^-

The reduction equation is:

MnO₄^-(aq) + 8H⁺(aq) + 5e^-→Mn²⁺(aq) + 4H₂O(l)

From these you can work out the balanced redox equation:

MnO₄^-(aq) + 5Fe²⁺(aq) + 8H⁺(aq)→Mn²⁺(aq) + 5Fe³⁺ + 4H₂O(l)

Answer:

• C_x = ?
• V_x = 20 cm³
• n_x = 1
• C_y = 0.02 mol l⁼¹
• V_y = 23.5 cm³
• n_y = 5

Where, x refers to Fe²⁺(aq) and y refers to MnO₄^-(aq)

C_x = (C_y × V_y × n_y) ÷ (V_x × n_x)

C_x = (0.02 × 23.5 × 5) ÷ (20 × 1)

C_x = 0.12 mol l^-1

Answer: The concentration of the iron(II) sulphate solution is 0.12 mol l^-1

Example:

In an experiment to find the concentration of sodium sulphite solution, 25 cm³ samples of the solution were titrated with 0.01 mol l^-1 acidified potassium permanganate solution. The results were as follows:

The redox equation for the reaction is:

2MnO₄^-(aq) + 6H⁺(aq) + 5SO₃^2-(aq)→2Mn²⁺(aq) + 5SO₄^2-(aq) + 3H₂O(l)

Calculate the concentration of the sulphite solution.

Step 1

In order to work out n _x and n _y the oxidation and reduction equation have to be written - they are found on page 11 of the SQA Higher Chemistry data booklet.

Reduction: MnO₄^-(aq) + 8H⁺(aq) + 5e^-→Mn²⁺(aq) + 4H₂O(l)

Oxidation: SO₃^2-(aq) + H₂O(l)→SO₄^2-(aq) + 2H⁺(aq) + 2e^-

From this n _x = 2 and n _y = 5

Step 2

To find V _y the average titre must be calculated. The first volume is always omitted. The other two titres are then averaged:

(15.3 + 15.2) ÷ 2 = 30.5 ÷ 2 = 15.25 cm³

So, V _y = 15.25 cm³

Step 3

From the question, V _x = 25 cm³ and C _y = 0.01 mol l^-1

Step 4

C _x can be calculated from:

C_x = (C_y × V_y × n_y) ÷ (V_x × n_x)

C_x = (0.01 × 15.25 × 5) ÷ (25 × 2)

C_x = 0.015 mol l^-1

Answer: The concentration of sodium sulphite solution is 0.015 mol l^-1