Volumetric Problems in Redox Reactions (Leaving Cert Chemistry): Revision Notes

Worked Example: Standardising a Permanganate Solution

The problem: A student prepares a 0.12 M ammonium iron(II) sulphate solution and uses this to standardise a potassium permanganate solution. In the titration, 25 cm³ of the ammonium iron(II) sulphate solution is titrated against the potassium permanganate solution. The equation for the reaction is:

$\text{MnO}_4^- + 8\text{H}^+ + 5\text{Fe}^{2+} \rightarrow \text{Mn}^{2+} + 5\text{Fe}^{3+} + 4\text{H}_2\text{O}$

The average titration figure was 23.75 cm³. Calculate the concentration of the potassium permanganate solution in both moles per litre and grammes per litre.

Solution approach:

Step 1: Calculate the molarity of the KMnO₄ solution (oxidising agent)

From the balanced equation, we can see that one mole of MnO₄⁻ reacts with 5 moles of Fe²⁺, so $n_1 = 1$ and $n_2 = 5$.

Given information:

• $V_1 = 23.75$ cm³
• $M_1 = ?$ (what we're finding)
• $n_1 = 1$
• $V_2 = 25$ cm³
• $M_2 = 0.12$ M
• $n_2 = 5$

Using our formula: $V_1 \times \frac{M_1}{n_1} = V_2 \times \frac{M_2}{n_2}$

$23.75 \times \frac{M_1}{1} = 25 \times \frac{0.12}{5}$

$M_1 = \frac{25 \times 0.12}{5 \times 23.75} = 0.0253 \text{ moles per litre}$

Step 2: Convert to grammes per litre First, we calculate the relative molecular mass of KMnO₄ = 158.04 g/mol

Concentration in g/L = $0.0253 \times 158.04 = 3.998$ g/L

Answer: The concentration of KMnO₄ solution is (a) 0.0253 moles per litre, (b) 3.998 grammes per litre.

Worked Example: Analysing Iron(II) Sulphate Crystals

The problem: A sample of iron(II) sulphate crystals FeSO₄·xH₂O was analysed to find the value of x. 6.42 g of the crystals were dissolved in deionised water, and the resulting solution was made up to 250 cm³ in a volumetric flask. 25 cm³ samples of this solution were titrated against 0.018 M KMnO₄ solution. The balanced equation for the reaction is:

$\text{MnO}_4^- + 8\text{H}^+ + 5\text{Fe}^{2+} \rightarrow \text{Mn}^{2+} + 5\text{Fe}^{3+} + 4\text{H}_2\text{O}$

The average titration figure was 25.5 cm³. Calculate the value of x in the above formula.

Solution approach:

Step 1: Calculate the molarity of the FeSO₄ solution (reducing agent)

Given:

• $V_1 = 25.5$ cm³ (KMnO₄)
• $M_1 = 0.018$ M
• $n_1 = 1$
• $V_2 = 25$ cm³ (FeSO₄)
• $n_2 = 5$

Using $V_1 \times \frac{M_1}{n_1} = V_2 \times \frac{M_2}{n_2}$:

$25.5 \times \frac{0.018}{1} = 25 \times \frac{M_2}{5}$

$M_2 = \frac{25.5 \times 0.018 \times 5}{25} = 0.0918 \text{ moles/litre FeSO}_4$

This means we have $\frac{0.0918}{4} = 0.023$ moles in 250 cm³ of solution.

Step 2: Calculate the relative molecular mass of FeSO₄·xH₂O

From Step 1, 0.023 moles of FeSO₄·xH₂O has a mass of 6.42 g.

Therefore: 1 mole FeSO₄·xH₂O = $\frac{6.42}{0.023} = 279.13$ g So: 1 mole FeSO₄·xH₂O = 279.13 g

Step 3: Calculate the value of x

We know that:

• $M_r$ of FeSO₄ = 151.92
• $M_r$ of H₂O = 18
• $M_r$ of FeSO₄·xH₂O = 279.13

Therefore: $151.92 + (18x) = 279.13$ $18x = 279.13 - 151.92 = 127.21$ $x = \frac{127.21}{18} = 7$

Answer: x = 7. The formula is FeSO₄·7H₂O