The percentage by mass of iron in a steel wire is determined by a student - AQA - A-Level Chemistry - Question 5 - 2019 - Paper 3

To calculate the mean titre, we take the values from the titre readings:

• Titre 1: 22.90 cm³
• Titre 2: 22.70 cm³
• Titre 3: 22.60 cm³

First, we sum the titres:

Mean titre = ( \frac{22.90 + 22.70 + 22.60}{3} = \frac{68.20}{3} = 22.73 \text{ cm}^3 )

The overall ionic equation for the oxidation of Fe²⁺ by manganate(VII) ions can be described as:

$5Fe^{2+} + MnO_4^{-} + 8H^{+} \rightarrow 5Fe^{3+} + Mn^{2+} + 4H_2O$

The colour change seen at the end point of the titration is from green (due to Fe²⁺) to a pale pink or purple, indicating the presence of MnO₄⁻.

For taking the 25.0 cm³ portions, a pipette should be used. For adding the potassium manganate(VII) solution, a burette is needed.

To calculate the percentage uncertainty, we can use the following formula:

( ext{Percentage Uncertainty} = \left( \frac{\text{Uncertainty}}{\text{Mass}} \right) \times 100 % )

Given the mass is 680 mg (or 0.680 g) and the uncertainty is ±0.005 g,

[ \text{Percentage Uncertainty} = \left( \frac{0.005}{0.680} \right) \times 100 % \approx 0.735% ]