1. The Fe²⁺ content of iron tablets was determined by titration with a freshly standardised solution of potassium manganate(VII), KMnO₄ - Leaving Cert Chemistry - Question 1 - 2009

Potassium manganate(VII) solutions must be standardised due to their susceptibility to changes in concentration caused by factors such as evaporation and degradation upon storage. It is necessary to standardise just before use to ensure the accuracy of titration results. The reagent used for this purpose is ammonium iron(II) sulfate, which acts as a reducing agent that allows for precise determination of the KMnO₄ concentration.

To prepare the 250 cm³ of Fe²⁺ solution, the five iron tablets were crushed to a fine powder and then dissolved in water. The total mass of iron from the five tablets is calculated as:

$5 \times 0.325 \text{ g} = 1.625 \text{ g}$

This mixture was then transferred to a volumetric flask and diluted with water to reach the 250 cm³ mark.

Dilute sulfuric acid was added to ensure that the iron(II) ions remain in the ferrous state during the titration and to avoid oxidation, which would convert them to ferric ions.

Dilute sulfuric acid is added to ensure that the iron(II) ions do not oxidize to iron(III) during the titration process. This reaction requires an acidic medium for the reduction of MnO₄^- to Mn²⁺. The presence of acid helps to maintain the required pH level for the reactions to proceed without any interference.

The molarity (M) of the Fe²⁺ solution can be calculated using the titration data provided. Since 18.75 cm³ of 0.01 M KMnO₄ was used:

1. Calculate moles of KMnO₄:

   $ext{Moles of KMnO}_4 = 0.01 \times \frac{18.75}{1000} = 0.0001875$

2. The stoichiometry shows that 1 mole of KMnO₄ reacts with 5 moles of Fe²⁺:

   $\text{Moles of Fe}^{2+} = 5 \times 0.0001875 = 0.0009375$

3. Since these moles are in 25 cm³:

   $\text{Molarity} = \frac{0.0009375}{\frac{25}{1000}} = 0.0375 ext{ M}$

To find the mass of iron in the solution:

1. Moles of Fe²⁺ in 250 cm³:

   $0.0375 \times \frac{250}{1000} = 0.009375 ext{ moles}$

2. Using the molar mass of iron (approximately 56 g/mol):

   $\text{Mass of Fe} = 0.009375 \times 56 = 0.525 ext{ g}$

To find the percentage by mass of iron in the tablets:

1. Total mass of iron from five tablets:

   $\text{Total mass of tablets} = 5 \times 0.325 = 1.625 ext{ g}$

2. Using the mass of iron calculated:

   $\text{Percentage by mass} = \left( \frac{0.525}{1.625} \right) \times 100 = 32.3\%$