Iron tablets may be used in the treatment of anemia - Leaving Cert Chemistry - Question 1 - 2003

1. Weigh four iron tablets accurately to determine their total mass (4 x 0.360 g).
2. Crush the tablets into a fine powder using a mortar and pestle.
3. Transfer the powdered tablets into a beaker.
4. Measure approximately 15 cm³ of dilute sulfuric acid and add it to the beaker with powdered tablets to dissolve them.
5. Use a wash bottle to rinse any residue from the mortar into the beaker.
6. Once dissolved, transfer the solution into a 250 cm³ volumetric flask.
7. Rinse the beaker with deionised water and add this wash to the volumetric flask.
8. Add deionised water to the volumetric flask until the total volume reaches the 250 cm³ mark. Invert the flask several times to mix thoroughly.

More dilute sulfuric acid was added before the titrations to ensure that the iron(II) ions remain in their reduced state during the reaction. The additional acid provides sufficient H⁺ ions required for the complete reduction of manganese(VII) ions to manganese(II) ions in acidic medium, preventing interference by oxygen or other oxidizing agents.

The end-point of the titration was detected by the first permanent change in color of the solution from green to pink/purple. This color change indicates that all the iron(II) ions have been oxidized to iron(III) ions, and any excess potassium manganate(VII) is present, resulting in the pink color.

Using the titration data:

1. Each mole of KMnO₄ reacts with 5 moles of Fe²⁺.
2. The moles of KMnO₄ used are calculated from its concentration and volume: $\text{Moles of KMnO₄} = 0.010 imes \frac{13.9}{1000} = 0.000139.$
3. Moles of Fe²⁺ can be calculated as follows: $\text{Moles of Fe²⁺} = 5 \times 0.000139 = 0.000695.$
4. Since this was from 25 cm³ of iron(II) solution, the concentration is: $C = \frac{0.000695}{25/1000} = 0.0278 \text{ mol/L}.$

To find the mass of iron(II) in one tablet:

1. We already calculated the concentration of iron(II) in the solution to be 0.0278 mol/L.
2. Molar mass of iron(II) is approximately 55.85 g/mol. Therefore, the mass of iron(II) in 250 cm³ is: $\text{Mass} = 0.0278 \times 0.250 \times 55.85 \, \text{g} = 0.3866 \, \text{g}.$
3. The mass of iron(II) in one tablet is then: $\frac{0.3866}{4} = 0.09665 \, \text{g}.$

To find the percentage by mass of iron(II) in each tablet:

1. The mass of iron(II) in one tablet is approximately 0.09665 g.
2. The mass of each tablet is 0.360 g. Thus, the percentage is: $\text{Percentage} = \left( \frac{0.09665}{0.360} \right) \times 100 = 26.83\%.$
3. Rounding to two decimal places, the percentage by mass of iron(II) in each tablet is approximately 26.83%.