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1. The Fe<sup>2+</sup> content of iron tablets was determined by titration with a freshly standardised solution of potassium manganate(VII), KMnO<sub>4</sub> - Leaving Cert Chemistry - Question 1 - 2009

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1.-The-Fe<sup>2+</sup>-content-of-iron-tablets-was-determined-by-titration-with-a-freshly-standardised-solution-of-potassium-manganate(VII),-KMnO<sub>4</sub>-Leaving Cert Chemistry-Question 1-2009.png

1. The Fe<sup>2+</sup> content of iron tablets was determined by titration with a freshly standardised solution of potassium manganate(VII), KMnO<sub>4</sub>. The e... show full transcript

Worked Solution & Example Answer:1. The Fe<sup>2+</sup> content of iron tablets was determined by titration with a freshly standardised solution of potassium manganate(VII), KMnO<sub>4</sub> - Leaving Cert Chemistry - Question 1 - 2009

Step 1

Why are iron tablets sometimes medically prescribed?

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Answer

Iron tablets are prescribed primarily to prevent anemia, particularly to address conditions such as hemoglobin deficiency which is crucial for oxygen transport in the blood. They are essential for maintaining adequate iron levels in the body, thereby preventing symptoms related to blood disorders.

Step 2

Why must potassium manganate(VII) solutions be standardised? Why was it necessary to standardise the potassium manganate(VII) solution immediately before use in the titration? What reagent is used for this purpose?

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Answer

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 KMnO<sub>4</sub> concentration.

Step 3

Describe how exactly 250 cm<sup>3</sup> of Fe<sup>2+</sup> solution was prepared from five iron tablets, each of mass 0.325 g. Why was some dilute sulfuric acid used in making this solution?

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Answer

To prepare the 250 cm<sup>3</sup> of Fe<sup>2+</sup> 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:

5×0.325 g=1.625 g5 \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 cm<sup>3</sup> 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.

Step 4

Explain why dilute sulfuric acid must be added to the titration flask before each titration is carried out.

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Answer

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 MnO<sub>4</sub><sup>-</sup> to Mn<sup>2+</sup>. The presence of acid helps to maintain the required pH level for the reactions to proceed without any interference.

Step 5

Calculate the molarity of the Fe<sup>2+</sup> solution.

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Answer

The molarity (M) of the Fe<sup>2+</sup> solution can be calculated using the titration data provided. Since 18.75 cm<sup>3</sup> of 0.01 M KMnO<sub>4</sub> was used:

  1. Calculate moles of KMnO<sub>4</sub>:

    extMolesofKMnO4=0.01×18.751000=0.0001875 ext{Moles of KMnO}_4 = 0.01 \times \frac{18.75}{1000} = 0.0001875

  2. The stoichiometry shows that 1 mole of KMnO<sub>4</sub> reacts with 5 moles of Fe<sup>2+</sup>:

    Moles of Fe2+=5×0.0001875=0.0009375\text{Moles of Fe}^{2+} = 5 \times 0.0001875 = 0.0009375

  3. Since these moles are in 25 cm<sup>3</sup>:

    Molarity=0.0009375251000=0.0375extM\text{Molarity} = \frac{0.0009375}{\frac{25}{1000}} = 0.0375 ext{ M}

Step 6

Calculate the mass of iron in the 250 cm<sup>3</sup> solution.

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Answer

To find the mass of iron in the solution:

  1. Moles of Fe<sup>2+</sup> in 250 cm<sup>3</sup>:

    0.0375×2501000=0.009375extmoles0.0375 \times \frac{250}{1000} = 0.009375 ext{ moles}

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

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

Step 7

Calculate the percentage by mass of iron in the tablets.

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Answer

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

  1. Total mass of iron from five tablets:

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

  2. Using the mass of iron calculated:

    Percentage by mass=(0.5251.625)×100=32.3%\text{Percentage by mass} = \left( \frac{0.525}{1.625} \right) \times 100 = 32.3\%

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