Calculating the Volume of a 0.4 M FeCl3·6H2O Solution Required to Contain 600 mg of Fe3

When working with chemical solutions in a laboratory setting, it is often necessary to determine the volume of a given reagent needed to achieve a certain concentration of an ion of interest. This article delves into the process of calculating the volume of a 0.4 M iron(III) chloride hexahydrate (FeCl3·6H2O) solution required to contain 600 mg of Fe3 ion. Understanding this calculation is crucial for chemists and students involved in quantitative chemical analysis.

Step-by-Step Calculation

The first step in the calculation is to determine the number of moles of Fe3 ions present in 600 mg (0.6 g) of iron(III) chloride hexahydrate. Here is the step-by-step procedure:

1. Calculate the Moles of Fe3 in 600 mg of FeCl3·6H2O

Fe (iron) has a molar mass of approximately 55.85 g/mol. We start by converting the given mass in milligrams to grams:

600 mg 0.6 g

Now, we can determine the moles of Fe:

moles of Fe3 (mass / molar mass) (0.6 g / 55.85 g/mole) ≈ 0.01073 moles

2. Determine the Moles of FeCl3·6H2O Needed

Each mole of FeCl3·6H2O contains one mole of Fe3 ions. Therefore, the moles of FeCl3·6H2O required are also 0.01073 moles.

3. Calculate the Volume of the 0.4 M Solution Needed

The concentration (C) of the solution is given as 0.4 M, which means 0.4 moles of solute per liter of solution. We can use the formula for concentration:

C n / V

Rearranging the formula to find the volume (V) gives:

V n / C

Substituting the values:

V 0.01073 moles / 0.4 moles/liters 0.026825 liters

To convert this to milliliters:

V ≈ 0.026825 liters × 1000 milliliters/liter ≈ 26.83 milliliters

Therefore, you will need approximately 26.83 milliliters of 0.4 M FeCl3·6H2O to contain 600 mg of Fe3 ions.

Alternative Calculations for Clarity

For a clearer understanding, let's consider an alternative approach:

1. Molar Mass Calculation

The molar mass of FeCl3·6H2O can be calculated as:

Molar mass of FeCl3·6H2O 55.85 g/mol (Fe) 3 × 35.45 g/mol (Cl) 6 × 18 g/mol (H2O)

55.85 106.35 108.0 270.2 g/mol

Thus, 1 liter of a 1 M solution contains 270.2 g of FeCl3·6H2O.

For a 0.4 M solution:

Amount of the salt in 1 liter of 0.4M solution 270.2 g × 0.4 108.12 g

Since one mole of FeCl3·6H2O contains only one mole of Fe3 , 108.12 g of the salt will contain:

108.12 g / 270.2 g × 55.85 g 22.34 g of Fe3

Therefore, 1 liter (1000 ml) of a 0.4 M FeCl3·6H2O solution contains 22.34 g of Fe3 .

2. Volume Calculation for 600 mg of Fe3

To determine the volume needed to contain 600 mg (0.6 g) of Fe3 :

V (0.6 g / 22.34 g) × 1000 ml ≈ 26.9 ml

This aligns with the initial calculation, providing further confirmation of the required volume.

Conclusion

Understanding the calculation of the volume of solution required to contain a specific amount of a given ion is fundamental in quantitative chemical analysis. The steps outlined in this article ensure precision and accuracy in achieving the desired concentration. Whether you are a chemist, a student, or a professional in the laboratory, these calculations are indispensable for conducting precise experiments and analyses.

Keywords: FeCl3·6H2O, Molarity, Volume Calculation