How do I prepare a solution of 0.0025M Fe3+ by dissolving the appropriate amount of ferric ammonium sulfate in 500 cm^3 of 0.0025M sulfuric acid?
Ferric ammonium sulfate is Fe3(NH4)2(SO4)3. Look up the molar mass,multiply by 0.0025 to get the moles needed for a liter of solution. Since there are 3 moles Fe^+3 per mole of Fe3(NH4)2(SO4)3, divide by 3 to find the amount necessary for moles Fe^+3, then divide by 2 since you only want to prepare 500 cc solution.
To prepare a solution of 0.0025M Fe3+ by dissolving ferric ammonium sulfate in sulfuric acid, you will need to follow these steps:
1. Determine the molecular weight of ferric ammonium sulfate (Fe(NH4)2(SO4)2) and ferric ion (Fe3+). The molecular weight of ferric ammonium sulfate is 392.14 g/mol, and the ferric ion has a molecular weight of 55.85 g/mol.
2. Calculate the moles of ferric ion required by using the molarity (M) and volume of the solution. Given that the desired molarity is 0.0025M and the volume is 500 cm^3 (which is equivalent to 0.5 L), use the formula:
Moles of Fe3+ = Molarity * Volume
Moles of Fe3+ = 0.0025 mol/L * 0.5 L
Moles of Fe3+ = 0.00125 moles
3. Determine the stoichiometric ratio between ferric ammonium sulfate and ferric ion. By examining the chemical formula of ferric ammonium sulfate (Fe(NH4)2(SO4)2), you can see that one mole of ferric ammonium sulfate contains one mole of ferric ion. Therefore, you will need 0.00125 moles of ferric ammonium sulfate.
4. Calculate the mass of ferric ammonium sulfate required by multiplying the number of moles by its molecular weight:
Mass of ferric ammonium sulfate = Moles of ferric ammonium sulfate * Molecular weight
Mass of ferric ammonium sulfate = 0.00125 mol * 392.14 g/mol
Mass of ferric ammonium sulfate = 0.49 g (rounded to two decimal places)
Therefore, to prepare a 0.0025M Fe3+ solution, you would need approximately 0.49 grams of ferric ammonium sulfate dissolved in 500 cm^3 of 0.0025M sulfuric acid.