a ferric phosphate solution is prepared by adding 2.4 g of ferric phosphate to a volumetric flask and bringing the final volume to 1.00L by adding water having a phosphate concentration of 1.0 mg/L. what is the concentration of soluble iron in this solution? (Temperature is 25C)
To find the concentration of soluble iron in the ferric phosphate solution, we need to consider the solubility equilibrium of ferric phosphate.
The solubility equilibrium is given by the equation:
FePO₄(s) ⇌ Fe³⁺(aq) + PO₄³⁻(aq)
According to the balanced equation, 1 mol of ferric phosphate dissolves to give 1 mol of Fe³⁺ ions and 1 mol of PO₄³⁻ ions.
We are given that 2.4 g of ferric phosphate is added to a volumetric flask and brought to a final volume of 1.00 L. The molar mass of ferric phosphate (FePO₄) is 150.82 g/mol.
To find the moles of ferric phosphate, we can use the formula:
moles = mass / molar mass
moles = 2.4 g / 150.82 g/mol ≈ 0.0159 mol
Since 1 mol of ferric phosphate dissociates to give 1 mol of Fe³⁺ ions, the concentration of Fe³⁺ ions is also 0.0159 mol/L.
However, we need to take into account the dilution of the water added to the volumetric flask, which has a phosphate concentration of 1.0 mg/L.
To find the concentration of phosphate ions in the final solution, we can use the formula:
concentration = mass / molar mass
concentration = 1.0 mg/L / 94.97 g/mol = 0.0105 mmol/L
Since 1 mol of ferric phosphate dissociates to give 1 mol of PO₄³⁻ ions, the concentration of PO₄³⁻ ions is also 0.0105 mmol/L.
Given that the stoichiometry of the balanced equation is 1:1, the concentration of soluble iron (Fe³⁺) in the solution is also 0.0105 mmol/L.
Please note that mmol/L is the same as μM (micromolar) concentration unit.
Therefore, the concentration of soluble iron in the ferric phosphate solution is approximately 0.0105 mmol/L or 0.0105 μM.