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 determine the amount of iron (Fe) that dissolves and forms a soluble compound.
First, let's calculate the moles of ferric phosphate added to the solution:
Molar mass of ferric phosphate (FePO4) = (55.845 g/mol) + (94.971 g/mol) = 150.816 g/mol
Moles of ferric phosphate = mass / molar mass = 2.4 g / 150.816 g/mol ≈ 0.01591 mol
Since 1 mole of ferric phosphate contains 1 mole of iron (Fe), the moles of iron in the solution will also be approximately 0.01591 mol.
Next, we need to calculate the volume of water containing phosphate that was added to the solution to reach a final volume of 1.00 L. Since the phosphate concentration is given in mg/L, we'll need to convert it to g/L:
Phosphate concentration = 1.0 mg/L = 1.0 × 10^-3 g/L
Let V be the volume of water added to the solution in liters.
The equation for calculating molarity is:
Molarity (M) = moles of solute / volume of solution (in liters)
We can rearrange this equation to solve for the volume of solution:
Volume of solution (in liters) = moles of solute / Molarity
Since the moles of solute (phosphate) are not given, we'll assume it's the same as the moles of iron (since ferric phosphate contains 1 mole of iron for every mole of phosphate).
Therefore, we can write the equation as follows:
V = 0.01591 mol / (1.0 × 10^-3 mol/L)
V ≈ 15.91 L
To reach the final volume of 1.00 L, we must have added 15.91 L - 1.00 L = 14.91 L of water.
Now, we can calculate the concentration of soluble iron in the solution:
Concentration of soluble iron = moles of iron / final volume of solution
Concentration of soluble iron = 0.01591 mol / 1.00 L
Therefore, the concentration of soluble iron in this solution is approximately 0.01591 mol/L.
Note: The temperature (25°C) does not affect the concentration calculation for this specific scenario.