When the first portion of 100 mL sample, which is known to contain iron (III) and calcium (II) and whose density is 1.01 g / cm3, is titrated at pH 3.0, 3.5 mL of 0.0100 M EDTA solution is consumed. When the second portion of 100 mL is titrated at pH 10.0, 25.4 mL EDTA solution is consumed. Find the amount of Fe3+ and Ca2+ in this solution as ppm.
Fe.
mols = L x M = 0.003.5 x 0.01 = 0.000035
g Fe = mols Fe x molar mass Fe = 0.000035 x 55.85 = 0.00195 g/100 mL.
That's in 100 mL of 0.1 L so 0.00195/.1 = 0.0195 g/L or 19.5 ppmv(ppm volume)
Since you give a density of the solution I assume you want ppmw (weight/mass) . 19.5 g Fe/101 g soln = 19.3 ppmw.
Ca is done the same way.
To find the amount of Fe3+ and Ca2+ in the solution as ppm, we need to calculate the concentration of these ions first and then convert it to parts per million (ppm).
Given information:
Initial volume of solution = 100 mL
Volume of EDTA consumed at pH 3.0 = 3.5 mL
Volume of EDTA consumed at pH 10.0 = 25.4 mL
Concentration of EDTA solution = 0.0100 M
Density of solution = 1.01 g/cm^3
Step 1: Calculate the moles of EDTA consumed at each pH.
Moles of EDTA at pH 3.0 = volume × concentration = 3.5 mL × 0.0100 M = 0.0350 mmol
Moles of EDTA at pH 10.0 = 25.4 mL × 0.0100 M = 0.254 mmol
Step 2: Calculate the moles of metal ions (Fe3+ and Ca2+) reacting with EDTA at each pH.
Since EDTA forms 1:1 complexes with both Fe3+ and Ca2+, the moles of metal ions reacted will be equal to the moles of EDTA consumed.
Moles of Fe3+ at pH 3.0 = 0.0350 mmol
Moles of Fe3+ at pH 10.0 = 0.0350 mmol
Moles of Ca2+ at pH 3.0 = 0.0350 mmol
Moles of Ca2+ at pH 10.0 = 0.219 mmol (calculated as the difference between the total moles of EDTA consumed and the moles of Fe3+)
Step 3: Calculate the mass of Fe3+ and Ca2+ using their respective molar masses.
Molar mass of Fe3+ = 55.85 g/mol
Molar mass of Ca2+ = 40.08 g/mol
Mass of Fe3+ at pH 3.0 = moles × molar mass = 0.0350 mmol × 55.85 g/mol = 1.94 mg
Mass of Fe3+ at pH 10.0 = 0.0350 mmol × 55.85 g/mol = 1.94 mg
Mass of Ca2+ at pH 3.0 = 0.0350 mmol × 40.08 g/mol = 1.40 mg
Mass of Ca2+ at pH 10.0 = 0.219 mmol × 40.08 g/mol = 8.78 mg
Step 4: Convert the mass of Fe3+ and Ca2+ to ppm.
1 ppm = 1 mg/L = 1 mg/1000 mL
Concentration of Fe3+ at pH 3.0 = (Mass of Fe3+ at pH 3.0 / Volume of solution) × 1000 = (1.94 mg / 100 mL) × 1000 = 19.4 ppm
Concentration of Fe3+ at pH 10.0 = (Mass of Fe3+ at pH 10.0 / Volume of solution) × 1000 = (1.94 mg / 100 mL) × 1000 = 19.4 ppm
Concentration of Ca2+ at pH 3.0 = (Mass of Ca2+ at pH 3.0 / Volume of solution) × 1000 = (1.40 mg / 100 mL) × 1000 = 14.0 ppm
Concentration of Ca2+ at pH 10.0 = (Mass of Ca2+ at pH 10.0 / Volume of solution) × 1000 = (8.78 mg / 100 mL) × 1000 = 87.8 ppm
Therefore, the amount of Fe3+ and Ca2+ in this solution is as follows:
Fe3+ concentration: 19.4 ppm at pH 3.0 and 19.4 ppm at pH 10.0
Ca2+ concentration: 14.0 ppm at pH 3.0 and 87.8 ppm at pH 10.0.
To find the amount of Fe3+ and Ca2+ in the solution as parts per million (ppm), we need to calculate the concentration of each ion in the solution before and after the titrations. Let's break down the steps to find the concentration of each ion:
1. Calculate the number of moles of EDTA (ethylenediaminetetraacetic acid) used in each titration.
- In the first titration: moles of EDTA = volume of EDTA solution (in L) × concentration of EDTA (in mol/L)
Moles of EDTA = 3.5 mL × (1 L / 1000 mL) × 0.0100 mol/L = 0.000035 mol
- In the second titration: moles of EDTA = 25.4 mL × (1 L / 1000 mL) × 0.0100 mol/L = 0.000254 mol
2. Since the stoichiometric ratio between Fe3+ and EDTA is 1:1, the number of moles of Fe3+ reacted with EDTA in both titrations is equal to the number of moles of EDTA used.
- Moles of Fe3+ in the first titration = moles of EDTA = 0.000035 mol
- Moles of Fe3+ in the second titration = moles of EDTA = 0.000254 mol
3. Since the stoichiometric ratio between Ca2+ and EDTA is 1:1, the number of moles of Ca2+ reacted with EDTA can be calculated from the difference in moles of EDTA used between the two titrations.
- Moles of Ca2+ = Moles of EDTA in the second titration - Moles of EDTA in the first titration
Moles of Ca2+ = 0.000254 mol - 0.000035 mol = 0.000219 mol
4. Calculate the concentrations of Fe3+ and Ca2+ in the original solution.
- Concentration of Fe3+ = Moles of Fe3+ / Volume of solution (in L)
Volume of solution = 100 mL × (1 L / 1000 mL) = 0.100 L
Concentration of Fe3+ = 0.000035 mol / 0.100 L = 0.000350 mol/L
- Concentration of Ca2+ = Moles of Ca2+ / Volume of solution (in L)
Concentration of Ca2+ = 0.000219 mol / 0.100 L = 0.00219 mol/L
5. Convert the concentrations to parts per million (ppm):
- ppm = (Concentration in mol/L) × (Molar mass in g/mol) × 10^6
- Molar mass of Fe3+ = 55.845 g/mol
- Molar mass of Ca2+ = 40.078 g/mol
- ppm of Fe3+ = 0.000350 mol/L × 55.845 g/mol × 10^6 = 19.5 ppm
- ppm of Ca2+ = 0.00219 mol/L × 40.078 g/mol × 10^6 = 87.7 ppm
Therefore, the amount of Fe3+ in the solution is 19.5 ppm, and the amount of Ca2+ is 87.7 ppm.