15.00 mL of 0.01200 M EDTA were used to titrate 50 mL aliquot solution containing iron (II). How much iron (II) is present in the solution?
The answer in the book is 201 ppm. I tried getting the moles of EDTA (n,EDTA = 1.8 x 10^-4 mol). I think they're in 1:1 ratio, or I guess this isn't a simple titration with only one reaction involved?? Please help, thank you!
Yes they are in the 1:1 ratio.
1.8E-4 mols EDTA = 1.8E-4 mols Fe
1.8E-4 mols Fe x atomic mass Fe = grams Fe.
Thus you have that many grams Fe in 50 g (50 mL solution) = 2.01E-4 g/g or 201 g/1 million grams = 201 ppm.
Thank you! I understand it now. :)
To determine the amount of iron (II) present in the solution, you need to use the stoichiometry of the reaction between iron (II) and EDTA. You are correct that this is not a simple 1:1 titration.
The balanced chemical equation for the reaction between iron (II) and EDTA is:
Fe2+ + EDTA4- → Fe(EDTA)2-
From the equation, we can see that 1 mole of iron (II) reacts with 1 mole of EDTA to form 1 mole of the ferroin complex, Fe(EDTA)2-.
First, calculate the moles of EDTA used in the titration:
moles of EDTA = volume of EDTA solution (in L) × concentration of EDTA (in mol/L)
moles of EDTA = 15.00 mL × (1 L / 1000 mL) × 0.01200 mol/L
moles of EDTA = 1.8 x 10^-4 mol
Since the stoichiometry of the reaction is 1:1, the moles of EDTA are equal to the moles of iron (II) in the solution.
Now, calculate the concentration of iron (II) in the original solution using the following equation:
concentration of iron (II) = moles of iron (II) / volume of original solution (in L)
You have a 50 mL aliquot of the original solution, but you need it in liters, so convert it:
volume of original solution = 50 mL × (1 L / 1000 mL)
volume of original solution = 0.05 L
concentration of iron (II) = 1.8 x 10^-4 mol / 0.05 L
concentration of iron (II) = 0.0036 mol/L
To convert this concentration to parts per million (ppm), multiply by 1,000,000:
ppm = concentration of iron (II) × 1,000,000
ppm = 0.0036 mol/L × 1,000,000
ppm = 3600 ppm
Therefore, the concentration of iron (II) in the original solution is 3600 ppm, not 201 ppm as stated in the book. Please double-check your calculations or the given answer.
To determine the amount of iron (II) present in the solution, we need to understand the reaction that occurs between EDTA (ethylenediaminetetraacetic acid) and iron (II). In this reaction, EDTA forms a complex with iron (II) ions.
The balanced chemical equation for the reaction is:
Fe2+ + EDTA → Fe-EDTA complex
From the given information, we know that 15.00 mL of 0.01200 M EDTA was used to titrate a 50 mL aliquot solution containing iron (II).
First, let's calculate the moles of EDTA used:
Moles of EDTA = Volume of EDTA (in liters) x Concentration of EDTA
= 15.00 mL x (1 L / 1000 mL) x 0.01200 M
= 1.8 x 10^(-4) moles of EDTA
As you correctly mentioned, the stoichiometry of the reaction suggests a 1:1 molar ratio between EDTA and iron (II) ions. Therefore, the moles of iron (II) present in the solution would be equal to the moles of EDTA used, which is 1.8 x 10^(-4) moles.
Now, let's calculate the concentration of iron (II) in the solution:
Concentration of iron (II) = Moles of iron (II) / Volume of solution (in liters)
= 1.8 x 10^(-4) moles / (50 mL x (1 L / 1000 mL))
= 0.0036 M
Finally, to express the concentration of iron (II) in parts per million (ppm), we can use the equation:
ppm = (Concentration of iron (II) / Total volume of solution) x 10^6
Plugging in the values, we get:
ppm = (0.0036 M / 0.050 L) x 10^6
= 72 x 10^3
= 72000 ppm
Therefore, the concentration of iron (II) in the solution is 72000 ppm, which is different from the answer provided in the book (201 ppm). It's possible that there may be additional information or calculations required to reach the correct answer.