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?

Question

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!

Answer 1

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.

Answer 2

Thank you! I understand it now. :)