Calculating EDTA Titration Volume for CoSO4 Solution

[LakeMountD]

Homework Statement

A solution contains 1.694mg of CoSO4 (155.0 g/mol) per mL. Calculate the volume of 0.08640 M EDTA needed to titrate a 25 mL aliquot of this solution.

Homework Equations

Not really sure, the book is no help but this may help:

K(CoSO4) = [CoSO4] / ([CO2+] [SO42-])
K'CoSO4 = alpha4 * KCoSO4

The Attempt at a Solution

My book is absolutely horrible which is why I had to come to the forums for help but I am assuming I find the mol. of CoSO4, which was 2.73 x 10^-4 then assume that [CoSO4] ~= 4.37x10^-4

I think you use K'CoSO4 = alpha4 * KCoSO4, but again I can't figure it out :(

 

[LakeMountD]

Okay I think I might have figured it out. Get moles of CoSO4, then multiply that by the molar ratio of EDTA to CoSO4 then multiply by the concentration of EDTA?

 

[symbolipoint]

Superficially, best guess is that this should be a fairly uncomplicated 1 to 1 mole titration of the EDTA specie to the Co specie. If that "alpha" stuff is important in the exercise, then the superficial idea is not enough, and that some pH control might be needed. If you have the time, check an alternative quantitative book and ask your teacher.

 

[LakeMountD]

Superficially, best guess is that this should be a fairly uncomplicated 1 to 1 mole titration of the EDTA specie to the Co specie. If that "alpha" stuff is important in the exercise, then the superficial idea is not enough, and that some pH control might be needed. If you have the time, check an alternative quantitative book and ask your teacher.

Yeah that is the problem I am having. How do I know what the molar ratio of EDTA to any metal is? The book doesn't explain it in the examples, it just throws the ratio in there. It gives a 1:1 ratio to EDTA:Mg(NO2)2 but not sure about the others.

Thanks.

 

[symbolipoint]

How do I know 1:1 ? Habit! Not remember exactly how. Just assume 1:1 mole ratio. The EDTA is not just a ligand; it is a CHELATOR. One EDTA will coordinate bond with one metal ion. (mole-wise, that is)