[rproctor]

A 50.0 mL sample containing Cd2 and Mn2 was treated with 45.4 mL of 0.0600 M EDTA. Titration of the excess unreacted EDTA required 18.3 mL of 0.0300 M Ca2 . The Cd2 was displaced from EDTA by the addition of an excess of CN–. Titration of the newly freed EDTA required 14.3 mL of 0.0300 M Ca2 . What were the molarities of Cd2 and Mn2 in the original solution?

I am trying to figure out this problem, but I am having some difficulty. I know to first calculate the mmol of Cd2 and Mn2 in the mixture by subtracting the number of mmol of EDTA consumed in the back titration with Ca2 from the total mmol of EDTA. I get 2.17 mmol.
I cannot figure out where to go from here. Can anyone help me?

[Borek]

Start writing all reaction equations.