posted by Chris on .
A 50.0 mL sample containing Cd2 and Mn2 was treated with 49.1 mL of 0.0700 M EDTA. Titration of the excess unreacted EDTA required 17.5 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 10.9 mL of 0.0300 M Ca2 . What were the molarities of Cd2 and Mn2 in the original solution?
By taking the total amount of EDTA (3.44 mmol), and subtracting the consumed portion (0.525 mmol), I got 2.91 mmol as my Cd2 + Mn2 mmol.
How do I take this, and calculate Cd2 by itself, and the Mn2 from that?
Total = 3.44-0.525 = ?
Mn^2+ alone = 10.9 x 0.03 = ? mmol.
I took that answer (0.327 mmol) and divided it by 50ml to get my Molarity. But this does not seem correct. What am I doing wrong?
Probably not anything. I didn't read the problem right. The 10.9 mL x 0.03 is "the newly freed EDTA from the CN^- complex with Cd^2+) so this is the Cd and the difference is Mn.
That is, Cd alone is the 10.9 x 0.03 mmol and convert that to M as usual. Mn is the difference between total and Cd.