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May 27, 2015

Homework Help: Chemistry

Posted by Chris on Thursday, October 25, 2012 at 11:26pm.

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?

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