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A 50.0 mL sample containing Cd2 and Mn2 was treated with 51.4 mL of 0.0600 M...

A 50.0 mL sample containing Cd2 and Mn2 was treated with 51.4 mL of 0.0600 M EDTA. Titration of the excess unreacted EDTA required 11.1 mL of 0.0190 M Ca2 . The Cd2 was displaced from EDTA by the addition of an excess of CN–. Titration of the newly freed EDTA required 11.3 mL of 0.0190 M Ca2 . What were the molarities of Cd2 and Mn2 in the original solution?

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Answer #1

Moles of EDTA = volume x molarity

= 0.0514 L X 0.0600 mol/L

= 0.003084 mol
Moles of Ca2+ = volume x molarity

= 0.0111 L X 0.0190

= 0.0002109 mol


moles Cd2+ - moles of Mn2+ = 0.003084 - 0.0002109

= 0.0028731 mol

In the last titration
Moles of Cd2+ = 0.0113 L x 0.0190 mol/L

= 0.0002147 mol

Moles of Cd2+ in original solution = 0.0002147 mol
Moles Mn2+ = 0.0028731 - 0.0002147

= 0.0026584 mol

Volume of sample = 0.0500 L

Molarity of Mn2+ = moles of Mn2+ / Volume of sample

[Mn2+] = 0.0026584 mol / 0.0500 L = 0.053168 M

Molarity of Cd2+ = moles of Cd2+ / Volume of sample

[Cd2+] = 0.0002147 mol / 0.0500 L = 0.004294 M

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