Question

A sample containing Cd²⁺ and Mn²⁺ was treated with 60.1 ml of 0.0400M EDTA. Titration of the excess unreacted EDTA required 16.8ml of 0.013M Ca²⁺. The Cd²⁺ was displaced from EDTA by the addition of an excess of CN^-. Titration of the newly freed EDTA required 10.4ml of 0.013M Ca²⁼. What are the concentrations of Cd²⁺ and Mn²⁺ in the original solution?

Answer 1

EDTA reactions with metal ions are 1:1 :

Cd²⁺ + EDTA <==> (Cd-EDTA)

Mn²⁺ + EDTA <==> (Mn-EDTA)

Ca²⁺ + EDTA <==> (Ca-EDTA)

This experiment involves back-titration as masking.

Sample containing Cd²⁺ and Mn²⁺ was treated with 60.1 ml of 0.0400M EDTA.

moles of EDTA =( 0.0400 mol/L*60.1 ml) / 1000 ml/L = 2.404*10^-3 mole

Titration of the excess unreacted EDTA required 16.8ml of 0.013M Ca²⁺.

So, moles of Ca²⁺ of reacted = ( 0.013 mol/L* 16.8 ml) / 1000 ml/L =2.184*10^-4 mole

so moles of ( Cd²⁺ +Mn²⁺ ) = 2.1856*10^-3 mole

addition of excess of CN- , displaced Cd²⁺ from Cd-EDTA complex and liberated EDTA, free EDTA was back titrated with   Ca²⁺ :

moles of Ca²⁺ reacted = moles EDTA liberated = moles of Cd²⁺ present

=  ( 0.013 mol/L* 10.4 ml) / 1000 ml/L =1.352*10^-4 mole

so, moles of Cd²⁺ present in initial solution : 1.352*10^-4 moles

and moles of Mn²⁺ present in initial solution : = 2.1856*10^-3 mole​​​​​​​ - 1.352*10^-4 mole​​​​​​​

= 2.0504*10^-3 moles

(you can calculate respective concentration if you volume of sample initially reacted )