A50.0 ml sample containing cd2+ and mn2+ was treated with 58.4 ml of 0.0400 m edta . titration of the excess unreacted edta required 15.9 ml of 0.0130 m ca2+ . the cd2+ was displaced from edta by the addition of an excess of cn− . titration of the newly freed edta required 22.4 ml of 0.0130 m ca2+ . what are the concentrations of cd2+ and mn2+ in the original solution?

[Cd²⁺] = 5.82x10⁻³ M

[Mn²⁺] = 0.037M

Explanation:

The titration of EDTA with Cd²⁺, Mn²⁺ and Ca²⁺ are based in the formation of complexes with a relation 1:1, EDTA: metal, so from the titration of the excess unreacted EDTA with Ca²⁺ we can calculate the EDTA moles by:

The moles of EDTA are:

     (1)

where : is the total moles of EDTA, : is the EDTA excess moles and : is the EDTA moles that react with Cd²⁺ and Mn²⁺

Hence the EDTA moles that react with Cd²⁺ and Mn²⁺ are equal to the moles of both metals:

  (2)

From equation (1) we can find the :

    (3)

Now, from the titration of the EDTA free of Cd²⁺ with Ca²⁺, we can calculate the moles of EDTA that react with the Cd²⁺ and hence the Cd²⁺ moles:

Having the Cd²⁺ moles we can now calculate the Mn²⁺ moles using equation (3):

Now, with the Cd²⁺ moles and Mn²⁺ moles we can find their concentrations in the original solution:

I hope it helps you!

it would be a gas formation

i do not get it   at all