Question

A 50.0 mL sample containing Cd2+ and Mn2+ was treated with 57.2 mL of 0.0700 M EDTA. Titration of the excess unreacted EDTA required 14.6 mL of 0.0150 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 0.0150 M Ca2+. What are the concentrations of Cd2+ and Mn2+ in the original solution? concentration: M Mn2+ concentration: M Cd2+

Answer 1

First calculate total moles of EDTA added to the solution.

We have , [ EDTA ] = No. of moles of EDTA / volume of solution in L

$\therefore$ Total no. of moles of EDTA added to the solution = [ EDTA ] ( volume of solution in L )

We have, [ EDTA ] = 0.0700 M , Volume of EDTA solution = 57.2 ml = 0.0572 L

$\therefore$Total no. of moles of EDTA added to the solution = 0.0700 mol / L $\times$ 0.0572 L = 0.004004 mol ----->( 1)

Now, calculate moles of Excess EDTA in the solution.

Consider reaction, Ca ²⁺ + EDTA  $\rightleftharpoons$ [ CaEDTA ] ^2-

From reaction, 1 mol Ca ²⁺$\equiv$ 1 mol EDTA

$\therefore$ No. of moles of Ca ²⁺ = 0.0150 mol / L $\times$ 0.0146 L = 0.000219 mol

$\therefore$No. of moles of excess EDTA = 0.000219 mol -------------------> (2)

From above calculations we can calculate moles of EDTA consumed by ( Cd ²⁺ + Mn ²⁺ ) .

Moles of EDTA consumed by ( Cd ²⁺ + Mn ²⁺ ) =  Total no. of moles of EDTA added to the solution - No. of moles of excess EDTA

Moles of EDTA consumed by ( Cd ²⁺ + Mn ²⁺ ) = 0.004004 mol - 0.000219 mol = 0.003785 mol ------> ( 3)

i e No. of moles of ( Cd ²⁺ + Mn ²⁺ ) = 0.003785 mol -----------------> ( 4)

Now, we can calculate no. of moles of Cd ²⁺ present in the solution.

we can write correlation, No. of moles of Cd ²⁺ made free by adding CN ^- = No. of moles of Ca ²⁺

No. of moles of Ca ²⁺ = 0.0150 mol / L $\times$ 0.0113 L = 0.0001695 mol

Therefore, no. of moles of Cd ²⁺ present in the solution = 0.0001695 mol

0.0001695 mol Cd ²⁺ is present in initial 50.0 ml solution.

Hence [ Cd ²⁺ ] = 0.0001695 mol / 0.0500 L = 0.00339 M

We have calculated , No. of moles of ( Cd ²⁺ + Mn ²⁺ ) = 0.003785 mol

No. of moles of  Mn ²⁺ = No. of moles of ( Cd ²⁺ + Mn ²⁺ ) - no. of moles of Cd ²⁺

No. of moles of  Mn ²⁺ = 0.003785 mol - 0.0001695 mol = 0.003616 mol

Hence [ Mn ²⁺​​​​​​​ ] = 0.003616 mol / 0.0500 L = 0.0723 M

ANSWER : Concentration : 0.0723 M Mn ²⁺

Concentration : 0.00339 M Cd ²⁺