Question

A 50.0 mL sample containing Cd2+ and Mn2+ was treated with 64.0 mL of 0.0600 M EDTA . Titration of the excess unreacted EDTA required 16.1 mL of 0.0310 M Ca2+ . The Cd2+ was displaced from EDTA by the addition of an excess of CN− . Titration of the newly freed EDTA required 14.2 mL of 0.0310 M Ca2+ . What are the concentrations of Cd2+ and Mn2+ in the original solution?

Answer 1

Answer:

the concentration of $Cd^{2+}$  in the original solution= 0.0088 M

the concentration of $Mn^{2+}$ in the original solution = 0.058 M

Explanation:

Given that:

The volume of the sample  containing Cd2+ and Mn2+ =  50.0 mL; &

was treated with 64.0 mL of 0.0600 M EDTA

Titration of the excess unreacted EDTA required 16.1 mL of 0.0310 M Ca2+

i.e the strength of the Ca2+ = 0.0310 M

Titration of the newly freed EDTA required 14.2 mL of 0.0310 M Ca2+

To determine the concentrations of Cd2+ and Mn2+ in the original solution; we have the following :

Volume of newly freed EDTA = $\frac{Volume\ of \ Ca^{2+}* Sample \ of \ strength }{Strength \ of EDTA}$

= $\frac{14.2*0.0310}{0.0600}$

= 7.3367 mL

concentration of  $Cd^{2+}$ = $\frac{volume \ of \ newly \ freed \ EDTA * strength \ of \ EDTA }{volume \ of \ sample}$

= $\frac{7.3367*0.0600}{50}$

= 0.0088 M

Thus the concentration of $Cd^{2+}$ in the original solution = 0.0088 M

Volume of excess unreacted EDTA = $\frac{volume \ of \ Ca^{2+} \ * strength \ of Ca^{2+} }{Strength \ of \ EDTA}$

= $\frac{16.1*0.0310}{0.0600}$

= 8.318 mL

Volume of EDTA required for sample containing $Cd^{2+}$   and  $Mn^{2+}$  = (64.0 - 8.318) mL

= 55.682 mL

Volume of EDTA required for $Mn^{2+}$  = Volume of EDTA required for

                                                                sample containing  $Cd^{2+}$   and  

                                                             $Mn^{2+}$ --  Volume of newly freed EDTA

Volume of EDTA required for $Mn^{2+}$  = 55.682 - 7.3367

= 48.3453 mL

Concentration  of $Mn^{2+}$ = $\frac{Volume \ of EDTA \ required \ for Mn^{2+} * strength \ of \ EDTA}{volume \ of \ sample}$

Concentration  of $Mn^{2+}$ =  $\frac{48.3453*0.0600}{50}$

Concentration  of $Mn^{2+}$  in the original solution=   0.058 M

Thus the concentration of $Mn^{2+}$ = 0.058 M