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?

the concentration of  in the original solution= 0.0088 M

the concentration of 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 =

=

= 7.3367 mL

concentration of   =

=

= 0.0088 M

Thus the concentration of in the original solution = 0.0088 M

Volume of excess unreacted EDTA =

=

= 8.318 mL

Volume of EDTA required for sample containing   and    = (64.0 - 8.318) mL

= 55.682 mL

Volume of EDTA required for  = Volume of EDTA required for

                                                                sample containing     and  

                                                              --  Volume of newly freed EDTA

Volume of EDTA required for  = 55.682 - 7.3367

= 48.3453 mL

Concentration  of =

Concentration  of =  

Concentration  of  in the original solution=   0.058 M

Thus the concentration of = 0.058 M

the density of gold = 19.3 g/cm3

the relation between density and mass and volume is

density = mass / volume

given :

mass of gold = 13 g

density = 13g / volume = 19.3

volume = 13 /19.3 = 0.67 cm^3

thus volume of gold nugget = 0.67 cm^3