Question

A 50.0 mL sample containing Cd2+ and Mn2+ was treated with 56.3 mL of 0.0500 M EDTA. Titration of the excess unreacted EDTA required 13.4 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 28.2 mL of 0.0310 M Ca2+. What are the concentrations of Cd2+ and Mn2+ in the original solution?

Answer 1

The concentration of Cd2+ is 0.0175 M

The concentration of Mn2+ is 0.0305 M

Step-by-step explanation:

Step 1: Data given

A 50.0 mL sample contains Cd2+ and Mn2+

volume of 0.05 M EDTA = 56.3 mL

Titration of the excess unreacted EDTA required 13.4 mL of 0.0310 M Ca2+.

Titration of the newly freed EDTA required 28.2 mL of 0.0310 M Ca2+

Step 2: Calculate mole ratio

The reaction of EDTA and any metal ion is 1:1, the number of mol of Cd2+ and Mn2+ in the mixture equals the total number of mol of EDTA minus the number of mol of EDTA consumed in the back titration with Ca2+:

Step 3: Calculate total mol of EDTA

Total EDTA = (56.3 mL EDTA)(0.0500 M EDTA) = 0.002815 mol EDTA

Consumed EDTA = 0.002815 mol – (13.4 mL Ca2+)(0.0310 M Ca2+) = 0.002815 - 0.0004154 = 0.0023996 mol EDTA

Step 4: Calculate total moles of CD2+ and Mn2+

So, the total moles of Cd2+ and Mn2+ must be 0.0023996 mol

Step 5: Calculate remaining moles of Cd2+

The quantity of cadmium must be the same as the quantity of EDTA freed after the reaction with cyanide:

Moles Cd2+ = (28.2 mL Ca2+)(0.0310 M Ca2+) = 0.0008742 mol Cd2+.

Step 6: Calculate remaining moles of Mn2+

The remaining moles must be Mn2+: 0.0023996 - 0.0008742 = 0.0015254 moles Mn2+

Step 7: Calculate initial concentrations

The initial concentrations must have been:

(0.0008742 mol Cd2+)/(50.0 mL) = 0.0175 M Cd2+

(0.0015254 mol Mn2+)/(50.0 mL) = 0.0305 M Mn2+

The concentration of Cd2+ is 0.0175 M

The concentration of Mn2+ is 0.0305 M