Answer:
The concentration of Cd2+ is 0.0175 M
The concentration of Mn2+ is 0.0305 M
Explanation:
<u>Step 1:</u> 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+
<u>Step 2:</u> 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+:
<u>Step 3: </u>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
<u>Step 4:</u> Calculate total moles of CD2+ and Mn2+
So, the total moles of Cd2+ and Mn2+ must be 0.0023996 mol
<u>Step 5:</u> 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+.
<u>Step 6:</u> Calculate remaining moles of Mn2+
The remaining moles must be Mn2+: 0.0023996 - 0.0008742 = 0.0015254 moles Mn2+
<u>Step 7: </u>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
