A 50.0 mL sample containing Cd^2+ and Mn^2+ was treated with 49.1 mL of 0.0400 M
ID: 1049055 • Letter: A
Question
A 50.0 mL sample containing Cd^2+ and Mn^2+ was treated with 49.1 mL of 0.0400 M EDTA. Titration of the excess unreacted EDTA required 10.0 mL of 0.0110 M Ca^2+. The Cd^2+ was displaced from EDTA by the addition of an excess of CN^-. Titration of the newly freed EDTA required 13.7 mL of 0.0110 M Ca^2+. What were the molarities of Cd^2+ and Mn^2+ in the original solution? A 25.0-mL solution of 0.0530 M EDTA was added to a 53.0-mL sample containing an unknown concentration of V^3+. All V^3+ present formed a complex, leaving excess EDTA in solution. This solution was back-titrated with a 0.0460 M Ga^3+ solution until all the EDTA reacted, requiring 15.0 mL of the Ga^3+ solution. What was the original concentration of the V^3+ solution?Explanation / Answer
Analysis of sample containing Cd2+ and Mn2+
Total EDTA added initially = 0.04 M x 49.1 ml = 1.964 mmol
Excess EDTA left after reaction = 0.011 M x 10 ml = 0.11 mmol
EDTA reacted with Cd2+ and Mn2+ = 1.964 - 0.11 = 1.854 mmol
Cd2+ freed and reacted with CN-
free EDTA by this = 0.011 M x 13.7 ml = 0.1507 mmol
moles of Mn2+ in sample = 0.1507 - 0.11 = 0.0407 mmol
molarity of Mn2+ in solution = 0.0407 mmol/50 ml = 0.000814 M
moles of Cd2+ in sample = 1.854 - 0.0407 = 1.8133 mmol
molarity of Cd2+ in solution = 1.8133 mmol/50 ml = 0.036266 M
Analysis of sample containing V3+
Total EDTA added = 0.053 M x 25 ml = 1.325 mmol
Excess EDTA left after reaction = 0.046 M x 15 ml = 0.69 mmol
EDTA reacted with V3+ = 1.325 - 0.69 = 0.635 mmol
molarity of V3+ in solution = 0.635 mmol/53 ml = 0.012 M
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