Important information about Quant problems
EDTA I
Calculate pFe2+ at Ve in the titration of 25.00 mL of 0.02026 M EDTA by 0.03855 M Fe2+ at pH
6.00.

EDTA II
A sample contained Ni and Zn ions as the only metals that would bind to EDTA. A 50.0 mL solution of Ni2+ and Zn2+ was treated with 25.0 mL of 0.0452 M EDTA to bind all of the metal. The excess unreacted EDTA required 12.4 mL of 0.0123 M Mg2+ for complete reaction. An excess of the reagent 2,3-dimercapto-1-propanol was then added to the solution to displace the EDTA that was bound to zinc. Another 29.2 mL of 0.0123 Mg2+ was required for reaction of the EDTA that had been liberated from zinc. Calculate the molarity of Ni2+ and the molarity of Zn2+ in the original solution.

EDTA III
Consider the EDTA titration below:
Zn2+ + Y4- ↔ ZnY2- Kf = 3.2×1016
a.) How many mL of 0.1050 M EDTA solution are required to reach the equivalence point in the titration of 20.00 mL of 0.1433 M Zn2+, buffered at pH 10.00? (αY4- = 0.30 at pH 10.00) b.) What is the pZn at the equivalence point? c.) How many mL of EDTA solution would have been needed if the analyte solution was 20.00 mL of 0.1433 M Fe3+ instead of 20.00 mL of 0.1433 M Zn2+?

EDTA IV
A 50.0 mL sample containing Ni2+ was treated with 25.0 mL of 0.0500 M EDTA and buffered at pH 10.00. The excess EDTA was then back-titrated, requiring 5.00 mL of 0.0518 M Zn2+. What was the concentration of Ni2+ in the original solution? (At pH 10.00 αY4- = 0.30. KNiY = 2.5×1018. KZnY = 3.2×1016)