Explore BrainMass

Explore BrainMass

    Chemistry Quant Sample Problems

    This content was COPIED from BrainMass.com - View the original, and get the already-completed solution here!

    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

    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

    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+?

    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)

    © BrainMass Inc. brainmass.com October 10, 2019, 12:13 am ad1c9bdddf

    Solution Preview

    Dear Student,
    The initial mmole of 0.5065 for Fe+2 is based on the initial ...

    Solution Summary

    The expert examines quant titrations calculations.