EDTA titration--(Analytical Chem.) A mixture of Mn2+, Mg2+, and Zn2+ was analyzed as follows: The 25.00mL sample was treated with 0.25g of NH3OHCl (hydroxylammonium chloride)--a reducing agent that maintains manganese in the 2+ state. 10 mL of ammonia buffer (pH10) and a few drops of eriochrome black T indicator and then diluted to 100 mL. It was warmed to 40 degrees (C) and titrated with 39.98 mL of 0.04500 M EDTA to the blue end point. Then 2.5g of NaF was added to displace Mg2+ from its EDTA complex. The liberated EDTA required 10.26 mL of standard 0.02065 M Mn2+ for complete titration. After this second end point was reached, 5 mL of 15 wt% aqueous KCN was added to displace Zn2+ from its EDTA complex. This time the liberated EDTA required 15.47 mL of standard 0.02065 M Mn2+. Calculate the number of milligrams of each metal (Mn2+, Zn2+, and Mg2+) in the 25.00 mL sample of unknown. (note: EDTA reacts 1:1 with metal ions). --ok, I seriously have no idea where to start. I definately need a starting point. Am I right to assume since EDTA reacts 1:1, that I do not need to do redox for each metal ion? ---Please help!