A mixture of Mn2+, Mg2+ and Zn2+ was analyzed as follows: The 25000 mL sample was treated with 0.25 g of NH3OH+Cl- (hydroxylammonium cgloride, a reducing agent that maintains manganese in the +2 state), 10 mL of ammonia buffer (pH 10), and a few drops of erichrome black T indicator and then duluted to 100 mL. It was warmed to 40C and titrated with 39.98 mL of 0.04500 M EDTA complex. The liberated EDTA required 10.26 mL of standard 0.20 65 M Mn2+ for complete titration. After this second end point was reached, 5 mL of 15 wt% aqueous KCN was added to displace Zn 2+ from its EDTA complex. This time the liberated EDTA required 15.47 mL of standard 0.020 65 M Mn2+. Calculate the number of miligrams of each metal (Mn2+, Zn2+, and Mg 2+) in the 25.00 mL sample of unknown.© BrainMass Inc. brainmass.com March 22, 2019, 2:07 am ad1c9bdddf
Total number of moles of all three metals = Total number of moles of EDTA
= 39.98 mL x 0.045 moles/L
=39.98 mL x 0.045 mmoles/mL
Totsl of Mn2+, Zn2+ and Mg2+ =1.7991 mmoles.
Then the Mg2+ was removed ...
This solution deals with the determination of metals in a mixture of
Mn2+,Mg2+ and Zn2+