EDTA titration-(Analytical Chem.)

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Discussion Overview

The discussion revolves around the analysis of a mixture containing Mn2+, Mg2+, and Zn2+ using EDTA titration in an analytical chemistry context. Participants explore the calculations required to determine the concentrations of each metal ion based on titration data and stoichiometric relationships.

Discussion Character

  • Homework-related
  • Mathematical reasoning
  • Technical explanation

Main Points Raised

  • One participant expresses uncertainty about the initial steps in calculating the concentrations of metal ions, questioning whether redox reactions are necessary for each ion since EDTA reacts 1:1 with metal ions.
  • Another participant confirms that EDTA reacts with metal ions in a 1:1 stoichiometric ratio and suggests converting volumes and concentrations to millimoles as a starting point.
  • A participant shares their calculations for Mn2+, obtaining a value of 5.95 mg, and seeks confirmation on the correctness of their approach and results.
  • Another participant discusses the formation of precipitates and complexes involving Mg2+ and Zn2+, suggesting that writing out the formation reactions could help in determining the moles of reactants needed.
  • Subsequent posts provide revised calculations for the amounts of each metal ion, with one participant reporting values of 7.035 mg Zn2+, 2.14 x 103 mg Mg2+, and 5.95 mg Mn2+, while another participant corrects their values to 5.150 mg Mg2+, 20.89 mg Zn2+, and 70.21 mg Mn2+.
  • Clarification is provided regarding the source of the total moles of EDTA used in the titration, which is calculated based on the volume and molarity of the EDTA solution.

Areas of Agreement / Disagreement

Participants generally agree on the stoichiometric relationship of EDTA with metal ions and the method of calculating millimoles. However, there is no consensus on the final values for the amounts of each metal ion, as participants present differing results based on their calculations.

Contextual Notes

Some calculations presented rely on specific assumptions about the stoichiometry of the reactions and the completeness of the titrations. There may be dependencies on the definitions of the complexes formed and the conditions under which the reactions occur.

Who May Find This Useful

Students and practitioners in analytical chemistry, particularly those interested in titration methods and metal ion analysis, may find this discussion relevant.

tipton12
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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 definitely 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!
 
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Yes, EDTA reacts with almost all metal ions in 1:1 stoichiometric molar ratio, as it has six donor atoms to bind to metal ions. Multiply all the volumes in mL with their corresponding concentrations in mol/L to convert to their millimoles. This may help you fro the beginning.
 
I did that already to get:

(10.26mL)(0.02065mmol/mL)=0.2112 mmol Mn2+
(15.47mL)(0.02065mmol/mL)=0.3195 mmol Mn2+

Then, I subtracted the two, to get 0.1083mmol Mn2+
I then took that number and multiplied it with the weight:
(.1083mmol Mn2+)(54.938mg/mmol Mn2+)=5.95mg Mn2+

Is this correct for the Mn2+? Or do I need to do something different? When I got to that point, I didn't know where else to go to get the Zn2+ and the Mg2+. Any suggestions?
 
It seems okay with Mn2+ ion.

In the beginning, you know that
39.98 mL of 0.04500 M EDTA
is the total of three ions.

MgF2 precipitate is formed by the action of fluoride to magnesium, if I am not wrong. So one mole of EDTA is released to the medium, and this is treated with the given amount of standard Mg2+ solution.

Zinc ion forms a tetracyanozincate(II) complex, as shown with the formula [Zn(CN)4]2-.

What you need to do is simply write the formation reactions of the compounds I mentioned, and find how many moles of reactants (fluoride or cyanide) is required.
 
I got 7.035mg Zn2+ , 2.14x10^3 mg Mg2+, and 5.95mg Mn 2+. I hope these values are correct,but if they are not then it is only an error in my math. thanks a lot for all your help! tipton12
 
I changed my answer last minute to get the actual correct answers:
mmoles Mg2+= (10.26)(.02065M)=.2119
mmoles Zn2+=(15.47mL)(.02065M)=.3195
mmoles Mn2+=1.799-.2119-.3195=1.278

(.2119mmolesMg2+)(24.3051mg/mmole)=5.150mg Mg2+
(.3195mmolesZn2+)(65.392mg/mmole)=20.89mg Zn2+
(1.278mmolesMn2+)(54.9380mg/mmole)=70.21mgMn2+
 
the 1.799 comes from (39.98mL)(.04500mmoles/mL EDTA)=1.799
 

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