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EDTA titration-(Analytical Chem.)

  1. Oct 17, 2004 #1
    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!
    Last edited: Oct 17, 2004
  2. jcsd
  3. Oct 18, 2004 #2


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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.
  4. Oct 18, 2004 #3
    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?
  5. Oct 18, 2004 #4


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    It seems okay with Mn2+ ion.

    In the beginning, you know that
    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.
  6. Oct 19, 2004 #5
    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!! Sincerely, tipton12
  7. Oct 21, 2004 #6
    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+
  8. Oct 21, 2004 #7
    the 1.799 comes from (39.98mL)(.04500mmoles/mL EDTA)=1.799
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