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Topic: Analytical Chemistry- A Question On EDTA Titrations and Direct/Back Titrations  (Read 4948 times)

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Offline Book_of_Problems

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I have been given this question in a sample set of questions. I was wondering if anyone could help in getting the solution. I will tell you the answer:

Ni2+ = 0.0124M and Zn2+ = 0.00718

The question is:

"A 50.0-mL solution containing Ni2+ and Zn2+ was treated with 25.0-mL of 0.0452M EDTA to bind all the metal. The excess unreacted EDTA required 12.4-mL of 0.0123M Mg2+ for complete reaction. An excess of reagent 2,3-dimercapto-1-propanol was then add to displace the EDTA from zinc. Another 29.2-mL of Mg2+ were required to for reaction with the liberated EDTA. Calculate the molarity of Ni2+ and Zn2+ in the original solution.

If anyone can help that would be great seeing as i can't even get the proper mols of EDTA needed.
« Last Edit: December 13, 2009, 01:44:17 AM by Book_of_Problems »

Offline Book_of_Problems

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Ok So i ended up getting what i think the answer is. And i just want a confirmation if possible please and thanks.

I got the total mols of EDTA and then subtracted that mol amount from the First amount of Mg2+ added.

nEDTA = (0.0452MEDTA*25.0-mL) - (0.0123MMg2+*12.4-mL)
nEDTA = 1.13mmol

I then realized the amount of Mg2+ used to react with the liberated EDTA would be equal to the amount of mols of Zn2+ in my solution seeing as the EDTA was displaced off of all the metal Zn2+. So i then said that:

nMg2+ = nZn2+

nZn2+ = nMg2+ = (0.0123MMg2+*29.2-mL)

nZn2+ = 0.359mmol

[Zn2+]= 0.359/50-mL = 0.00718M

I then took the nZn2+ and subtracted it from the total EDTA used for both the Ni2+ and Zn2+ as to be able to get just the mols of Ni2+.
 
0.953-mmol - 0.359-mmol = 0.594-mmol

[Ni2+] = 0.594/50-mL = 0.0118M not the 0.0124M so i am not sure if this would be right or wrong. Or if there is a typo in the book.

Edit: tags corrected.
« Last Edit: December 13, 2009, 05:08:09 AM by Borek »

Offline Borek

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