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Topic: Unknown Sample Analysis for Hg poisoning  (Read 11641 times)

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Jacquita

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Unknown Sample Analysis for Hg poisoning
« on: March 30, 2005, 10:42:13 PM »
 ???

Hello everyone,

I have a question which I'm totally confused about.

QUESTION:

42g of yoghurt was supposedly implicated in the posisoning of a victim. the LD50 value of HgCl2 is 1mg/kg. Victim was 65kg

QUESTION:   How to calculate the concentration of Hg in the digest solution?
                  How to calculate the mass of Hg in the sample of yoghurt?
                   How to calculate the maximum amount of Hg that may have
                    been ingested?

A sample of yoghurt (18.7g from a 100g tub) (supposedly poisoned with mercury) was mixed with 20ml of conc nitric acid, boiled gently til the solution clarified. The solution was diluted and brought to pH of 3.5 by the careful addition of NaHCO3 before a final dilution into a 250.00mL volumetric flask with DI water. It is standardized with a standard solution of EDTA of 0.0520M.

An aliquot of the dilute nitric acid solution 24.9332 (+/- 0.14%)mL required 8.02 +/- 0.45%mL of EDTA standard to reach visible end point.  Blank samples had no significant titre.

WHAT I TRIED TO SOLVE WITH:

I understand that it is a back titration. What I don't understand is trying to find the amount of supposed Hg contained in the yoghurt.

1) I worked out the number of moles for EDTA.
2) I couldn't find the stoichiometric ratio for dil HNO3 and EDTA reactions.
3) From the pH of 3.5, I managed to find the [H+]
4) I don't understand how to tie up all the reactions.

1st reaction : yoghurt reacts with conc HNO3
2nd reaction: HNO3 mixture + NaHCO3 + H2O ----> I can't figure out if i wrote the equation properly and I've no idea what the products could be
3rd reaction : dil HNO3 + EDTA ----->  ?
                    I can't balance any of the reactions.

My problem now is in figuring out how the reactions look like, What is supposed to react with what, and figure out the stoichiometric ratios. How do I go about trying to discover the amount of Hg in the sample?

Any help? Please? I'm so stumped with this question.

Any help is greatly appreciated.

Thanks very much in advanced.
« Last Edit: April 01, 2005, 10:45:54 AM by Jacquita »

savoy7

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Re:Unknown Sample Analysis for Hg poisoning
« Reply #1 on: March 31, 2005, 01:18:43 AM »
Jacquita,

This is a little warning - I'm just guessing - this maybe completely wrong.

I've never done this type of problem before.

Okay now that's out of the way, maybe I could say something that might help.

I also start problems backwards.  42 g of yogart must contain 65 mg of HgCl2 in order to kill a 65 Kg man.

Basically, it must have at least 1.54 mg HgCl2/g of yogart.

Now if we can figure out how much HgCl2 is in the 18.7g of sample yogart - we are home free.


I may be wrong, but I don't think the nitric acid has a lot to do with the calculation.  I think that it allows for the digestion of the yogart to free the Hg+2.  EDTA is a chelating agent that "steals" divalent cations like Hg.  

0.0520 M EDTA * 8.02 ml = 24.9332 ml of the diluted digested yogart * it's conc in M

I get 0.0167 M Hg

that was the conc of Hg in the diluted yogart digest.  Previously the yogart was diluted to 250 ml.  

so 0.0167 M Hg = moles of Hg
                          .250 L of solution

that gives me 0.004175 moles of Hg+2

For every 1 mole of Hg+2 there is 1 mole of HgCl2 - there are 0.004175 moles of HgCl2

moles of HgCl2 can be converted to grams of HgCl2
the test sample was 18.7 gram of yogart

with that ratio you can determine grams of HgCl2 in the 42 g sample.



I wouldn't go to the moon on what I said, but I think the chelating part is correct.

savoy

I apologize if this steers you on a wrong path

Jacquita

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Re:Unknown Sample Analysis for Hg poisoning
« Reply #2 on: March 31, 2005, 01:36:04 AM »
 :-\

Savoy,

Thank you for answering me.  I agree that EDTA is a chelating agent, but couldn't it have reacted with some of the NaHCO3?  

I'm quite sure that either the nitric acid or the NaHCO3 was there to mislead us, but I need to understand the concept.

Hmmm, I'll try to work out the answer using your theory and see if it makes sense.

Jacquita

savoy7

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Re:Unknown Sample Analysis for Hg poisoning
« Reply #3 on: March 31, 2005, 02:03:20 AM »
I am a little confused with the pH of the solution.

I've used EDTA to titrate for Mg or Ca ions.  In doing so, the pH had to be at 10 to allow EDTA to be fully deprotonated.  It can then bind in a 1:1 ratio with metal ions.  I can understand the need to digest with the acid to free the ions, but I would have thought that the pH would have need to be in the basic realm for the EDTA to be deprotonated.


as for Na - the Kf for it is 45.7 while Hg2+ is 6.3 x 1021.  This means that EDTA will more likely bond to mercury than sodium


hmmm,
I'll think about it some more.

good luck.
savoy

Jacquita

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Re:Unknown Sample Analysis for Hg poisoning
« Reply #4 on: March 31, 2005, 04:42:25 AM »
I see where you're going with the pH. It's bothering me quite a bit as I couldn't figure out why is it kept the solution so acidic.

EDTA is a very strange standard.

I'll try working out. Can't sleep til I solve this question.

Offline hmx9123

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Re:Unknown Sample Analysis for Hg poisoning
« Reply #5 on: April 01, 2005, 07:09:26 AM »
Are you sure you're not working with disodium dicalcium EDTA (the form in which is is most usually encountered)?

Jacquita

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Re:Unknown Sample Analysis for Hg poisoning
« Reply #6 on: April 01, 2005, 10:15:57 AM »
Well, it's written as

"An EDTA solution was prepared from a VOLUCON (trademark) capsule giving a standard solution of 0.0520M."

That's it.  Not too sure if the brand is a disodium dicalcium type. I thought it would've been like the disodium EDTA that we use for Complete Blood Count samples for Haematological analysis.

I'm not sure which info was designed to mislead us poor students.


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Re:Unknown Sample Analysis for Hg poisoning
« Reply #7 on: April 01, 2005, 10:19:20 AM »
I see where you're going with the pH. It's bothering me quite a bit as I couldn't figure out why is it kept the solution so acidic.

IIRC Hg2+ has to be titrated in slightly acidic solution to avoid side reactions with OH- ions.
ChemBuddy chemical calculators - stoichiometry, pH, concentration, buffer preparation, titrations.info, pH-meter.info

Online Borek

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Re:Unknown Sample Analysis for Hg poisoning
« Reply #8 on: April 01, 2005, 10:21:40 AM »
Are you sure you're not working with disodium dicalcium EDTA (the form in which is is most usually encountered)?

I always thought it is sold (at least for analytical purposes) as disodium salt. Any calcium added will spoil EDTA complexing ability.
ChemBuddy chemical calculators - stoichiometry, pH, concentration, buffer preparation, titrations.info, pH-meter.info

Jacquita

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Re:Unknown Sample Analysis for Hg poisoning
« Reply #9 on: April 01, 2005, 10:32:47 AM »
I've always known EDTA to be a calcium "lover" meaning that it loves to attract calcium into it's "cage"-like structure. I tried working out the problem using Savoy's method, but it doesn't seem to click. Somehow I felt something was missing.

I'm so troubled.

Jacquita

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Re:Unknown Sample Analysis for Hg poisoning
« Reply #10 on: May 07, 2005, 08:54:58 AM »
Hey all that have attempted to help me,  I managed to find the solution for this question and luckily I passed that assignment.

The EDTA had nothing to do with the pH or whatever type it was sold as commercially.  All I had to do was to find the number of moles of Hg using the number of moles of EDTA which was pretty straight forward.  The rest of the question worked on rules of proportion to determine how much Hg was in the tub of yoghurt in beginning.  The formation constant thing helped, Savoy.

Thanks heaps for throwing ideas at me.

Cheers!

Jacquita

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