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Topic: Cuantification of Si, Mn, Fe in alloy  (Read 8267 times)

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Offline Lectos

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Cuantification of Si, Mn, Fe in alloy
« on: April 13, 2009, 09:11:51 AM »
I need to cuantify Si, Mn, Fe in a sample. The sample contains only those metals plus P. I have a tentative procedure and I want to know if it is correct and I also have a couple of questions.

I have already separated the Mn and Fe from the other elements by disolving the sample with HNO3. The idea is to separate the Mn by forming a complex ion with NH3Cl and NH4OH so that the Fe will precipitate. Then the solution and the precipitate will be separated by filtartion with a filter paper. The paper will be calcinated and the Fe redisolved and trinitrated. Afterwards the solution of Fe will be redisolved and tritrated. The complex ion of Mn will be broken by teh addition of HCl and then the solution wll be also tritrated.

¿Is the procedure is correct?
¿The amount of Mn that will be absorved by the filter paper will produce a significant error?
¿What is the color of the Mn complex ion?
¿Is there a diferencie in the order of addition of NH3Cl and NH4OH?
¿What is the color of the complex ion?

Now the Si and P are in a filter paper. The idea is to add calcium calronate and calcinate to form silicates. Afterwards the silicates will be solved with HCl and filtered and the filter paper calcinated, the solution should contain Si. The change of weight of the calcinated filter papers should be the weight of the Si.

Note: I dont have the proceduire of this las part with me right now, and I have the senation taht i am missing a step. I will clarify later this.

¿Will this procedure work?
¿All the Si is converted to silicates?
¿Is all the Si is disolved?

Thanks in advance.
« Last Edit: April 14, 2009, 04:01:43 AM by Borek »

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Re: Cuantification of Si, Mn, Fe in alloy
« Reply #1 on: April 13, 2009, 10:24:43 AM »
The idea is to separate the Mg by forming a complex ion with NH3Cl and NH4OH so that the Fe will precipitate. Then the solution and the precipitate will be separated by filtartion with a filter paper. The paper will be calcinated and the Fe redisolved and trinitrated. Afterwards the solution of Fe will be redisolved and tritrated.

Why not titrate Fe directly (permanganate), without separation?
« Last Edit: April 14, 2009, 04:01:56 AM by Borek »
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Offline 408

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Re: Cuantification of Si, Mn, Fe in alloy
« Reply #2 on: April 13, 2009, 11:29:41 AM »
Any P in a metal sample being dissolved in acid is going to be reduced by nascent hydrogen into phosphine, which other than affecting your analysis is also really toxic.

Why not EDX?
« Last Edit: April 14, 2009, 04:02:06 AM by Borek »

Offline Lectos

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Re: Cuantification of Si, Mn, Fe in alloy
« Reply #3 on: April 14, 2009, 12:14:15 AM »
Why not titrate Fe directly (permanganate), without separation?

My apologies. I dont know what I was thinking  :-\ it is not Mg it is Mn. So permanganate will not work because of the Mn.

Any P in a metal sample being dissolved in acid is going to be reduced by nascent hydrogen into phosphine, which other than affecting your analysis is also really toxic.

Why not EDX?

We can only use traditional methods.  :(

So if the P in the metal was solubilized with HNO3 ¿Can I safetly asume that in the precipitate left there is only Si?

And I should mention that the current state of the analysis is the dissolved metal and a solution that I think contains: Mn, Fe and I guess that PH3. And the presipitate of the dissolved metal currently is on a filter paper and I'm guessing that it only contains Si.

PS.

If a nice mod could change my mistake in the title Mg to Mn and also in the first post as I cannot see how to do it ;D
« Last Edit: April 14, 2009, 04:02:19 AM by Borek »

Offline 408

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Re: Cuantification of Si, Mn, Fe in alloy
« Reply #4 on: April 14, 2009, 10:36:34 AM »
Phosphine is a gas and its formation is not quantitative

Offline 408

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Re: Cuantification of Si, Mn, Fe in alloy
« Reply #5 on: April 14, 2009, 01:25:37 PM »
Because I cannot modify the post, once you have the Si/P, dry it and heat it to a few hundred C sublimating off the P. Si can then be massed.  I do not beleive these conditions will give n-doped Si.  Why make silicates when you can get a mass of Si?

Then can control precipitation of Fe vs Mn based on Kb and slow base addition.

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Re: Cuantification of Si, Mn, Fe in alloy
« Reply #6 on: April 14, 2009, 02:23:21 PM »
Si can then be massed.

My bet is that it will be contaminated with SiO2.

Quote
Why make silicates when you can get a mass of Si?

Assuming identical mass error, the higher the molar mass of the subtsance, the lower the relative error.

Quote
Then can control precipitation of Fe vs Mn based on Kb and slow base addition.

Won't work, coprecipitation.

I still think permanganate titration will work. Mn2+ won't interfere, it is even deliberately added in the form of Zimmermann-Reinhardt solution to lower permanganate oxidation potential, so that chlorides don't get oxidized to chlorine.
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Re: Cuantification of Si, Mn, Fe in alloy
« Reply #8 on: April 14, 2009, 02:56:37 PM »
I don't have access to the sciencedirect, so can't comment on the method. At least in the abstract it doesn't say anything about precipitates. Coprecipitation is quite often a problem in the analytical procedures that otherwise look logical.

I can be wrong, but I would not risk reinventing the wheel, steel analysis is a well known teritory.
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Offline Lectos

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Re: Cuantification of Si, Mn, Fe in alloy
« Reply #9 on: April 15, 2009, 01:08:18 AM »
I already separated the Mn and Fe, what I ended up doing was adding to the solution NaBiO3 so the Mn stayed in the solution in the form of permanganate and the Fe precipitated. The Fe was filtered though paper and calcinated and weighted. The idea is to titrate the Mn with something...

Now I just hope that the Si procedure works  ;D

Thanks for your help

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