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Author: Subject: How do you make 10% HNO3 from 1M HNO3?
TheDrivah
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[*] posted on 15-11-2008 at 23:29
How do you make 10% HNO3 from 1M HNO3?


Trying to do this experiment:

For semi quantitative determination, 100 mL of sample was heated to boiling temperature with 10 mL of HNO3, as recommended by JUNGREIS (1984), during 15 min, using a reaction flask similar to the one shown in Figure 1. After cooling, the solution was transferred to a determination flask (Figure 2), and 10 mL of the reducing solution were added (50 % w/v of tin chloride in hydrochloric acid 50 %). The mercury vapour is forced to pass through a detecting paper covered with an emulsion containing cuprous iodide. A colourful complex is formed (Equation 1), with a characteristic reddish colour, whereas the colour intensity is proportional to the mercury concentration in the sample (Figure 3).

Link:
http://www.cetem.gov.br/publicacao/CTs/CT2004-044-00.pdf

So for the 10 mL of HNO3, the HNO3 needs to be 10%. How would I make 10% HNO3 from 1 M HNO3?
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kclo4
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[*] posted on 15-11-2008 at 23:49


Um, your testing for Hg poisoning or something?
Nice.. Why?

http://en.wikipedia.org/wiki/Molarity
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TheDrivah
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[*] posted on 16-11-2008 at 01:01


Im designing a research project for class. I will be comparing the cocnentrations of mercury in the urine of those with and without amalgam fillings, which are approx. 50% mercury.

Still wasn't able to figure out what to do from the link you posted. I need to make 10 mL of 10% HNO3, how would one do this from a 1 M concentration of HNO3?
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Formatik
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[*] posted on 16-11-2008 at 02:05


This should be in beginnings.

M = moles per Liter.

Look at the wiki example.
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TheDrivah
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[*] posted on 16-11-2008 at 10:06


wiki example isn't doing anything for me.

Im really lost here.

Im not making 10% of 1 M HNO3. I need to make 10% HNO3 from 1 M HNO3.

[Edited on 16-11-2008 by TheDrivah]
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chemoleo
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[*] posted on 16-11-2008 at 10:40


I'm surprised that this gives you trouble, but the rest of the practical doesn't.
1M HNO3 means 1 mole of HNO3 per litre, which is 63 g per litre, which is 6.3% (weight per volume).
So either you concentrate it or you get a 2 M stock, 1 M is not enough.




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watson.fawkes
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[*] posted on 16-11-2008 at 10:48


Quote:
Originally posted by chemoleo
I'm surprised that this gives you trouble, but the rest of the practical doesn't.
1M HNO3 means 1 mole of HNO3 per litre, which is 63 g per litre, which is 6.3% (weight per volume).
The original paper doesn't specify "w/v" for the nitric acid. The experienced will just infer it, but the less so will have an underspecified substance, since the percentage could be w/v, w/w, or any of the other percentages that get used. Such sloppiness about detail always gets me going, because it's so easy to avoid and so damaging when it's a problem.
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chemoleo
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[*] posted on 16-11-2008 at 11:16


Any reasonable designer of practicals, scientist, and advanced student would choose weight per volume or weight, weight or volume being identical for the diluant (water). If it's anything else the practical maker should be sacked! :o



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watson.fawkes
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[*] posted on 16-11-2008 at 11:34


Quote:
Originally posted by chemoleo
Any reasonable designer of practicals, scientist, and advanced student would choose weight per volume or weight, weight or volume being identical for the diluant (water). If it's anything else the practical maker should be sacked! :o
Well, they're only identical for the units "mL" and "g". And this is exactly my point: for a competent outsider, such missing information, however minor, can present a large obstacle, far larger than the minimal effort it takes to be complete. And I don't know of a reference work where you can look up how to supply missing information of this nature.

Since I'm constantly doing cross-disciplinary work, I'm constantly an outsider when I need to learn something about an unfamiliar specialty. This happens all the time for me.
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TheDrivah
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[*] posted on 16-11-2008 at 15:27


Thanks for the replies you guys.

So then for the 10 mL reducing solution (50 % w/v of tin chloride in hydrochloric acid 50 %) it would be 5 grams SnCl2 in 10 mL of 12+ M HCl? is that correct?
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[*] posted on 18-11-2008 at 16:54


Let me know what you determine. Do you expect there to be a measurable quantity of Hg in the urine of subjects with amalgam fillings?



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brew
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[*] posted on 21-11-2008 at 07:26


how do you get 50% HCl acid. Id be checking the reference lists in that file for sure. Perhaps I am confused. It is late. must be v/v

[Edited on 21-11-2008 by brew]
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[*] posted on 21-11-2008 at 19:37


IIRC, conc. HCl can go over 70% under 0°C, so that would mean cooling the acid and gassing it again. But I doubt that's what is refered to, rather H2SO4, no?



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[*] posted on 22-11-2008 at 07:34
Might be able to use 1M HNO3 instead


If it is a 10% HNO3 vol to vol. of reagent grade 70% HNO3
i.e. 10 ml of 70% Conc. reagent to 90 ml of water.
Then it is a 7% conc. of HNO3.
Since 1M is 6.3% HNO3 it is probably close enough to use the 1M HNO3. With this type of problem you can only guess at what the author used. That is what sucks about % solutions.
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Klute
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[*] posted on 22-11-2008 at 12:06


I think that's a bit too far streched. You never express the concentration of a solution as a % of a concentrated solution. A dilution factor would be used in that case, and it would be expressed as such ("conc HCl diluted 10 times").

I'm thinking that the 50% HCl might be a typo, and they are simply referring to conc acid, that 50% could have ended up at the end of the sentence by mistake.




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[*] posted on 22-11-2008 at 16:41


I've seen in workplace a protocol requiring 10% HNO3, from there I got a pen and a paper to make calculation then the guy supervising me just said 10% mean a factor 10 dilution. So sometime it can be 10% of 63% (6.3%) or really 10% which would be a factor 6.3 dilution. But from what I understand when they use % like that it mostly mean that it's qualitative more than quantitative. So just got around 10%, 6.3% is close enough. Also I only see two significative number on their 10ml so it's probably also around 10mLm, just go in excess.



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