alchemistcanada
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concentration Hydrogen fluoride gas
Hi
I would like to prepare HF gas of known concentration say 5 ppm by using the following equation such as
CaF2 + H2SO4 ---> CaSO4 + 2HF
How much CaF2 and H2SO4 is required to produce HF gas of 5 ppm.
Thanks
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blogfast25
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Firstly search this forum (using the search facility) to get aquainted with the dangers of HF anf HF generators. Then decide if you want to proceed
with your plan. 
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alchemistcanada
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Hi
I am well aware of the potential hazards of HF and using all necessary PPE to run the experiments.
Thanks.
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NurdRage
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my question is what s/he wants to use it for. 5 ppm is rather specific, so they must have a very specific use.
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DDTea
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This is actually a very basic calculation. However, because it is so basic, I'd expect you to be able to look that up yourself. It's simple
stoichiometry + some work with the ideal gas law (PV = nRT). At the very least, that would be a good start (HF isn't really an ideal gas, but come
on: show us SOME work!)
Regarding the ppm concentration, the amount you use depends on the pressure you're trying to run whatever reaction you're performing at. An easier
way might be to make HF and dilute the fumes with some other gas, but it's difficult to say anything definitively without knowing what you are trying
to do as NurdRage mentioned.
In any case, HF gas is one of those chemicals that are exceedingly dangerous and it's hard to overstate its hazards. Beyond the acute toxicity, it
can cause serious long-term effects. If you can feel it irritating your eyes (and it's a potent lachrymator), it's in a concentration way beyond what
you should be exposed to. More than that, it's corrosive to just about everything. If you don't have equipment dedicated to working with HF, I would
advise you not to proceed. Glassware will not last more than 5 minutes against HF fumes.
[Edited on 3-2-11 by DDTea]
"In the end the proud scientist or philosopher who cannot be bothered to make his thought accessible has no choice but to retire to the heights in
which dwell the Great Misunderstood and the Great Ignored, there to rail in Olympic superiority at the folly of mankind." - Reginald Kapp.
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alchemistcanada
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Hi
Let us assume that one mole of CaF2 reacts completely with concentrated H2SO4, two moles of HF
Hence if 100g of CaF2 is taken then 100 / 78.07 = 1.280 mole
If all the CaF2 reacts 2 mole of HF is produced. Hence 2.56 moles of HF is produced.
The density of HF is 0.922 g /L then the volume of HF will be 2.36 L/mole (molar volume).
Let us say if it is mixed with nitrogen gas ...........,
since the gas is non-ideal it does not obey the PV = nRT.
Any help further to calculate the concentration of HF in ppm ?
Thanks
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Eclectic
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Seems like it would be simpler making your HF and then dealing with mixing to a known low concentration.
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Tsjerk
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Doesn't ppm stand for mass percentage? So 5 ppm would be 5 grams HF per a million grams of gas? You don't need the gas law at all. You only need to
know the density of the gas your mixing the HF with.
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Jor
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When you are talking about concentrations of substances in solids or liquids, for example fluoride in toothpaste, you're talking about mass, so 1 ppm
is 1 gram per 10^6 grams.
However in air, we talk about particles per millions. That's why most MSDSs and articles give a conversion factor to calculate mg/m3 from ppm and vica
versa. To obtain 1 ppm, you need t release 1mL of gas for every cubic meter of gas.
To get 5ppm HF in one cubic metre of gas, you need to release 5mL of gas.
1 mole of HF is 20 grams. 1 mole of gas is about 24,8 L of gas IIRC. This means that (5,0x10^-3 / 24,8) x 20 = 4x10^-3 grams, or 4 mg HF.
However, this is following theory of ideal gasses. I can imagine some gasses, especially ones that form hydrogen bonds (including HF) have a higher
density than this calculation suggests (for example wiki gives a density of HF as 1,15g/L of gas).
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DDTea
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You can always *start* by modeling it as an ideal gas and making the necessary corrections. Why not look up just how "ideal" it is at room
temperature/1 bar? Otherwise, you could always use the Van Der Waal's equation of state for a slightly more sophisticated calculation.
You say you want 5 ppm. What would you consider an acceptable error? Because even if CaF2 and H2SO4 react stoichiometrically and completely, who's
to say that you won't lose some of the HF that's formed? Knowing what your tolerance limits are could really simplify things--maybe the ideal gas law
would be "good enough" here.
Start simple and complicate things only as necessary 
"In the end the proud scientist or philosopher who cannot be bothered to make his thought accessible has no choice but to retire to the heights in
which dwell the Great Misunderstood and the Great Ignored, there to rail in Olympic superiority at the folly of mankind." - Reginald Kapp.
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