Safety in handling of concentrated ammonium fluoride

Question

I recently saw a protocol for extracting organic materials that are bound in the silica matrix formed by diatoms. The extraction protocol used 10 molar ammonium fluoride ($\ce{NH4F}$).

I briefly considered using the protocol and then endeavoring to remove the ammonium fluoride (and water) by lyophilization or with a rotary evaporator.

However, I am concerned about the safety of doing this. My first worry is that any glass equipment or vessels in the lyophilizer or rotary evaporator systems would be damaged by the fluoride. My second worry is that even if the equipment used did not contain any glass, dangerous solutions of hydrogen fluoride (i.e. hydrofluoric acid) could be formed in the cold trap of the equipment.

My first question is, am I right that it is a bad idea to evaporate ammonium fluoride in glass equipment?

My second (but related) question is, even if glass is not used, what are the safety risks of evaporating ammonium fluoride under vacuum? Imagining that a glass-free apparatus could be found, is there a way to do the evaporation safely?

Answer 1

In $\ce{SiO2}$ etching, there are two active fluorine-containing species, $\ce{HF}$ and $\ce{HF2-}$.

The rate of etching at 25 degrees C is proportional to :

$$9.66[\ce{HF2-}] + 2.50 [\ce{HF}] -0.14$$

according to A Study of the Dissolution of SiO2 in Acidic Fluoride Solutions J. Electrochem. Soc. vol. 118, pp. 1772-1775.

The relevant equilibrium constants at 25 degrees C and ionic strength of 1M are

$$\frac{[\ce{H+}][\ce{F-}]}{[\ce{HF}]} = 0.0013M $$

$$\frac{[\ce{HF}][\ce{F-}]}{[\ce{HF2-}]} = 0.104M $$

From the second equilibrium, it can be seen that if the fluoride concentration is 10M as in the question, the more active $\ce{HF2-}$ will be present at a concentration of about 100 times $[\ce{HF}]$.

Multiplying the two equilibrium equations together, and substituting $[\ce{F-}] = 10M$:

$$\frac{[\ce{H+}]}{[\ce{HF2-}]} = 1.4 \times 10^{-6} $$

Therefore at pH = $\log (1.4 \times 10^{-6}) = 5.9$, there will be about 1M $\ce{HF2-}$, diminishing by a factor of 10 for each pH unit increase.

Experimentally, concentrated Ammonium Fluoride has a pH of 7.2 according to http://www.solvaychemicals.us/SiteCollectionDocuments/sds/P20117-USA_CN_EN.pdf (which also says to avoid glass).

At pH 7.2, [$\ce{HF2-}$] is 0.05M.

In conclusion, there will be a significant concentration of the most active etching species, $\ce{HF2-}$, present in 10M ammonium fluoride and I would be concerned about glassware.

Answer 2

Note that I don't have first hand experience on this - I never used ammonium fluoride to release organic material from diatomeous earth, but I frequently used tetrabutylammonium fluoride (TBAF) in dichloromethane THF to cleave silyl ethers.

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Cleaving silyl ethers with TBAF

To a solution of 3.8 mmol of a cyclopentylsilylether in 50 mL THF at 0 °C under argon, 4.5 mL of a TBAF solution (1.0 M in THF) were added. The mixture was stirred for 10 min, allowed to come to RT and the stirred for another 10 min. Then, 100 mL diethyl ether were added.

The organic layer was extracted twice with each 20 mL water and then dried over sodium sulfate.

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To my knowledge, release of $\ce{HF}$ from $\ce{NH4F}$ (we are not talking about $\ce{NH4HF2 =\ NH4F*HF}$ here) is only to be expected under the following two conditions:

• acidification
• heating $\ce{NH4F}$ up to its melting point - sublimation is accompanied by decomposition.