I am a physicist who has run into some chemistry in the lab (and unfortunately, I've completely evaded any formal training in chemistry). I have exposed some samples (I think details of the sample are not important here, but would be happy to provide further information if necessary) to a solution of 30% ammonia and 70% water, and am trying to calculate approximately how much ammonia (in ppm, for example) reached my sample.

My samples are placed in an 3″×3″×1″ (L×W×H) container with the top exposed to the air. Until now, I've just been marking down the volume of the solution that I've placed (about 15 mm) directly below my samples in the container (e.g., 5 μL, 10 μL, …, 200 μL). I think it's good enough to assume room temperature and standard atmospheric pressure for my labs. I've looked into Henry's law and related relationships, but am not sure that these are the appropriate tools. I would greatly appreciate any help or guidance!

Maybe these tables will help (I've stumbled upon them trying to figure this out in the context of Henry's law):

P.S. If a closed container is a more tractable problem, I also collected some data with a closed container (same dimensions), and would very much appreciate insight into that problem too (if it is very different, or the only solvable problem). I should also mention that the solution was removed from the container in about 10 seconds in all cases (if that makes a difference).

  • $\begingroup$ "Ammonia has a very strong odor that is irritating and that you can smell when it is in the air at a level higher than 5 ppm" - did you smell it? Depending whether your nose or your sample is closer to the solution (and the geometry of the setup), your nose will experience a higher or lower concentration than your sample. $\endgroup$
    – Karsten
    Commented Mar 26, 2020 at 16:05
  • 1
    $\begingroup$ It's not doable even with closed container, unless you were real careful to have equilibrium vapor pressure in there. $\endgroup$
    – Mithoron
    Commented Mar 27, 2020 at 0:14
  • $\begingroup$ It is a good question which relates to liquid-liquid extraction equilibria and head space GC. I would argue that the question should stay open, a person who knows the thermodynamics of liquid-liquid extraction or vapor / liquid systems should address it $\endgroup$ Commented Apr 10, 2020 at 7:31

2 Answers 2


If the container is open, as shown below, then you have a total mess.

enter image description here

Obviously eventually all the ammonia and water will evaporate into the room eventually. There is absolutely no way to tell how much ammonia gas would be with the volume of the container at any given time, and the ammonia concentration within the container wouldn't ever be homogeneous.

If the container is closed, then you have a somewhat more tractable problem.

enter image description here

However there are two equilibriums to consider. First the gas phase must be saturated with water vapor, and second the ammonia must be distributed between the gas and liquid phases (if any liquid phase remains).

Just out of curiosity, 3x3x1 inchs has a volume of 9 cubic inches, or about 147 mL. So if the air in the lab was at $\pu{20 ^\circ C}$ and completely dry then about $\pu{2.5 \mu L}$ of water would be needed to saturate the volume to 100% humidity.

water vapor pressure at $\pu{20 ^\circ C} = \pu{17.5 torr}$
atmospheric vapor pressure = 760 torr
molar volume at $\pu{20 ^\circ C}$ and 760 torr = 24.1 liters
Moles water = $\dfrac{17.5}{760}\cdot \dfrac{0.147}{24.1} = \pu{1.4\cdot10^{-4} moles}$
water is $\pu{18 g/mole}$ and $\pu{1 g/ml} = \pu{0.001 g/\mu L}$
$ \pu{\mu L water} = \dfrac{\pu{1.4\cdot10^{-4} moles}\cdot \pu{18 g/mole}}{\pu{0.001 g/\mu L}} = \pu{2.5 \mu L} $

NOTE: If the container is sealed there is still the problem of estimating how long it takes the system to reach equilibrium.

  • 1
    $\begingroup$ At STP, 147 mL gas contain about 6 mmol particles. Only 2% or so of that are water (at 100% humidity), i.e. 0.12 mmol, or about 2 microliter. Huh? Did you start with vacuum, and the container is pure water vapour? $\endgroup$
    – Karsten
    Commented Mar 26, 2020 at 16:03
  • $\begingroup$ @KarstenTheis - Missed your comment. Thanks for the correction, which I've made. I've also shown the calculations, which I should have done before. $\endgroup$
    – MaxW
    Commented Apr 25, 2020 at 23:14

Careful, if you have samples that contain transition metals and/or their salts, in the presence of ammonia fumes, air (a source of oxygen), water (or even moisture) and an electrolyte (any soluble salts), you may have some electrochemistry afoot as well. In particular, for example, is the leaching of copper ores (removing the copper ions) with NH3/air with a spontaneous (after an inception period) electrochemical reaction mixed with some more standard chemistry.

See this source, as an example of some possible unexpected chemistry: "Kinetics and Mechanism of Copper Dissolution In Aqueous Ammonia", (source link).

Now, if the ammonia is being consumed in a reaction with air exposure, then equilibrium reactions relating to NH3 exposure are going to be invalid.


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