# Dissociation of water into H+ and OH-: Does the law of mass action hold at nanoscale?

Suppose water under neutral conditions is confined in a virtual spherical nanocontainer with a radius of 25 nm. To calculate the number of hydronium ions, one uses water dissociation constant which is derived from the law of mass action. In this case, considering the temperature at 25°C I come up with a number which is smaller than one. A logical suggestion here is probability. Is there any limitation to the law of mass action, i.e. does it break down at the nanoscale level and quantum mechanics and thus probability kick in? If yes, is there any boundary to specify things with respect to the size of the system?

There is definitely a breakdown of mass action at very low numbers of molecules. This can happen either because concentrations are tiny, or because the system of interest is tiny.

In either case, your idea to turn to probability is an excellent one, and is how these systems are modeled. But the reason doesn't necessarily have to do with quantum mechanics. It's just a consequence that matter is not really a continuum that can have any concentration, but is in fact granular. At the scale of atoms, there can either be an atom, or not be an atom. You can't have 0.5 of an atom.

The right boundary at which mass action breaks down depends a bit on the particular system, in particular, if reactions are occurring that depend on H+. In the less complicated case of no reactions depending on $\ce{H+}$, then the boundary between mass action being valid and not is actually fairly easy to calculate. If you know the average or expected value of $\ce{H+}$ ions in your water sphere (which you do, since you calculated it) the actually observed number of ions will follow a Poisson distrubtion. If the mean number of atoms is $N$, then the standard deviation of atoms is $\sqrt{N}$, so the relative uncertainty in the mean number of atoms is proportional to $\frac{\sqrt{N}}{N}=\frac{1}{\sqrt{N}}$. According to your calculation, $N$ is less than one, so the relative uncertainty is greater than 100%! But if you enlarged the sphere so the expected number of ions was $N=10$, then the uncertainty would be about 32%, still pretty bad. If you enlarged the sphere so $N=10000$, then the relative uncertainty is only 1%, meaning that mass action predictions for this system are pretty close to accurate! Getting $N$ tht high requires a sphere about 10,000-times bigger in volume, which means the radius would have to be about 38 times bigger. But keep in mind this result is only true for the water dissociation problem. So each chemical system has its own scale where mass action breaks down.

For systems where reactions are occurring that depend on $\ce{H+}$, this simple logic breaks down. Complex probabilistic simulations need to be done. Depending on the details of the reactions, the chemical master equation or the chemical Langevin equation would have to be used.

One time I calculated the concentration of frogs in an aquarium was about 1 zeptomolar. I don't think we could assume frogs followed the law of mass action!

• Isn't Heisenberg's uncertainty principle stating the exact opposite of what you say in paragraph two? – Martin - マーチン Feb 28 '15 at 19:48
• Martin, I'm not sure what you mean. Can you elaborate? – Curt F. Feb 28 '15 at 20:38
• It may be worth saying that the limit at which deterministic kinetics represent the system reasonably well is called the Thermodynamic Limit. Also, van Kampen's System Size Expansion is a work worth citing. – WYSIWYG Sep 11 '15 at 8:29

A detailed analysis of the significance of the mass action law at the limit of very few molecules can be found in the book The Law of Mass Action. Section 3.7 ('Breakdown' of the law of mass action) on page 99 even specifically analyses the case of water dissociation in very small droplets, though the discussion and some of the results derived within are general.

It takes statistical thermodynamics to prove, but as Curt points out well, it's a consequence of molecules/ions coming in discrete (integer) amounts. We're used to doing calculations in systems involving many billions of species, so we can get away with treating the number of any species $A$ as continuous, even though it's an approximation; we unconsciously equate the most probable amount of $A$ with the expected amount of $A$.

One of the consequences of making the distinction clear is that systems with zero molecules of a reagent or product can still be in equilibrium, even though a naive application of the mass action law would result in undefined equilibrium constants. What differentiates then an equilibrium constant of $10^{100}$ from $10^{200}$ (assuming the same amount of species are involved) is how often the equilibrium can be found with zero molecules of the products; the higher the equilibrium constant, the less likely any given measurement will indicate the presence of at least one molecule of reactant.

This can be an unintuitive concept to grasp, and since its derivation is somewhat involved, it can be hard to explain. However, there is a simpler case which shows up in physics, based on the granularity of light (photons) rather than the granularity of matter (atoms), which is useful as a stepping stone. I present it below:

Imagine a point light source in a perfect vacuum with a power of $100\ W$, emitting light at a certain wavelength, say $500\ nm$. Via the formula $E=\frac{hc}{\lambda}$, it is possible to calculate that the light source is therefore emitting about $2.52 \times 10^{20}$ photons per second. As is well known, light intensity falls off with the square of the distance $r$ from the source, since the photons are spread out over a spherical shell with area $4\pi r^2$. Thus, for example, at a distance of $r=100\ m$ from the source, the $2.52 \times 10^{20}$ photons per second are spread out over an area of approximately $31416\ m^2$, which means the photon flux is that distance is $\Phi = \frac{2.52 \times 10^{20}\ photons\ s^{-1}}{4\pi (100\ m)^2} = 2.00\times 10^{15}\ photons\ m^{-2} s^{-1}$.

Now we place a perfect photon detector with a $1\ m^2$ cross section facing the light source. At a distance of $100\ m$, as previously calculated it will detect $8.02\times 10^{15}$ photons arriving in one second. The distance between the light source and detector is then increased greatly to $10^7\ km$. At this new distance, we calculate a photon flux of $\Phi = \frac{2.52 \times 10^{20}\ photons\ s^{-1}}{4\pi (10^{10}\ m)^2} = 0.200\ photons\ m^{-2} s^{-1}$. Thus, the detector, with a $1\ m^2$ cross section at $10^7\ km$ away from the source, will count 0.2 photons per second. There is no such thing as a fifth of a photon, so what does this number mean? Of course, this number is meant as a statistical time average. If the detector is left open for $10^5$ seconds, then we can expect it to detect $2\times 10^4$ photons.

But what happens if we force the detector to stay open for only one second at a time? Once again, there is no such thing as a fraction of a photon, so the detector must display an integer number of counts. How can the expected amount of 0.2 photons per second possibly be met in this case? Easy: sometimes the detector will measure photons, and sometimes it will not. In fact, if the test is repeated many times, we will find that the detector will count exactly 0 photons with 81.9% probability, exactly 1 photon with 16.4% probability, exactly 2 photons with 1.6% probability, and so on, such that the probabilities obey a Poisson distribution with an expected value of 0.2.

In much the same way as photons are discrete, molecules are also discrete. Thus, when a reaction at equilibrium contains less than one molecule of a given species, what is actually meant is that there will be times when the system will be measured to have zero molecules, and other times more than zero. The smaller the fraction of expected molecules, the less probable it is to measure the system while containing more than zero molecules of the species.