Dissolution of Carbon Dioxide in water It may sound like an old problem but I don't find a good reference.
Gases are physically soluble in water. Given a water column without any salts solved in it. Air with a given concentration $c_0$ (maybe as mol $\text{CO}_2$ per litre or whatever unit) of carbon dioxide gas is given in the water. The gas solves (and a part of it reacts with the water to carbonic acid). Finally the process comes to an end when equilibrium is (asymptotically?) reached. Then the chemical potential in any high is balanced and Henrys's law at the surface is fulfilled.
The effective chemical potential depends from high $h$ above the ground of the column $\mu\left(h\right)=\mu_{0}+mgh$, where $m$ is the mass of one molecule.
So one would expect a carbon dioxide concentration $c(h)$ that depends from $h$.
The question is:

*

*Is this right an then, what is this $c(h)$-dependence?
I would guess formally the same as the barometric density distribution in the atmosphere?

*Are there any measurements?

As said, this sounds nearly like a textbook example. But where could I search for it?
 A: At equilibrium the entire column will have the same concentration and that concentration will satisfy Henry's law. The only way to have a concentration gradient is if you have a steady removal of CO$_2$ inside the column. The reaction with water you mention cannot be run for ever because they depend on the finite amount of water in the column and eventually will reach equilibrium.
The process you are discussing sounds like a standard problem of diffusion and reaction and is discussed in chemical engineering textbooks.
A: Yes, if the system is in thermodynamic equilibrium, gravity will cause gradient of concentration of the dissolved molecule in water, the same as in atmosphere (with concentration being exponentially decreasing function of height). This is easy to see both from statistical physics laws and from general thermodynamics.
Given the system is in thermodynamic equilibrium, then it has the same temperature everywhere, and the molecules should be distributed in accordance with the Boltzmannian  probability proportional to $e^{-\frac{mgh}{k_BT}}$. That  implies concentration profile that decreases exponentially with height $h$. The highest concentration is at the bottom.
The other argument is as follows. Let's assume $\text{CO}_2$ in atmosphere outside the column is described by the familiar exponential function of height. Let the water column of height $H$ be held in place due to a capillary dipped in water reservoir: the top is open and exposed to the atmosphere, and the bottom is also open and surrounded by the water at the ground level. In equilibrium, the top layer of water (with the shape of concave meniscus, or the "U shape") has to be in equilibrium with the thin concentration of $\text{CO}_2$ in the atmosphere at height $H$, which we can denote $c_a(H)$; so the concentration in the water is $c_w(H) = kc_a(H)$, where $k$ is the constant of proportionality for $\text{CO}_2$ and $\text{H}_2\text{O}$ (due to Henry's law). And the bottom has to be in equilibrium with concentration of $\text{CO}_2$ in the reservoir, which in equilibrium is $kc_a(0)$. Thus concentration of $\text{CO}_2$ in the capillary has to decrease with height. Now, imagine we close the capillary at the bottom so it becomes a thin and tall vessel with a closed bottom and an open end. This will change nothing in the equilibrium state of the system: the water height in the capillary will stay the same, and the concentration of the $\text{CO}_2$ molecules at every height will stay the same.
You may say, but this only fixes two points, the top and the bottom, what about the intermediate heights? Maybe the concentration there is different from the exponential law?
We can put any water layer in the column at any height $h$ in contact with the atmosphere via very small opening in the wall, so the capillary effect won't allow the atmosphere to be sucked in and destroy the equilibrium. Then the same argument applies: the water in the opening is in thermodynamic equilibrium with the atmosphere at that height, so concentration of $\text{CO}_2$ in the water has to be $c_w(h) = kc_a(h)$.
It is hard to find a water column high enough to observe this easily. Ocean has a high water column, but it is not in thermodynamic equilibrium. Nevertheless, at great depths >1km, concentration of atmospheric gases increases slightly with depth, but this has a lot to do with the fact water on the surface gets oxygenated and cooled by the atmosphere, then it falls down due to its higher density. The increasing concentration with depth is thus not a manifestation of thermodynamic equilibrium.
