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How do precipitation reactions behave in the absence of gravity, say on the International Space Station (ISS)? I have seen water taking the shape of a sphere and not that of the container in space due to its surface tension. It takes a spherical shape in order to minimize the energy due to forces of surface tension. In space all directions are equivalent and hence there is no notion of up or down.

In precipitation reactions, generally, the precipitate formed either floats on the surface or settles down. I know the solid particles will be formed in space due to the chemical reaction between the reactants. What will happen to the solid particles (precipitate), will they concentrate at the centre of the sphere of the reaction mixture (solution) or will they come to the surface, or will they remain suspended, or is there any other possibility? Has such an experiment been done on the ISS to date?

For reference, this is how water looks in the absence of gravity:

enter image description here

Please note: Water is coloured to enhance visibility

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    $\begingroup$ @Poutnik, I thought of that only for particles which form at the centre of the sphere where all the forces balance each other to give no net force. But, I think if they were formed somewhere between then they must come to the surface. So, I think some particles may be at the centre and most on the surface. But, I don't know what will happen in reality. And that's why I posted the question here :) $\endgroup$ – Intellex Oct 20 at 4:47
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    $\begingroup$ @Poutnik, Thank you. So are you trying to say all the particles will be uniform throughout? If so will the surface be entirely water molecules or does have some precipitate? $\endgroup$ – Intellex Oct 20 at 4:54
  • $\begingroup$ OK, I have summarised the comments in the answer. $\endgroup$ – Poutnik Oct 20 at 5:19
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In gravityless environment, all directions are equivalent save the very vicinity of the surface(*), where just few water molecules are toward the surface.

Therefore, in the bulk volume, the precipitate would stay where it is, unless some currents exist for whatever reason.

At the very surface, it depends on nature of precipitate.

If it is rather hydrophilic, water molecules would have tendency to surround it, so it would be rather pulled from surface inwards.

If it is rather hydrophobic, water molecules would have tendency to expelled it, so it would be rather pushed outwards to the surface.

If it is far enough ( tens molecules ) from surface, all effects will get spherically symmetric, but one....

(*) ..... as in a very very long term, precipitate denser than water would be more concentrated near the centre by the own gravity of the "bubble" and vice versa.

I am not aware of such an ISS experiment, but it makes little sense to do it for me. In the microgravity context, it is very very long term experiment.

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    $\begingroup$ Sure, but just in the center. :-) There will be microgravity force toward the centre, strongest at the surface, zero at the centre. $\endgroup$ – Poutnik Oct 20 at 5:26
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    $\begingroup$ Rather, they will, not might, but very very slow. $\endgroup$ – Poutnik Oct 20 at 5:30
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    $\begingroup$ I am not, but it makes little sense to do it. It is very very long term experiment. $\endgroup$ – Poutnik Oct 20 at 5:33
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    $\begingroup$ When it comes to the self-gravity of the drop vs. Brownian motion, I suspect that the gravity will only ever become apparent statistically - the particles of precipitate won't hang around near the centre for very long. This random motion may also mean that over time particles will get close enough to the surface for them to stay there if hydrophobic. I wonder how we could model this in the lab - acoustic levitation probably creates internal forces in the drop; electrostatic perhaps less - A nice undergrad project (if only I had the kit) $\endgroup$ – Chris H Oct 21 at 10:57
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    $\begingroup$ @Intellex Precipitation of supersaturated solutions has been studied on the ISS and Space Shuttle. Protein crystals for X-ray diffraction are the goal. ncbi.nlm.nih.gov/pmc/articles/PMC5515504 $\endgroup$ – DavePhD Oct 21 at 13:46
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Inspired by @Jan's Answer.

Following image considers the case when the liquid drop is under rotation (shown by blue curved arrows). Let us consider the frame of reference as one which is rotating with the same angular velocity as that of the droplet. Since we are in a non-inertial frame, we need to consider the pseudo forces - and here it is the centrifugal force (acts away from the axis or centre of rotation). Due to this, there is some kind of artificial gravity inside the sphere as mentioned in @Jan's answer. This force is represented by the orange arrow in the diagram. Due to this, and due to the light nature of bubbles of dissolved gas, the bubbles form in the central region as denoted by the dotted circle region. Here the precipitate will collect on the inner surface of the sphere if it is denser than water (shown by the light green portion along the circumference in this cross-sectional view).

enter image description here

The above image is a view from the top of the axis. The following image is a side view (cross-sectional view again!) of the rotating system. See the sphere is slightly compressed at the poles and elongated at the equator. The precipitates accumulate near the equatorial region of the sphere.

enter image description here

For Both Images:

Image Source: My own work :) (Hope you like it!)

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In addition to Poutnik’s answer, it is possible to manipulate a drop of water in space to generate a force similar to gravity within it: by applying gentle airflow to make the drop of water rotate.

Water in a rotating drop is subject to centrifugal forces, meaning that there is now an ‘along the force’ and an ‘opposite to the force’ direction which act in a very similar manner as ‘up’ and ‘down’ on Earth. Therefore, a precipitate that would collect at the bottom of the flask on Earth collects on the outside of a rotating drop in space.

I read that an experiment of this type was performed with either carbonated soft drinks or beer in space. The amusing part is that carbon dioxide is, of course, lighter than water and thus collects in the middle of a rotating drop.

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    $\begingroup$ Now, the drop will be compressed at the poles (or along the axis) and elongated at the equator if you consider it to be rotating. $\endgroup$ – Intellex Oct 21 at 5:21
  • $\begingroup$ With reference to the previous comment kindly see this answer. $\endgroup$ – Intellex Oct 21 at 8:37

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