[scientizscht (OP)]

Hello

I want to create a solution and I plan to dissolve an amount of the substance into a bottle of water.

I will stir the solution well for the substance to dissolve but I will then leave it for hours.

Will the substance deposit at the bottom or since it is in a lower concentration, it will stay diluted?

Thanks

[Bored chemist]

If you stir a solid in water to make a solution then that solution will be stable (at least as long as the temperature doesn't change)

[scientizscht (OP)]

If you stir a solid in water to make a solution then that solution will be stable (at least as long as the temperature doesn't change)

Won't the dissolved substance drop into deposit due to gravity with time?

[chiralSPO]

If you stir a solid in water to make a solution then that solution will be stable (at least as long as the temperature doesn't change)

Won't the dissolved substance drop into deposit due to gravity with time?

Not if it's dissolved...

[AustinnEp]

Hi all

When making some comps with water and AL that need Boric Acid is there an easier way to get a water / boric acid solution made up before for the mixing? Most comps I see measure it out by weight and then you have to be very precise about getting the right amount of water with it dissolve din first.

I was thinking that I could simply make up a spray bottle first that I use for the purpose. What concentration would be good for this? If it is a good idea at all?

[JazzHandsMafia]

Okay, the question here concerns equilibrium.
If you create a solution by pouring some presumably soluble compound, then the general equilibrium will follow
A ↔ A(aq), if A is some non-ionizing compound that still enters into solution.
You are probably picturing something like
AB ↔ A⁺ + B^-, where AB is some ionic salt that is water soluble.

In either case, the degree to which the compound dissolves is a thermodynamic property depending on temperature, pressure, and composition. I.e, 'K', the equilibrium constant, depends on the system.

If K is a large positive number, the compound you are trying to dissolve will readily enter into its dissolved forms. If the value is between 0 and 1, this process is not considered favorable.
However, K is merely refers to where the continuous process of going back and forth between the dissolved and undissolved forms tends to rest (more dissolved vs. undissolved or vice versa)

Therefore, there is always some amount of the compound in either form.
Unless you supersaturate the solution, the amount of undissolved material might be (very) small, approaching none.

It is possible some compound (either form) will adsorb to the container walls, thus decreasing the effective solution concentration, and creating another layer of the total system equilibrium.

Note: Equilibrium is treated as a thermodynamic property, and is generally separated from the kinetic system view. In other words, it will tell that something will happen, not how fast it will happen.
Because of that, even if your compound is noted as being water soluble, it might take some time to actual dissolve.

Summary: Unless you add more compound than can be readily dissolved in the system, then once it dissolves, you should be fine. Keep stirring the solution at a constant temperature to make sure it will stay dissolved and homogeneous (same composition throughout).
Either way, it will eventually reach equilibrium throughout the system.