How does the dissolution of salt affect the solution density?

Question

Suppose you have a container of water as a solvent and you a certain amount of salt as a solute sitting at the bottom of the container that has yet to start dissolving. Supposing temperature and pressure are kept at a constant.

So at this stage: Water has density $\rho_w(x, y, z, t)$. Salt has density $\rho_s(x, y, z, t)$. And as the solution at the beginning is just comprised of the water itself it it has density $\rho_t(x, y, z, t) = \rho_s(x, y, z, t)$.

Now as the salt begins to dissolve the solution becomes a mixture of salt and water and we have $\rho_t = \alpha \rho_w + \beta \rho_s$.

I have a few questions about the state of the system now:

1. Is the density of water still the same as it was at the beginning now that salt has begun dissolving? Ie. is the density of water constant? Or will the dissolved salt molecules "squeeze" the water molecules into a smaller volume thus increasing the density of water? And actually we should be talking about concentration of water now instead of density of water as we are dealing with a mixture comprised of two substances?
2. Is the concentration/density of salt in the solution constant? This questions seems to have the obvious answer of "no it isn't because the salt is dissolving and thus the concentration of the salt in the solution is changing over space and time as it spreads out"?
3. The density of the solution itself, $\rho_t$...as the concentration of salt seems to be non-constant it therefore implies that the density of the solution is non-constant...I.e. it will vary at different points in the fluid over space and time as the salt dissolves until it reaches an equilibrium when all the salt is dissolved?
4. So if the solution has non-constant density it means it is compressible? And it will be governed by the compressible Navier-Stokes equations?

I'm trying to program a numerical method for a similar situation to what I described above so I appreciate any help with these questions!

Answer 1

Question 1: Is the density of water still the same as it was at the beginning now that salt has begun dissolving? Ie. is the density of water constant? Or will the dissolved salt molecules "squeeze" the water molecules into a smaller volume thus increasing the density of water? And actually we should be talking about concentration of water now instead of density of water as we are dealing with a mixture comprised of two substances?

If you don't mix the water, the salt will slowly dissolve into it, starting at the bottom of the container. This will give you a concentration gradient in the container, where the highest density corresponds to the solubility limit of salt at that temperature, on the bottom of the container, and the density decreases as you go up in the container. For a container that is sufficiently deep, you should have fresh water at the surface for a certain time, but salt will slowly diffuse from lower in the container, so I doubt that the surface will stay totally fresh as you observe the container for long time periods. There is a practical device, known as a solar pond, that operates on the principle that the density of water at the bottom of the pond is so high that absorbed solar radiation will not heat the bottom of the container enough to induce convection currents, effectively enabling the "top" water to insulate higher temperature "bottom" water. See https://en.wikipedia.org/wiki/Solar_pond for details. Note that the concentration gradient remains stable, even though the bottom of the pond is substantially warmer than the top of the pond.

Regarding your other sub-questions, the sodium ions and chlorine ions from the salt crystal become "solvated" with water molecules. From a chemistry viewpoint, I doubt that it is correct to assume that the different molecular species remain separate when the salt dissolves in water.

Question 2: Is the concentration/density of salt in the solution constant? This questions seems to have the obvious answer of "no it isn't because the salt is dissolving and thus the concentration of the salt in the solution is changing over space and time as it spreads out"?

As mentioned previously, the concentration of salt is not constant, IF you don't mix the water. Even if you let a lot of time go by, there will be a concentration gradient in the water column that is enough to allow solar ponds (mentioned above) to work.

Question 3: The density of the solution itself, ρt...as the concentration of salt seems to be non-constant it therefore implies that the density of the solution is non-constant...I.e. it will vary at different points in the fluid over space and time as the salt dissolves until it reaches an equilibrium when all the salt is dissolved?

If you carefully set up this experiment and do not stir the water, it is possible to have undissolved salt at the bottom of the container. Whether or not this is the case depends on how much salt you add.

Question 4: So if the solution has non-constant density it means it is compressible? And it will be governed by the compressible Navier-Stokes equations?

Unless you intend to impose VERY high pressures on your container, you can consider the liquid to be imcompressible. The density profile is caused by the concentration gradient, not compressibility.

Assuming that you want equations to estimate density vs. height in the water column, you may want to start with solar pond design. I have no doubt that designers had to work some of the same problems you are dealing with.

Answer 2

I think there are a few misunderstandings that led your question to be rather obscure.

I believe your first 2 questions amount to: "Is there a volume change between the fresh water+crystal salt system and the solution" ? Of course the mass is constant. I don't have the answer to this question, but put in this way you'll be able to look for it.

Question 3 is about diffusion. Look for Fick's law.

Question 4 is interesting. Actually, it is also a misunderstanding but a very common one in fluid dynamics. The equation $$\nabla \cdot \vec{u}=0$$ in Navier-Stokes equations is stronger than just incompressibility: it is incompressibility and iso-density. If you don't have isodensity, then the full continuity equation is $$\partial{\rho}/\partial{t} + \nabla\cdot(\rho\vec{u})=0.$$ You can complement that with a Fick's law for the diffusion of one species (e.g. salt concentration $c$) and a law relating $\rho$ and $c$ locally.

Answer 3

The question you are asking is actually a very complicated one, so I'll give a partial answer and some pointers to where you can find out more. It might be more productive to work in terms of "partial molar volumes" rather than densities. In any case, it is always easy to convert from partial molar volumes to partial densities. Generally the partial molar volume of species A, $v_A(x_A)$, is a function of the concentration of species A, $x_A$. For salt and water it is probably possible to look up functions which are fits to experimental data on partial molar volume as a function of concentration.

A key idea is that partial molar quantities, such as the partial molar volume, are always related via relations like the Gibbs-Duhem equation. It shows how, if you know how the partial molar quantity of one component of a mixture varies with the mixture composition, you can infer how that quantity varies for the other component of the mixture. So, in your case, you only need a function for the partial molar volume of salt and then you will be able to use a Gibbs-Duhem equation to get the corresponding function for water.

A good place to start learning about this would be a physical chemistry textbook, such as Atkins.

Answer 4

If you don't mix the water, the salt will slowly dissolve into it, starting at the bottom of the container. This will give you a concentration gradient in the container, where the highest density corresponds to the solubility limit of salt at that temperature, on the bottom of the container, and the density decreases as you go up in the container. For a container that is sufficiently deep, you should have fresh water at the surface for a certain time, but salt will slowly diffuse from lower in the container, so I doubt that the surface will stay totally fresh as you observe the container for long time periods. There is a practical device, known as a solar pond, that operates on the principle that the density of water at the bottom of the pond is so high that absorbed solar radiation will not heat the bottom of the container enough to induce convection currents, effectively enabling the "top" water to insulate higher temperature "bottom" water.