1
$\begingroup$

Goals

There are a few things that I'm looking to do:

  1. Separate salt from seawater via some evaporation technique.
  2. Separate the each of the major salts from the resulting sea water to get each into their purest possible form.
  3. In particular--and if possible--I want to try and separate any iodine-containing salts from the salt water.

General Information

As I've found, salt water is composed of much more than just $\ce{NaCl}$. According to Britannica, 99% of all sea salts are composed of the following ions by weight: $\ce{Ca^2+, Mg^2+, K+}$, and $\ce{Na+}$ cations, as well as $\ce{Cl-}$ and $\ce{SO4^2-}$ anions (they have a table showing concentrations of all of these ions). There are many more molecules and ions to find to find in the ocean, but these are the major ones in terms of weight. With respect to iodine (you'll see why this is relevant later), I've found other sources that say that iodine is one of the more abundant micronutrients in seawater.

These dissolved ions, without the presence of water, can combine with themselves and other substances found within seawater to create salts including calcium carbonate, gypsum (hydrate of calcium carbonate), sodium chloride (halite), potassium chloride and magnesium chloride (source). Note the way that the salts are separated by solubility here.


My Plan (WIP)

First, I'm going to get a boat out into the middle of the ocean and get a few liters of sea water--ideally, as much as I can take.

Then, I'll reduce the solution until only particulate matter is left, either by evaporating the sea water outside with the sun and/or boiling the seawater in batches over a stove. Once the particulate matter is isolated, I could dry it further in an oven. I'm not sure about any of the advantages or disadvantages with these methods aside from power consumption, so I'll make more than one batch and subject each of them to different procedures so as to see what kinds of yields I get for each batch.

Finally, I'll have to separate the salts somehow, and this is where I'm lost. I have a couple of ideas for doing this:

  1. Using a microscope, I could separate crystals with visually different shapes from others. I would like to avoid this, since it seems both inefficient, imprecise, and time-consuming. Plus, I don't own a microscope.
  2. In one of the sources above, they say to separate the salts by solubility. I have never separated soluble solids before by differing degrees of solubility, so this would be new for me, since I wouldn't be able to just dissolve a completely insoluble solution in water and filter the concentrated particulate from the water. However, this seems like the most reliable method I could find for separating sea salts, so I think this would be the best method for doing this.

Tl;dr, I plan to make my own sea salt by getting salt water from the ocean and evaporating the water out (either via boiling or leaving it out in the sun) until only salt is left. Although, instead of using the sea salt as is, I want to see if I can separate it to see what particular kinds of salts are left behind. In particular--and if possible--I'm looking for iodine salt, to see if I can make my own iodized salt from just seawater. However, I don't really know how to separate water-soluble solid substances from each other, and I don't know if I'll even be able to obtain any salt with iodine in it, so I'm looking for any tips, guides or procedures for doing this.

$\endgroup$
1
  • 1
    $\begingroup$ Not all sea water is the same. If you want iodine salts, you might need to start with some that is high in iodine. OR, perhaps better, start your experiments by acquiring some sea salt (ie flakes produced by evaporating sea water which saves a step in your experiment). $\endgroup$
    – matt_black
    Commented Feb 20, 2023 at 21:47

2 Answers 2

1
$\begingroup$

Sounds like you're on the right track -- but rather than separating crystals by hand, use the difference in solubility of salts to make the work easier.

In commercial salt evaporation ponds, with a concentration gradient, the least soluble salts will precipitate first. This is perhaps the least work-intensive way to separate the various salts. So you might try letting the brine slowly trickle down an incline in the sun, and see what precipitates out at various places. This Wikipedia solubility table might give you an idea of what will precipitate, knowing the concentrations of species in sea water, for example. BTW, this may be used to recover the now-valuable lithium from brines.

$\endgroup$
-1
$\begingroup$

I would suggest that you look up the solubility products of salts such as CaSO4 and all the other things which you can form from the cations and anions which are present in seawater.

I would heat the sea water to reduce the volume of water present and thus increase the concentration of the salts. Rather than boiling to dry salt, I would want to reduce the volume slightly. This might cause the least soluble salts to come out of solution.

There is a big problem which the solubility tables cannot help you with, this is the fact that the activity coefficents (or activity functions) of the species in the sea water will change as the concentration of the sodium chloride (and other salts) changes.

The solubility product of a salt is defined in terms of activities, so we can write for calcium sulfate.

Ksp = [SO42-][Ca2+]fSO42-fCa2+

We can get the values of fCa2+ and fSO42- using either a Pitzer equation or the SIT equation. I hold the view that the vast polynomial in the Pitzer equations will allow them to fit anything but as the polynomial increases in size while its ability to fit things increases the fit is becoming more and more abstract and less grounded in chemical theory.

The SIT equation is a good compromise it would be for calcium cations

Log (fCa2+) = -A z+ z- sqrt(I)/(1+(1.5 sqrt(I))) + B[Cl-] + B'[SO42-]

We can ignore the B' term as the concentration of sulfate in the mixture will be very low.

If you read Effect of Chloride Salts and Bicarbonate on Solubility of CaSO4 in Aqueous Solutions at 37°C by Yuexia Zhang, Zhenhua Yang, Dan Guo, Hong Geng and Chuan Dong, Procedia Environmental Sciences, Volume 18, 2013, Pages 84-91 then you will see how the solubility of calcium sulfate changes as a function of sodium chloride concentration and temperture.

Using the data from another paper, I was able to make this graph.

Graph of calcium sulfate solubility at 298 K as a function of sodium chloride concentration

As a result it should be clear that if the solubility product is expressed in concentrations (an apparent solubility product) then it will change as a function of the sodium chloride concentration.

The A constant in the Debye Huckel term in the SIT equation at 25 oC will be 0.509. I was able to fit the data in the paper using the solver function in excel where I used a slightly more complex activity function equation.

I computed the average activity function for calcium and sulfate using

log f = -2A sqrt(I)/(1+(Q sqrt(I))) + B[NaCl] + C[NaCl]^2

Where A was 0.509 Q was 1.55 C was 0.02796 B was -0.1316 Ksp = 0.000165

If you want to make a rational design of a process for making pure sodium chloride from sea water by partial evapouration of the water then you will have to do a lot of modelling of the mixtures of the different salts in sodium chloride solutions. If you do this then you will find it much more easy.

$\endgroup$

Not the answer you're looking for? Browse other questions tagged or ask your own question.