Would this Desalination by Ion Exchange method work?

Factual Questions
1

I’m really interested in the problem of cheaply desalinating water for some reason. =) To me, it’s one of those problems that can be stated so simply and is so specific yet has so many possible solutions. I’ll try to motivate this topic with water softening systems…

We’ve all heart of water softening and why most homes, at least in America, have water softeners. It’s because even small concentrations of Mg2+ and Ca2+ cations can lead to precipitate forming in pipes and drains, although I think it wouldn’t be too bad if you drank “moderately” hard water. The way that these small concentrations of Mg2+ and Ca2+ are reduced even further to the point where they are negligible is by exchanging these two cations with Na+ using what is called an ion exchange resin. Ion exchange resins pretty much do what they say; two non-obvious facts are that the law of conservation of charge applies (i.e. two sodium cations are exchanged for one of either magnesium or calcium cation), and that the exchange process is reversible, so that the Mg2+ or Ca2+ can be re-exchanged for Na+ once more; upon the completion of the re-exchange process, the ion resin is said to be “rejuvenated.”)

Is there any reason we want to replace the Mg2+/Ca2+ with Na+ in particular? Somewhat. Na+ isn’t the only ion that we can use for water softening (K+ is also one that is commonly used); in general, any cation can be chosen to replace the Mg2+/Ca2+ by selecting the appropriate resin, but we wouldn’t use Pb2+/Hg2+ (too dangerous) or Ag+ (too expensive). Na+ is a good choice because first, having a slightly higher concentration of Na+ in our bathwater and drinking water is negligible (health, taste, etc.), and secondly, and more relevant on a chemical level is the fact that it has a solubility that is far higher than that of Mg2+/Ca2+, and so essentially eliminates the problem of precipitates and deposits forming in home piping systems.

Key point: If you don’t want precipitate to form in a multi-component ionic solution, then a possible solution is to exchange the less soluble ion(s) in that solution for a more soluble one(s).

Where my idea comes in, and adds a twist to the above key point:
If you do want precipitate to form in a multi-component ionic solution, then a possible solution is to exchange a more soluble ion in that solution for a less soluble one. Specifically in the context of desalination of seawater, use an ion exchange resin that originally contains a cation that is very insoluble in water in the presence of Cl- (which is the major anionic component of seawater), removing an equimolar amount of Na+ cation in exchange (which is the major cationic component of seawater).

The singly dissolved ionic specie of copper, namely Cu+, is extremely insoluble in water in the presence of Cl-, with its solubility being 0.00066 mol CuCl/L water (this is in pure water; because of the common ion effect, the solubility of CuCl in seawater would be drastically lower than even this).

The molarity (M = mol/L) of sodium chloride in seawater is 0.469 M, with the chloride ion having a slightly greater concentration than the sodium ion. If we place our ion exchange resin that donates a Cu+ ion and accepts an Na+ ion in our seawater sample, each small amount of Cu+ added would result in the two following consequences: first, an equimolar small amount of Na+ would be removed from solution through the ion exchange resin as expected, and secondly, provided that “small” is not at the level of individual ions or at the nano-scale, an equimolar small amount of CuCl would precipitate from the aqueous seawater solution as a result of the extremely low solubility of the Cu+ ions in the presence of the Cl-.

Since the second outcome will continue to occur even when most (i.e. almost all) of the Cl- has been precipitated; this being because the CuCl solubility is so very low, continuing to exchange the Cu+ for Na+ will ultimately lead to, given the ion resin exchange capacity is sufficient, a (direct) removal of most of the Na+ cations, with the (“indirect,” that is, as a result of precipitation) removals of both the Cu+ ions which replaced the Na+ ions and the Cl- ions, with the CuCl precipitate sinking down to the bottom of the container where it can easily be filtered out.

Finally, to rejuvenate the ion resin, one would place the recovered CuCl precipitate in ammonium hydroxide (NH4Cl) or concentrated HCl, in which CuCl is highly soluble, and then let the Na+ that has built up in the resin where the Cu+ once was diffuse into either of the two CuCl solutions, with the Cu+ reoccupying its former place in the resin.

It seems to me that the remaining sticky points seem to be whether this procedure can be repeated to remove ions other than Na+ and Cl-, and I’m guessing the answer would be a tentative yes, although the appropriate ion resins might not be manufactured.

The biggest problem seems to be whether this process can be first, scaled up to nontrivial amounts. Since the ion resins could be rejuvenated as mentioned above, the scalability problem wouldn’t necessarily be in that we would constantly have to be buying new ion exchange resins, but more so in the question of whether the rejuvenation process itself is scalable, how quickly it can be done, wear and tear, and whether there are resins that can deal with “standard” seawater, which is a rather highly concentrated solution. Of course, there’s the cost issue…

I’ve pretty much ruled out this process being viable for standard seawater at almost any scale larger than the laboratory or for personal use; that is, if this process is even viable in principle at any scale. Overall what do you think of this method/process? And specifically, what about its 1) adherence to scientific principles 2) practicability in niche water treatment/desalination areas (specifically with desalination of water with concentrations [brackish groundwater for example] falling between the very dilute hard water softening situation and the concentrated seawater.

Thanks very much for those of you who took the time to read this. =)

Cheers,
supery00n

2

If I understood this, the reaction you’re proposing is: (An is just some anion)

CuAn (solid) + NaCl (aq) -> CuCl (solid) + NaAn (solid)

Some problems might be:

  1. This reaction seems to be reducing entropy, so it needs a lot of driving force to go forwards. I think it might even be endothermic, meaning it won’t happen at all. NaCl is very stable when dissolved, that’s why it’s so hard to get out of water (and so good at taking the place of other cations).

  2. Rejuvenating the resin may require an equal amount of freshwater (ideally). The final concentration of Na+ in the resin won’t be less than in the HCl or NH4Cl solution you use.

  3. Even if the reactions work, this method probably uses more energy or requires more maintenance than, say, distillation.

3

That’s a really great point; I didn’t take into account the fact that what is in the resin both initially (prior to the ion exchange) and afterward is in the solid state; two solid species on the product side versus only one on the reactant side, in addition to the high stability of NaCl(aq). I tried to find a reaction that would, at 25 degrees Celsius, have a Gibbs free energy of reaction that was negative, and I think I might have found one.

I chose An to be F-, the reaction being
CuF (solid) + NaCl (aq) -> CuCl (solid) + NaF (solid).

The standard Gibbs free energy of formation (at 25 C) of these four species are:
CuF (solid) = -171.544 kJ/mol
NaCl (aq) = -393.17 kJ/mol
CuCl (solid) = -119.88 kJ/mol
NaF (solid) = -543.51 kJ/mol.

Therefore, the standard Gibbs free energy of this reaction is -98.676 kJ/mol. So the reaction as written is spontaneous at conditions of 25C, 1 atm, and with the concentration of NaCl being 1 M. The actual concentration of NaCl in seawater being 0.479 M (and if this process is to be used for desalination of seawater, the reaction must remain spontaneous not just at 0.479 M, but at every other concentration less than 0.479 M), if the deviations of the Gibbs free energies of formation, molar entropies of formation, and enthalpies of formation of NaCl (aq) away from their respective standard values are sufficiently large, and if that difference is in the direction that would make this reaction nonspontaneous or much less so over the range of molarities “passed through” throughout the desalination process, then the actual nonstandard Gibbs free energy could very well be positive and thus the reaction could be nonspontaneous, the spontaneity having been shown for the ideal standard situation that is very different from the actual situation.

Essentially, the above calculation, although seemingly conclusive provided the correct reaction has been written down, has an assumption that the concentrations of NaCl(aq) encountered, which are strictly nonstandard concentrations at any point throughout the desalination process, are “standard enough” for us to say that the standard Gibbs free energy of formation is not too far off from the actual Gibbs free energy of formation.

http://chemistrytable.webs.com/enthalpyentropyandgibbs.htm This list contains tables of standard state Gibbs free energies of formations, standard molar entropies of formation, and standard state enthalpies of formation of various compounds; the ones that are missing I referenced from the appendix in my general chem textbook (Principles of Modern Chemistry 7th edition, Oxtoby et. al).

I calculated the standard molar entropy and the standard enthalpy of the reaction, and found that they are -111.386 kJ/mol and -42.692 J mol-1K-1. Since both quantities are negative, this means that there will be a *maximum *temperature above which this reaction is no longer spontaneous. Converting the standard molar entropy’s units so that the J becomes a kJ (divide by 1000), this maximum temperature is T = deltaH/deltaS = 2609 K, which was surprisingly large. Since 2609 K > the normal boiling point of water, this reaction (if our assumption two paragraphs above was valid) will be spontaneous throughout the entire temperature range of liquid water (at one atmosphere).

The reaction seems to be exothermic but involves a decrease in entropy. I think I might have gotten lucky and found a substance (NaF(s)) that seems to be even more stable than NaCl(aq).

Let’s say that the reaction as written (with the general anion in the resin being a fluoride ion) is in fact spontaneous. Doesn’t that introduce another problem here, where the reverse rejuvenation reaction would then be non-spontaneous to the same magnitude, namely +98.676 kJ/mol, and therefore would make the “maintenance” energetically expensive (in addition to the equal amount of freshwater needed for the rejuvenation), in effect, making any gains from spontaneity in the forward reaction being neutralized and forgone by the losses due to the nonspontaneity in the reverse reaction?

4

In continuation of my last post, I’ll try to find a fluoride compound other than NaF that can react with the CuCl(s) to regenerate the CuF(s) ; I think the problem of the nonspontaneity of the reverse reaction with NaF is pretty clear…

5

Oops. It turns out that CuF probably doesn’t exist (Wikipedia). I think a better forward reaction is AgNO3 (s) + NaCl (aq) -> NaNO3 (s) + AgCl (s); the silver ion in the resin being exchanged for the sodium ion in the solution so that the silver ion and chloride ion then precipitate. I calculated the standard-state G_f, and it’s -51 kJ/mol.

6

Nice detailed analysis. I didn’t check your calculations, but your point about concentration of NaCl sounds correct. At high concentrations it’s going to be easy to remove it, but can the reaction continue to a concentration low enough to drink?

The difficulty is finding a reaction energetic enough to proceed, yet not too much to make the reverse difficult. Maybe you could nudge the reactions using temperature or pressure.

7

I think you are confusing two things:
1> Ion exchange as the chemical process where ions are exchanged between two electrolytes (whether or not a resin a present)
2> Ion exchange as a sorption (Adsorption and Absorption) process where one of the ions (usually the bigger one) is trapped preferentially.

You are addressing part 1 only but part 2 is a mass transfer process.

Looking at AgNO3 (s) + NaCl (aq) -> NaNO3 (s) + AgCl (s)

I did not get into your numbers - but it is highly unlikely that NaNO3 will precipitate out - (Any chemistry novice will tell you that most nitrates are higly soluble). So NaNO3 (s) is very unlikely. Also for solutions - Gibbs free energy estimates are difficult - the industry norm is to use the OLI software (http://www.olisystems.com).

Also AaronX - slight nitpick here - I am sure you meant evaporation when you said distillation.

8

I thought distillation means you condense and collect the solute