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I am conducting an experiment that involves reacting a weak acid and a weak base to form water and a salt with this overall equation

$$\ce{Mg(OH)2(s) + 2CH3COOH(l) <=> 2H2O(l) + Mg(CH3COO)2(aq)}$$ Currently, I am trying to filter out the salt to purify the water; however, there are specific restrictions.

The filtration process must be able to:

  • be done quickly. I can't wait a long time for the water to be distilled
  • be done at home. This filtration needs to be something that the average person can could do with household ingredients or ingredients they can buy at a drug store (Assume that the $\ce{Mg(CH3COO)2}$ solution is given to them pre-made)

So far, I have tried to use Gelatin as a clarifying agent, which is typically used to clear beer. However, I couldn't find much info on using gelatin with salt water. Does anyone have any insight about using gelatin as a clarifying agent in this solution? Is there a better way to do such a thing that still meets the requirements?

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    $\begingroup$ I'm not sure what you're trying to accomplish, so I'm not sure if this will help, but have you considered ion-exchange? You can't filter something soluble, but something like ion-exchange resin is pretty easy to come by and is very fast and effective at removing soluble salts. $\endgroup$ – SendersReagent Apr 24 '16 at 22:36
  • $\begingroup$ it seems like ion-exchange resin beads would work. Where could someone buy some of these beads? What type of resin would I use to remove this specific salt? Will it remove the Magnesium and Acetate ions completely or replace them with Sodium ions? $\endgroup$ – Abob Apr 25 '16 at 4:54
  • $\begingroup$ Actually, I found this pdf showing that I can use a cation and anion resin beads. All that remains is where I could find such a thing $\endgroup$ – Abob Apr 25 '16 at 17:30
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Ion-exchange resin can be made with $\ce{H+}$ cations on one type of resin and $\ce{OH-}$ anions on another type of resin. These would leave the water without the any of the magnesium or acetate ions effectively without replacing the ions with other ions. This is how we deionize for general chemical use. This happens according to the reactions below: $$\ce{2HSO3R + Mg^{2+} + 2C2H3O2- -> Mg(SO3R)2 + 2HC2H3O2}$$ $$\ce{R4N+OH- + HC2H3O2 -> [R4N+ +\ C2H3O2^{-}] + H2O}$$ In these examples, the sulfonates/sulfonic acids and ammonium cations are tethered to the resin so that they can't dissolve.

The downside to this is that if you are trying to measure something else in the system, if it is ionic, it will be removed as well.

I've never had to source ion-exchange resin myself, so I'm not sure of the easiest way to go about it. There's pleanty of deionization resin on Amazon. The mixed stuff is simple to use, just mix it with your water, agitate for a while and then filter it out and you'll have removed all the ions. That is expensive, though, because it's single use.

You can also buy regeneratable ion exchange resins. You don't want to mix the regeneratable stuff, and you have to do two separate ion-exchange steps, one with the cation-exchange and one with the anion-exchange. It's pretty easy to regenerate, though, as far as I know. You just use strong acid with the cation-exchange to redissolve the soluble cations and use a hydroxide solution to regenerate the anion exchange resin. It can't be done forever because there are some cations and anions that just won't want to redissolve, but if you're doing it a lot, might be worth it. I can't seem to find the anion stuff except in bulk.

I'm guessing the mixed type would be more appropriate for most applications. If you really need it quickly, I think you can buy the filters at Lowes/Home Depot and just run it through those or open them up and get the resin out.

Update: These filters are the only ones I saw that said they actually have the "dual ion-exchange system." Their website actually says it gets down to "000 total dissolved solids," which is pretty impressive relative to other water filters from what I've seen (depending on what "000 TDS" actually means).

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  • $\begingroup$ Thanks a lot, that solves the finding the resin part. Also, at first your equation seemed confusing to me compared to what i found here, but I then realized that the water from the second equation comes from the hydroxide and hydrogen ions from the resin beads. I'm still confused on which type of cation and anion resin beads would be best to use, but i think It would be better to use one that doesn't need filtering, since the pitcher won't work well since every person who wants to replicate this has to spend $20 extra $\endgroup$ – Abob Apr 25 '16 at 17:33
  • $\begingroup$ If the pitcher is the only possible and/or most efficient way, it might work, but is there anyway to use the resins without filtration? Could I just use beads that exchange the ions (if they're not too hard to come by) and then filter them out with filter paper? $\endgroup$ – Abob Apr 25 '16 at 17:40
  • $\begingroup$ @Abob78 That's exactly what my students do, all they do is put the resin in a vial with the water, shake it up and decant it off. Works fine. The beads aren't like grains of sand or anything, they're pretty easy to get out. The "mixed bed" resin I linked above should work, but I just don't think its as readily available at a local store. Could be wrong, but I didn't find anything online. If I were in a pinch, I'd use the filter pitcher or, because I'm destructive, I'd gut the filter, see how much was in one and decide if it was practical to do that way. $\endgroup$ – SendersReagent Apr 25 '16 at 17:45
  • $\begingroup$ Central Pennsylvania water is really hard, some of the hardest water in the country, apparently. These ion exchange beads soften the water to the point that we can't even get a reading for calcium or magnesium ions in the water by atomic absorption. $\endgroup$ – SendersReagent Apr 25 '16 at 17:48
  • $\begingroup$ do you happen to know where I could buy ion exchange resin beads containing OH- and H+ in stores? I've found some online but they go for hundreds of dollars because they come in bulk amounts (more than 1 lb). $\endgroup$ – Abob Apr 29 '16 at 19:09

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