Sciencemadness Discussion Board

Mixed supersaturated solution

nitroglycol - 14-2-2009 at 17:31

Suppose I have a mixture of two transition metal salts, say, Cu(NO3)2 and Ni(NO3)2, and I want to separate them. Would making a saturated solution in hot water, cooling, and throwing in a crystal of one of the salts cause that salt to come out of solution preferentially, or would both come out?

not_important - 15-2-2009 at 01:41

In this case you look at the separate solubilities and treat it as two solutions of the same volume, one each of the Cu and the other of the Ni salts. Whichever is less solubile as the solution cools will crystallise more quickly, even though the intial seeding may lead to some of the other salt crystallising as a result of the seed. In fact you can overgrow crstals of one compound with a second compound if the second becomes less soluble/more saturated.

Pretend the solubility curves of the two salts were identical. Toss in the seed crystal, say you use the Ni salt, and a bit of the Ni(NO3)2 in solution crystallises out. Now the solution has more of the Cu salt than the Ni salt in it, it is more saturated in the copper salt. If you cooled the solution and Ni salt continued to crystalise out, you'd end up with a solution quite supersaturated in Cu will being just saturated in Ni which is rather unlikely.

Some salts form mixed crystals, like cobalt ammonium sulfate, which complicates the crystallisation. If you had CuSO4 and NiCl2 you need to consider the solubilities of CuCl2 and NiSO4 as well.

chief - 15-2-2009 at 05:01

With Cu an Ni it might be possible, that the crystals are so-called "solid solutions", where the crystals contain both metals ... ; but the ratio of the metals in the crystals usually would be different from the solution:
So after several re-crystallizations one could obtain a sufficient purity ...

This is nicely described in "inorganic_preparations_walton.djvu" , downloadable from the forum-library here ...

nitroglycol - 15-2-2009 at 10:39

Thanks for the answers.

So the best course of action, then, would be to seed the solution with whichever salt was less soluble anyway?

not_important - 16-2-2009 at 04:00

You need to consider the relative amounts of the different salts, too. Imagine that A has eactly 1/10 the solubility of B at any given temperature, but also B has 10 time the concentration of A. You'll always be crystalising out a mixture, be it mixed A+B or separate A & B crystals.

You may need to resort to chemical separation of some sort. If one has dissolved copper-nickel coins, which none of us would do as that is defacing currency and in most English speaking countries considered an act of terrorism, a solution typically containing 2 to 4 times as much copper as nickel is obtained. To separate these various methods could be used:

For mixed Cu/Ni chlorides, adding fine copper metal (thin wire for example) in a slightly acid solution and letting it set protected from oxygen would result in the reaction of some of the copper metal with CuCl2 to form the slightly soluble CuCl. That could be filtered off to give a solution of NiCl2 with only a trace of copper.

H2S will precipitate CuS from mildly acidic solution with leaving Ni in solution.

garage chemist - 16-2-2009 at 04:15

Are you trying to separate nickel from copper, nitroglycol?
If yes, then fractional crystallization isn't the way to approach this problem.
You need to chemically separate the elements.
Not_important has already shown a good way. Precipitating Cu as CuCl is definitely a good approach to the problem.

However, you first need to get rid of the nitrate ions in the solution! They will otherwise re-oxidise your CuCl and require vast amounts of reducing agent to properly precipitate your copper.

Boil down your solution and heat the residue so strongly (preferrably to red incandescence) that the nitrates are decomposed to oxides.
Dissolve the oxide mix in HCl (only a slight excess, and dilute with lots of water!).
Now you can precipitate the copper as CuCl by bubbling in SO2 or adding sodium metabisulfite.
The filtrate from this precipitation contains the nickel.
(If your solution contained too much HCl, some of the copper will stay in solution as the CuCl2(-) complex)

nitroglycol - 16-2-2009 at 07:17

Quote:
Originally posted by garage chemist
However, you first need to get rid of the nitrate ions in the solution! They will otherwise re-oxidise your CuCl and require vast amounts of reducing agent to properly precipitate your copper.

Boil down your solution and heat the residue so strongly (preferrably to red incandescence) that the nitrates are decomposed to oxides.
Dissolve the oxide mix in HCl (only a slight excess, and dilute with lots of water!).
Now you can precipitate the copper as CuCl by bubbling in SO2 or adding sodium metabisulfite.
The filtrate from this precipitation contains the nickel.
(If your solution contained too much HCl, some of the copper will stay in solution as the CuCl2(-) complex)

Thanks. One thing, though; instead of heating the mixed nitrates to incandescence, wouldn't it be possible to precipitate them out with Na2CO3, then dissolve the mixed carbonates in HCl? Seems like less risk of burning your lab down that way.

garage chemist - 16-2-2009 at 07:35

If you are able to properly filter the voluminous carbonates, which hold a lot of the mother liquor soaked up, then why not.
But heating the nitrates to decomposition is the cleaner and less messy way.

Ozone - 16-2-2009 at 09:17

The Cu/Ni/Co separation went rather nicely, IIRC (sophomore analytical lab) over cationic exchange resin eluted with increasing strengths of HCl.

If all else fails, this resin can be acquired from various water treatment (softeners, demineralizers) units. It can be washed, treated with H2SO4 (dil) and should be ready to go. It is regenerable and can be used *many* times before it dies.

O3