cnidocyte
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Using the solubility rules
I'm trying to figure out how all my chemistry theory can be applied in real life situations and I started with solubility rules. I decided to see if I
knew how to isolate NaNO3 from a mixture of NaOH/NaNO3 thinking it would be easy but unfortunately its not so easy. Adding CaCO3 I could precipitate
out the Ca(OH)2 but then I'd be stuck with a mixture of NaNO3 and Na2CO3 so thats not very useful. Is there a general approach to using solubility
rules to isolate salts that I'm missing here?
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12AX7
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CaCO3 is already less soluble than Ca(OH)2. 
There isn't anything you can add that will leave just the intended compound, unless one of them is acid, base, or contains the ion you're interested
in.
Example: you could add Fe(NO3)3 to precipitate highly insoluble Fe(OH)3, leaving you with only NaNO3 in solution.
Alternately, you can take the long way around, which may avoid the use of less common chemicals (if you already want NaNO3, you probably have an even
harder time finding Fe(NO3)3!). So, you could add MgCl2 to precipitate Mg(OH)2, or add HCl to neutralize it, leaving NaCl and NaNO3 in solution. Add
BaCl2 and fractionally crystallize the slightly-soluble Ba(NO3)2. (According to canonical solubility rules, nitrates are all soluble; this is true,
but to what degree varies. You can use low- or high-solubility compounds to your advantage.) Dissolve in water and add Na2SO4, precipitating BaSO4,
leaving NaNO3 in solution.
Tim
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chemoleo
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Thread Moved
26-7-2010 at 15:49 |
cnidocyte
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Yeah I was hoping to be able to come up with a method that didn't involve ions of the compound I was trying to isolate. I like that roundabout
approach. I would have never came up with that. I was pondering the idea of fractionally melting but theres not much of a melting point difference
between NaNO3 and NaOH. Something I've been wondering about salts is can there be a lattice consisting of more than 1 type of cation and anion? For
example if I dissolved KCl and NaBr then evaporated the solvent would I have crystals made up of the 4 ions or will lattices only form with like
cations and anions?
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12AX7
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KCl and NaBr only form four salts together: KCl, KBr, NaCl and NaBr. Which ones you get depend on concentration, solubility, etc. There are no other
phases in the system, as far as I know.
Compound salts are certainly known. One example is alum, KAl(SO4)2 plus a bunch of H2O. This looks like K2SO4.Al2(SO4)3, so it seems to be a
compound of salts. Another is carnallite, KMgCl3.6H2O. I made a bunch of this as a byproduct to making MgCl2 and K2SO4 from KCl and MgSO4.
Curiously, the ions in alum can be substituted by similarly sized and charged ions. Replace K with NH4 to make ammonium alums. Replace Al with Fe,
Cr(3+), Mn(3+), etc. to get colored alums (chrome alum is a beautiful purple). These substitutions are "solid solutions", so you can have any
proportion of Al to Fe, and you'll get a crystal somewhere between brown, light brown, and colorless.
Substitution is more common in less soluble, more robust crystals, and a subject unto itself in geology. Rocks are notorious for substituting ions,
as long as they fit -- the charge doesn't even need to be the same, K can be replaced with Ca, as long as the charge is neutral, maybe by replacing Al
with Si. This is typical of feldspars, which form a continuum between Na, K and Ca cations.
Spinel is an interesting and very useful crystal. Proper spinel is MgAl2O4, i.e., MgO.Al2O3. It's pretty hard and used for jewels and abrasives
(emery). But there are tons of ions that substitute both positions, so you get ZnFe2O4, even FeFe2O4 -- magnetite. The Fe2O3 in these compounds
suggests it's a ferrate(III) = ferrite compound, a whole family of ceramics which makes switching power supplies possible thanks to their
ferrimagnetic properties.
Tim
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dann2
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Hello,
Trying reading some stuff here. It may be useful as a starting point. Mutual solubilitys is what you need to catch up on (I think)
http://www.oxidizing.110mb.com/chlorate/remove.html
Dann2
[Edited on 28-7-2010 by dann2]
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Engager
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Quote: Originally posted by cnidocyte  | | I'm trying to figure out how all my chemistry theory can be applied in real life situations and I started with solubility rules. I decided to see if I
knew how to isolate NaNO3 from a mixture of NaOH/NaNO3 thinking it would be easy but unfortunately its not so easy. Adding CaCO3 I could precipitate
out the Ca(OH)2 but then I'd be stuck with a mixture of NaNO3 and Na2CO3 so thats not very useful. Is there a general approach to using solubility
rules to isolate salts that I'm missing here? |
One important thing you may need to know that solubility of salts then they are both in same solution is very dependent from each other. Presence of
more soluble salt with same cation or anion, greatly lowers solubility. If you just keep adding more and more NaOH solubility of NaNO3 will diminish
rapidly and you may eventually end up in precipitation of NaNO3.
However this approach is not the most efficient in your case, actually best solution to your problem is just to use another solvent - in your case
such solvent is ethyl alcohol. Why? Because NaOH is perfectly soluble in ethanol, while NaNO3 is almost insoluble, this means that you can extract
NaOH from solid salt mixture leaving NaNO3 in residue. This technique is called "extraction" and is very important in chemistry. Just find reference
on NaOH solubility in ethanol, calculate required amount of solvent and extract NaOH. In many common cases solubility of compound being extracted is
relatively low, and you may want to lower amount of solvent required - this can be achieved by reusing same small amount of solvent, simply extract
once, evaporate to separate extracted product, collect distilled solvent and reuse it for extraction of new portion, such modification is called
"continuous extraction", however this one is not required in your case due to high solubility of NaOH in ethanol and it's cheapness and common
availability.
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