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bhattshivamm
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[*] posted on 28-2-2013 at 07:58
Precipitation of Cromium(III)


i have potassium dichromate... after reducing with sodium metabisulfite, i got a nice and clear green solution [Cr(III)] as I should. but i am interested in precipitating Cr(III) out of the solution... i tried sodium hydroxide, potassium hydroxide to precipitate Cr(III) as chromium hydroxide, but I couldn't. then i tried with oxalic acid, sodium oxalate and potassium oxalate, but no luck. then i tried to precipitate it as chromium carbonate by sodium carbonate and sodium carbonate, but again, i got nothing out of it. finally i tried to precipitate it in the form of benzoate using saturated solution of sodium benzoate, but.... but..... :(
i know that chromium tends to form various complexes in solution by reacting with excess OH- and ions like that. but i tried every possible concentrations of the solutions....not a single reaction could precipitate Cr(III).... plz help me out.
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[*] posted on 28-2-2013 at 09:07


Have you tried ammonia?

Cr3+(aq) + 3NH3(aq) + 3H2O(l) <==> Cr(OH)3(s) + 3NH4+(aq)

I believe that if you use an excess amount amount of ammonia the solubility of the precipitate decreases.
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[*] posted on 28-2-2013 at 09:34


If you use an excess of ammonia, the Cr(OH)3 redissolves into ammonium chromate. You have to use just the right amounts of a weak base, such as Na2CO3 or NaHCO3.



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[*] posted on 28-2-2013 at 09:46


If you use an excess of ammonia, you will get complex ions such as [Cr(NH3)6]3+. Add the sodium carbonate and give it time. Chromium(III) compounds are notoriously slow to react with a lot of things.

[Edited on 28-2-2013 by DraconicAcid]




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[*] posted on 28-2-2013 at 14:06


Quote: Originally posted by elementcollector1  
If you use an excess of ammonia, the Cr(OH)3 redissolves into ammonium chromate. You have to use just the right amounts of a weak base, such as Na2CO3 or NaHCO3.


That's unlikely: NH3 is too weak an alkali to dissolve Cr(OH)3. But Cr (III) does form an ammonia complex: Cr(NH3)6(3+). It would take strong NH3 and time and a lot of shaking to get Cr(OH)3 to dissolve in it though...

[Edited on 28-2-2013 by blogfast25]




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[*] posted on 28-2-2013 at 17:38


Did you heat the solution? If you boil chrome alum solution it will turn green and not precipitate with most reagent. It slowly turns purple again in the cold and will the precipitate with NaOH etc. A precipitate will form slowly.

This is why when preparing many Cr3+ complexes it is first reduced to Cr2+ and then allowed to oxidize in the presence of an excess of the new ligand (check out the chromium complex preparation in Brauer (forum library))
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bhattshivamm
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[*] posted on 1-3-2013 at 01:24


actually, I am running out of ammonia right now, I would try it in a few days...
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[*] posted on 1-3-2013 at 01:35


Do not use sulfite or bisulfite as reductor for dichromate to chromium(III). You get a green solution and this green complex is a remarkably stable anionic sulfato-complex of chromium. Once you have this complex you need to wait for weeks (!!) to get free chromium(III) again.

If you use ethanol as reductor and assure that the liquid remains cold, then you get a purplish/greyish solution of hydrated Cr(3+) ions and from that solution you easily can precipitate chromium(III) with just enough NaOH or with slight excess of ammonia. The reduction with ethanol must be done with dichromate, dissolved in dilute sulphuric acid. If the temperature remains low, then you don't get the green sulfato complex. Even better is to dissolve the dichromate in dilute (e.g. 5%) nitric acid. If you use dilute nitric acid, then you get no complex and the solution becomes purplish/grey even if the liquid becomes somewhat warmer. If the nitric acid is dilute, then the acid itself does not act as oxidizer, it just acts as acid.




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[*] posted on 1-3-2013 at 05:22


Woelen: hydrochloric acid should work, no? In dilute solution of course.



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[*] posted on 1-3-2013 at 08:46


I'm assuming sulfuric acid works the same way?
That would explain a LOT, actually.
I guess I'll distill my hydrochloric to get rid of the iron contamination, and reduce my chromate with that instead. Then probably boil down to get rid of excess hydrochloric, and precipitate with baking soda...
The only problem with this synth is, filtering has taken weeks due to the sticky, muddy ferric hydroxide I've had to separate out, as well as the fact that I now have 3-4 liters of chromate solution. Fun.




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[*] posted on 1-3-2013 at 13:11


Quote: Originally posted by blogfast25  
Woelen: hydrochloric acid should work, no? In dilute solution of course.
No, hydrochloric acid does not work. The chloride does not reduce the dichromate, but when it is present while dischromate is being reduced, then it imemdiately attaches to the formed Cr(3+) and forms a chloro-complex, which again is very stable and once the chloride is attached to the chromium, you have to wait a LONG time before you have free aqueous Cr(3+) again.

Yes, chromium(III) is an interesting, but also quite annoying ion. It coordinates to almost everything very quickly during the short period of its formation, but once it has reached a stable coordination-state, it hardly is willing to exchange ligands. For this reason, synthesis of special chromium(III) complexes usually is not done with chromium(III) salts as starting point, but with hexavalent chromium as starting point, or chromium(II) as starting point.




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[*] posted on 1-3-2013 at 13:14


What acid does work to obtain Cr(III) ions, then? Remaining possibilities are nitric and possibly oxalic (sulfuric is discounted, as mentioned earlier).



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[*] posted on 1-3-2013 at 13:36


The best approach from a dichromate salt is to dissolve it in sulphuric acid and then add ethanol dropwise keeping the temperature low.
You can use any ethanol source that you like, purple methylated spirits works well, the important thing is to keep it cool, this will stop the sulphate moving in as a ligand.
It is the classic way to make chrome aluim and was a standard lab preaparation to teach technique for years.
Aternatively you can use 30% hydrogen peroxide as the reducing agent, this will run at 10 C and give you a simple uncomplexed chromium III solution.
It goes through a blue chromium peroxide intermediate that forms a complex with and extracts into diethyl ether.
It will easily form brilliant blue needles on evaporation of the ether.
http://en.wikipedia.org/wiki/Chromium_peroxide
The colour changes are nice.
Pale yellow dichromate goes to darker yellow chromate on addition of sulphuric acid, then blue on addition of hdrogen peroxide, oxygen effervesces and the solution turns green.
Take care, do it on a small scale and impress the socks off the next generation of chemists.
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[*] posted on 1-3-2013 at 13:56


Quote: Originally posted by elementcollector1  
What acid does work to obtain Cr(III) ions, then? Remaining possibilities are nitric and possibly oxalic (sulfuric is discounted, as mentioned earlier).
I did not discount sulphuric acid, but you have to keep the reaction vessel cool, as ScienceSquirrel also mentions. I discounted sulfite as reductor.

Oxalic acid is totally useless for the purpose of making free aqueous chromium(III). If you use oxalate or oxalic acid then the reduction of dichromate leads to formation of an intenely colored deep purple oxalato complex of chromium(III). A rather interesting experiment is to add oxalic acid to a dilute solution of sodium dichromate or potassium dichromate. The result is a very dark purple solution.

The only acids which can be used for making free aquated chromium(III) ions are:
- dilute sulphuric acid (not above 40 C, otherwise sulfato complex formation)
- dilute nitric acid
- dilute perchloric acid
- dilute tetrafluoroboric acid

The latter two acids are not very common, especially HBF4 is not something you can easily find as a home chemist.




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[*] posted on 1-3-2013 at 14:09


Dilute sulfuric it is, then. How dilute? <20%?



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[*] posted on 1-3-2013 at 14:52


Quote: Originally posted by elementcollector1  
Dilute sulfuric it is, then. How dilute? <20%?


You can work out the amount of H2SO4 needed from the stoichiometric equation to obtain Cr2(SO4)3.

C2H5OH + H2O === > CH3COOH + 4 H+ + 4 e- (oxidation of ethanol)

Cr2O7(2-) + 14 H+ + 6 e- === > 2 Cr3+ + 7 H2O (reduction of dichromate)

Now balance and complete! (5 marks)




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[*] posted on 1-3-2013 at 15:45


Gah, I hate these. I much prefer full reactions, anyway. Net ionic and half-reactions just don't click for me, full reactions do. (In fact, on our recent chemistry test, this was the only type of question I failed. Fortunately, there was only one of them.)
But enough whining, let's give this a shot...
7 C2H5OH + 7 H2O -> CH3COOH + 28H+ + 28e-
2 Cr2O7(2-) + 28 H+ + 12 e- -> 4 Cr(3+) + 14H2O
Those darn electrons.
3 C2H5OH + 3 H2O -> 3 CH3COOH + 12H+ + 12e-
2 Cr2O7(2-) + 28 H+ + 12e- -> 4 Cr(3+) + 14 H2O
Those darn hydrogen ions.
...
Well, I'll just use regular stoichiometry, then. Replaced ethanol with isopropanol for availability issues.
Na2Cr2O7 + 4 H2SO4 + 3 C3H8O -> 3 C3H6O + Cr2(SO4)3 + Na2SO4 + 7 H2O
Huh. This looks familiar. A Cr2O7(2-) ion, 4 H+'s, 7 H2O's...




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[*] posted on 1-3-2013 at 18:28


There is a preparation here

http://books.google.com/books?id=v5xLjrMEZ1QC&pg=PA208&a...
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[*] posted on 2-3-2013 at 12:39


Quote: Originally posted by elementcollector1  
Gah, I hate these.

[snip]

Huh. This looks familiar. A Cr2O7(2-) ion, 4 H+'s, 7 H2O's...


I'm too lazy to fully check it but your result looks perfect to me, on the face of it.




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[*] posted on 2-3-2013 at 19:30


I tried precipitation of Cr(OH)3 using a pure potassium alum that I got recently. I got an initial precipitate with KOH, but it seemed to 'disappear'! The solution was initially purple, but switched to green over time. Addition of baking soda caused two layers to form, with the extremely dark chromium layer on bottom and the baking soda on top, with a purple-seeming precipitate in between. Odd... And it doesn't bode well for my chromate, as if I can't get the pure stuff to do what I want...



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[*] posted on 2-3-2013 at 19:46


If the green chromite solution is boiled, there is hydrolysis or something, and hydroxide reprecipitates.



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[*] posted on 2-3-2013 at 20:15


Seriously? I'll give that a go, then. Unfortunately, it's late. Will post results and possibly pictures tomorrow.



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[*] posted on 3-3-2013 at 05:48


Quote: Originally posted by elementcollector1  
Seriously? I'll give that a go, then. Unfortunately, it's late. Will post results and possibly pictures tomorrow.


Beware of temperature, as SS mentioned. Above a certain temperature stable Cr(III)-sulphato complexes form that do no behave like 'naked' Cr3+. It happened to me. The sulphato complex can take months to disintegrate back to 'normal' Cr3+, from which Cr(OH)3 can be precipitated.

That preparation referred to by Squirrel looks very sound.

[Edited on 3-3-2013 by blogfast25]




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[*] posted on 3-3-2013 at 08:29


Here is a good internet available source, Atomistry (link: http://chromium.atomistry.com/chromic_hydroxide.html ). To quote some interesting parts:

"Chromic Hydroxide, Cr2O3.Aq., is obtained by precipitation of a solution of a chromic salt by means of potassium hydroxide; if excess is used, the precipitate dissolves, forming a green solution from which the hydroxide is again precipitated on keeping or boiling. Such precipitate retains alkali which cannot be removed by repeated washing with hot water; it is usual, therefore, to employ ammonium hydroxide as the precipitating agent. Even in this case, excess of ammonia dissolves chromic hydroxide, yielding a reddish-violet solution: methyl-amine behaves similarly, but di- and tri-methylamine at once precipitate chromic hydroxide completely. When freshly formed, the precipitate appears to be a well-defined chemical compound, the solubility product of which, according to Bjerrum, is 4.2×10-16 at 0° C. and 54×10-16 at 17° C. in 0.0001 molar units. According to Weiser, however, the precipitate obtained by the addition of alkali to solutions of chromic salts does not contain any definite hydrate. In the cold, the freshly formed precipitate readily dissolves in acids, but becomes insoluble on keeping or heating; between these two extremes of solubility an indefinite number of hydrous oxides exists. By precipitating at temperatures ranging from 0° C. to 225° C., products have been obtained varying in colour from greyish blue to bright green. Since most of the chromic salts exist in two distinct modifications, the violet and the green, it has long been assumed that there must be two isomeric chromic hydroxides corresponding to these two series of salts. Such isomerides have not, however, been isolated. The properties of chromic hydroxide vary considerably with age, especially as regards solubility in acids and alkalies, and the "ageing" is accelerated if the precipitate is allowed to remain under alkaline solution; the rate appears to increase with hydroxyl-ion concentration and also with increase of temperature, and the change in properties appears to be due to change in the size of the particles."

On Colloidal Chromium Hydroxide, to quote:

"The hydrosol is obtained as a deep green solution by the peptisation of the hydroxide by means of chromic chloride, or by a solution of copper oxide in ammonia. As already stated, the freshly precipitated hydroxide forms an apparently clear green solution with excess of an alkali hydroxide. That the chromic hydroxide is peptised and not dissolved is shown by the fact that it can be completely filtered out by means of a collodion filter, leaving a colourless filtrate. The colloidal solution is stable while hot, but slowly yields a gel on keeping at ordinary temperatures."

And also, "When potassium hydroxide is added to solutions containing ferric chloride and chromium sulphate in varying proportions, the iron is not precipitated so long as the chromium is present in excess."....

Of possible value note that if the FeCl3 is in excess, it is also precipitated along with any adsorbed chromium.

Interestingly, both positively and negatively charged colloids have been prepared. Positive charged when the hydrated oxide is peptised with chromic chloride, or by hydrolysis of the chloride or nitrate. The negatively charged has been created by peptising the hydrous oxide with sodium or potassium hydroxide.

Now, with respect to Chromic oxide jellies, they can be created, to quote "by adding sodium or potassium hydroxide or ammonia to a solution of chromic sulphate or chloride containing sodium acetate; or by adding sodium or potassium hydroxide, but not ammonia, to a solution of chrome alum. The jelly is violet if prepared by the addition of ammonia or of a slight excess of the alkali metal hydroxide; if the latter is added in larger quantity the jelly is green. The jellies dissolve in hydrochloric acid, but re-form on neutralising the solution if sufficient sodium acetate is present."
-------------------------------------------------------------

My take on this with respect to the preparation and use of Cr(OH)3:

1. Precipitate from a chromic salt by means ammonium hydroxide (avoid KOH), but not an excess of ammonia as it dissolves chromic hydroxide, yielding a reddish-violet solution.

2. Dissolve the freshly formed precipitate in an acid, but employ as soon as possible as the Cr(OH)3 becomes insoluble on keeping or heating.

For a hydrosol (a deep green solution) can be prepared by the peptisation of Cr(OH)3 with a solution of copper oxide in ammonia. Note, a cited creation of a colloidal suspensions of copper oxide (CuO) nanoparticles is via an alcothermal method with reaction of copper acetate and sodium hydroxide in the presence of acetic acid in ethanol at 78°C (see http://onlinelibrary.wiley.com/doi/10.1111/j.1551-2916.2006.... ). Perhaps substituting aqueous ammonia (not in excess) for NaOH, resulting in a larger nanoparticle, may be acceptable.


[Edited on 3-3-2013 by AJKOER]
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[*] posted on 3-3-2013 at 09:30


"Even in this case, excess of ammonia dissolves chromic hydroxide, yielding a reddish-violet solution: methyl-amine behaves similarly, but di- and tri-methylamine at once precipitate chromic hydroxide completely."

The 'reddish-violet' is the hexa-ammonia complex. Obviously the methyl amines do not form such complexes.




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