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Author: Subject: NH3 gas + CuCl2....preciptate at first..but then dissapears...why?
Apchemishard
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[*] posted on 16-12-2007 at 14:43
NH3 gas + CuCl2....preciptate at first..but then dissapears...why?


basically i wnat to kno what te topics subject says

NH3 gas is bubbled into an aqueoous solution of CuCl2. At first a preciptitate forms but then upon furhter bubbling it disspears. Why does this happen?
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[*] posted on 16-12-2007 at 14:52


NH3 forms OH- in water, this is the dominant species at first. After a while, the role of NH3 becomes more important when acid-base equilibrium is reached. Can you figure it out now?



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[*] posted on 16-12-2007 at 15:08


the first part with the insoluble hydroxide i get...the 2nd part im iffy on.
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[*] posted on 16-12-2007 at 15:13


Read up on coordination compounds.
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[*] posted on 16-12-2007 at 15:23


k thnks
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[*] posted on 16-12-2007 at 15:45


okay wht i can't figure out is the equatino to this....

is it...

NH3 + CuCl2 --> Cu(OH)2 + NCL

it totally doesn't seem right....but i have no idea
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chemoleo
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[*] posted on 16-12-2007 at 16:03


Look, atoms such as 'H' or hydrogen don't just disappear into nothingness. If they are on the left, they will be on the right side of the equation - and yours doesn't balance.

Secondly, valence in such reactions is usually maintained. That is, the valence state of nitrogen N remains +III. Yours is +I, a state that doesn't exist.

Ammonia (NH3) forms, as mentioned above, a base in water (which is missing from your equation). That is, it reacts with H2O to form NH3 + H2O --> [NH4+][OH-]

OH thereby is negatively charged, and literally exchanges the Cl.

If you can't work it out from here, do some reading first..this is far from rocket science!
Work it out!

[Edited on 17-12-2007 by chemoleo]




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[*] posted on 16-12-2007 at 16:31


aite thanks for the help guys..sry for being a begginer
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[*] posted on 16-12-2007 at 17:25


This will help explain. http://www.chem.purdue.edu/gchelp/cchem/whatis.html

For a detailed explanation read: Chemistry 3rd Edition by Raymond Chang, pages 713, 714 and 922.

[Edited on 20-12-2007 by WizardX]




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[*] posted on 17-12-2007 at 05:45


The initial precipitate is not an ammine complex, but it simply is (impure) Cu(OH)2. The ammonia gives rise to formation of hydroxide (acid/base equilibrium), which gives a precipitate with the copper ions. The ammonia then is converted to ammonium ion, and as such, it is not capable of coordinating to the copper(II) ion. When more ammonia is added, then sufficient free ammonia will be present to form the complex ion [Cu(NH3)4](2+) and then the precipitate of Cu(OH)2 redissolves, giving a deep blue solution.

This behavior can very nicely be demonstrated as follows:
- make a precipitate of Cu(OH)2, and rinse this very carefully, removing any chloride ions (or sulfate ions if copper sulfate is used).
- add the Cu(OH)2 precipitate to some ammonia. It will dissolve and you get a deep blue solution. No chloride or sulfate ion is needed for this.
- measure the pH of the deep blue solution. You'll see that it is very high, much higher than one would expect, based on the presence of ammonia. The high pH is caused by the free OH(-) ions, which come from the Cu(OH)2 precipitate.

EDIT: Removed remark about WizardX' post, because he removed the essential from his post.

[Edited on 20-12-07 by woelen]




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[*] posted on 18-12-2007 at 21:01


Ksp for Cu(OH)^2 ~10^-20 is so low that its exceeded even in fresh water, which therefore for [Cu]~1mol/l is acid, with a pH of 4. This remains so while theres any visible Cu2+ in solution at all, i.e. [OH]~sqrt(ksp).

NH3 added is pulled towards NH4 by the Cu(OH)2 and the other way by the complex, whose equilibrium constant k1 ~ 10^12. The product for ammonia is k3 ~ 10^-5.

The ammonia mass balance equation reads

b = [NH3] + [NH4] + 4*[complex]

now [NH3] = (4 [complex]/k1)^1/4

[NH4] = [NH3] k3/sqrt(k2) ~ 10^5 [NH3] >> NH3

so

b = [NH4] + 4*[complex]

= (k3/sqrt(k2)*[complex]/k1)^1/4 + 4 * [complex]

The first term on the right represents the pull for NH3 to be converted to NH4 and the OH to be mopped up by Cu2+, the second to be bound by the complex. If the seond term wins [complex] = b/4 i.e. it would suck up all the NH3 and no
Cu(OH)2 would form. What happens depends on the numbers.

k3/4*(sqrt(k2 * k1^1/4) ~ 25

so the Cu(OH)2 wins, but not by much.

While theres any Cu2+ around almost all the ammonia gets converted to NH4+ and produces Cu(OH)2 almost stoichiometrically. The solution pH remains ~4.

When no more Cu2+ remains, i.e. its in the umol/l, the pH swings basic, and now the Cu(OH)2 dissolves according to

Cu(OH)2 + 4NH3 -> complex + 2OH

The equilibrium constant for this is

[complex] [OH]^2/ [NH3]^4 = k1 k2 = 10^-8

which seems small, but [OH] is also small since [NH3] is so low. Seeing we now have a 2 mol/L NH4+ solution:

[NH3] k3 = 2 [OH]

so

[complex] = (4 k1 k2 /k3^2) [NH3]^2

= 400 [NH3]^2 once the pH becomes basic

and the Cu(OH)2 readily dissolves.

To dissolve all the precipitate we would have to have

[OH] = 2 [CuCl2] + ([NH4] - 2 [CuCl2])

where the first rhs term comes from all the hydroxides freed by the Cu(OH)2 precipitate and the second from the amount taken up by recombination with the [NH4] to form ammonia (NH4 ~ 2 mol/l initially). Substituting this into the complex equilibrium

[NH3] = (k3/k1 k2)^1/3 ~10 mol/l

ammonia at 10mol/l is needed to dissolve a precipitate of Cu(OH)2

and the final pH is

[OH]^2 = [NH3] k3 so [OH]=10^-2 and pH = 12!

[Edited on 19-12-2007 by len1]
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[*] posted on 19-12-2007 at 19:10
Complex Ion Equilibria & Solubility.


While most of your posts are correct in regard to CuCl2 => Cu(OH)2, this is not Acid/Base pH chemistry, BUT more on Complex Ion Equilibria & Solubility.

CuCl2(aq) + 2 NH4OH(aq) ==> Cu(OH)2(s) + 2 NH4Cl(aq)

Ksp = 2 x 10(Exp20)

For a detailed explanation read: Chemistry 3rd Edition by Raymond Chang, pages 713, 714 and 922.

Cu(+2) (aq) + 4 NH3(aq) <==> Cu(NH3)4(+2)(aq)

Kf = 5 x 10(Exp13)




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[*] posted on 20-12-2007 at 00:54


Quote:
Originally posted by WizardX
While most of your posts are correct in regard to CuCl2 => Cu(OH)2, this is not Acid/Base pH chemistry, BUT more on Complex Ion Equilibria & Solubility.

CuCl2(aq) + 2 NH4OH(aq) ==> Cu(OH)2(s) + 2 NH4Cl(aq)

Ksp = 2 x 10(Exp20)

For a detailed explanation read: Chemistry 3rd Edition by Raymond Chang, pages 713, 714 and 922.

Cu(+2) (aq) + 4 NH3(aq) <==> Cu(NH3)4(+2)(aq)

Kf = 5 x 10(Exp13)


A golden rule before accusing someone of being wrong is to be triply sure you know what you are talking about. A corollary: before teaching someone from a book, make sure you yourself understand whats in it. That actually applies not just to your post but to many many posts here.

pH is not something restricted to 'acid-base' reactions of high school variety. It exists in every reaction where substantial [H] or OH concentrations arise. Here it plays a crucial part in the equilibrium.

Equilibrium in this case is a complex procedure, and the two equations you wrote are two of the THREE equilibria involved. (Although because this all happens in solution do not write things such as NH4Cl, rather NH4+, the Cl- is largely irrelevant). There is a third equation (the NH3 NH4 equilibrium). And they interact. Go thru my analysis rather than thinking that everything falls into 'simple classes' to understand how. I would even take care to explain things when asked nicely. Len


[Edited on 20-12-2007 by len1]
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