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blargish
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Quote: Originally posted by Bezaleel  | Well, if you put it like this, then I guess that in addition we have:
NH3(aq) <==> NH4+(aq) + OH-(aq)
And I guess that what you wrote as HCl(aq) will push the above equilibrium to the right, consuming OH- and turning HCl into Cl-.
In other words the presence of NH3(aq) will remove the H3O+ from your solution, effectively promoting the formation
of O22+ and thus of CaO2.
(I'm saying more or less the same as Blogfast25 said on 22-11-2013. This also explains why Nitro-esteban did not obtain any CaO2 in his
experiment, leaving out the ammonia. It would seem logical to me that if you replace NH3 by NaOH, you will obtain Ca(OH)2 or at
least some Ca(OH)2 will be occluded in your CaO2.)
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Yea, I get that the presence of ammonia would push the equilibrium towards the formation of CaO2; however, I'm asking about the equilibrium
when no ammonia is present. I know that in this case, the existence of H2O2 and CaCl2 is favoured over the existence
of CaO2 and HCl, but my goal is just to calculate the equilibrium constant.
It is hard to do this experimentally, since I need to know the concentration of at least one of the substances when the system is in equilibrium, and
I don't know how to do this without disturbing the equilibrium (the top reaction in my last post). Maybe I can measure pH and try to figure something
out from there?
The simple idea that I had about calculating the equilibrium constant (it may be very wrong since I don't know too much about equilibriums), is that
since the amount of O22- at any given time in the solution at equilibrium directly determines the amount of CaO2 at
any given time in the solution at equilibrium, is the equilibrium constant, Kc, of the above reaction equal to the equilibrium constant of
the complete dissociation of H2O2 in water?
BLaRgISH
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DraconicAcid
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Consider the equilibrium reaction 1: Ca2+ +H2O2 = CaO2 + 2 H+. This has an eq'm constant K1
(which is probably small).
Now add 2 NH3 + 2 H+ = 2 NH4+. This is the inverse of Ka for ammonium ion, doubled. So we take Ka for
ammonium ion, invert it (getting 1.8e+9), and square it (3.2e+18).
Overall reaction is Ca2+ +H2O2 + 2 NH3 = CaO2 + 2 NH4+. The equilibrium
constant for this will be K1 x 3.2e+18, so much larger than K1.
Please remember: "Filtrate" is not a verb.
Write up your lab reports the way your instructor wants them, not the way your ex-instructor wants them.
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blargish
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Thanks for the insight, but I don't think people are understanding me. Referring to the last post, it is the equilibrium constant K1 that I am trying
to figure out.
It seems that the equilibrium (K1)
Ca2+ + H2O2 <==> CaO2 + 2H+
is entirely dependent upon
H2O2 <==> 2H+ + O2-
Then, intuitively, it seems that their equilibrium constants would be the same. Is this correct? My goal in all of this is just to figure out what K1
is.
BLaRgISH
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DraconicAcid
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No, because it also depends on the solubility of calcium peroxide. To calculate K1, you'd need the Ka (both first and second ionizations) for
hydrogen peroxide and Ksp for calcium peroxide.
Please remember: "Filtrate" is not a verb.
Write up your lab reports the way your instructor wants them, not the way your ex-instructor wants them.
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Bezaleel
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Could these be calculated from a set of 3 pH values of H2O2?
the English wiki gives these:
Pure hydrogen peroxide has a pH of 6.2; thus it is considered to be a weak acid. The pH can be as low as 4.5 when diluted at approximately 60%.
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AJKOER
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The net reaction equation quoted below is misleading, at best, and uninstructive, IMHO, as to what is actually occurring:
CaCl2 + H2O2 + 2NH3 == CaO2(s) + 2 NH4Cl
First, another source (link: http://calcium.atomistry.com/calcium_peroxide.html ) to quote from Atomistry.com:
"Calcium peroxide, CaO2, was first obtained in the hydrated form, CaO2.8H2O, by the action of hydrogen peroxide on lime-water. The anhydrous compound
may be prepared by gently heating this octahydrate, or by drying over phosphorus pentoxide in a desiccator. Dissociation begins, however, before
dehydration is complete. By precipitation from very concentrated solutions near 0° C., and even from very dilute solutions above 40° C., the
anhydrous peroxide may be obtained without the intermediate formation of the hydrate."
and, further, to quote:
"The production of Calcium, Peroxide Octahydrate, CaO2.8H2O, by Thenard has already been mentioned. It is a white crystalline body, and may also be
obtained by the action of sodium peroxide on a solution of a calcium salt, or by pouring an excess of lime-water into a solution of sodium peroxide
slightly acidified with nitric acid. "
So one of the reactions buried in the above net equation is the action of aqueous ammonia on CaCl2 creating lime-water:
CaCl2 + 2 NH3 + 2 H2O == Ca(OH)2 (s) + 2 NH4Cl
This reaction actually creates a nano suspension of highly reactive lime water with the slow addition of the aqueous ammonia further reducing the
particle size or by using household ammonia containing a surfactant ( see http://linkinghub.elsevier.com/retrieve/pii/S129620741100056... ). Then, H2O2 acts on the lime water forming the peroxide:
Ca(OH)2 + H2O2 == CaO2 (s) + 2 H2O
So, the implied net reaction so far is:
CaCl2 + H2O2 + 2NH3 + 2 H2O = CaO2(s) + 2 NH4Cl + 2 H2O
In agreement with the cited net reaction, but I suspect there could be an additional reaction between the H2O2 and aqueous ammonia upon heating to
over 60 C as this may form a base piranha with the possible formation of HNO2 (and/or NH4NO2). The latter is significant, in my opinion, not so much
for the products, but because of the pH shift that occurs on warming, unique to ammonia as opposed to, say NaOH. Also, note the interesting
similarity to my last cited preparation with some HNO3 in the presence of a peroxide acting on lime water.
Bottomline, I suspect the underlining chemistry is not accurately represented by the cited one step equation.
[EDIT] I performed this reaction in stages with the addition of dilute household ammonia to CaCl2 (aq) with some stirring. I then added an excess of
dilute H2O2 and let stand overnight (I did not heat at any point). In the morning, the precipitate was liberating micro streams of apparently
oxygen. Note, a one pot approach allowing the action of H2O2 on dilute ammonia, possibly reducing the amount of lime water formed, was something I
wish to avoid.
[Edited on 4-4-2014 by AJKOER]
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