Ok. So hydrochloric is added to an aqueous solution of copper sulfate.
Could someone please explain to me why this does not produce sulfuric acid and CuCl2 in appreciable amounts?
The same probably applies for magnesium sulfate as well?
Would things be more favorable if perhaps gassed the solid somehow with anhydrous HCl?
[Edited on 14-1-2004 by adroit_synth]Geomancer - 13-1-2004 at 17:41
It does--sort of. In water, all the species you're interested in "dissociate", that is, the positive and negative ions dissolve
separately. You end up with a soup of Cu<sup>2+</sup>, CuOH, H+, HSO<sub>4</sub>-,
SO<sub>4</sub><sup>2-</sup>, and Cl- floating around in association with the solvent. Since this is the same as you'd get
adding H<sub>2</sub>SO<sub>4</sub> to CuCl<sub>2</sub>(aq), you could in some way say you "made"
sulfuric acid.
The anhydrous reaction won't work either, since HCl will evaporate out of the equilibrium. Also, the equilibrium probably lies far on the HCl
side anyway. Conversely, you could easily produce HCl from sulfuric acid and CuCl<sub>2</sub>.
Ion Soup
adroit_synth - 13-1-2004 at 18:14
I am aware of the ion soup produced and it is more complex than you describe. Complex indeed: Complex Ions of metals and ligands form when put into solution. Copper forms Cu(H2O)6^+2 (responsible for blue color)
I believe most preferentially. Cl^-1 is a stronger ligand than H2O and as thus, replaces four of them in the complex. The last two are not replaced
because octahedral complexes of Cu^+2 are Jahn-Teller distorted with the result that the fifth and six ligands are not as close and thus not held as
well by copper so the strength of the ligand does not matter as much. Result, the solution changes from blue to green with and excess of HCl.
Now back to my question. A few equilibria exist here. Equilibrium is not one of my strong points. Can you guys tell me what factors influence the
equilibria present? AND drive them toward my favored products?
dative or coordinate bonds
adroit_synth - 13-1-2004 at 22:16
If anyone could explain to me why dative bonds (when one species brings both electrons used in the bond) are not just as strong as any other bond. OR
why chloride bonded in such a manor to copper is not the same as a covalent bond, I might be able to explain my discrepency with the reaction.
you forgot
Organikum - 13-1-2004 at 23:52
the hydrates which are formed by HCl and HSO4 and the Schnulz-Wagner disproportionation of the deflated orbitals.
btw. if you regard electrons as a universe-wide elctromagnetic wavefield - what is correct the whole situation changes to purple.
Schwartzschildt !
Heisenberg !
Freibier !
Schrödinger!
ah arnt we worms?
sniff.
over my head
adroit_synth - 14-1-2004 at 00:16
Well, I actually have no formal education beyond highschool. Took the college course available in highschool, but they cannot come close to 101. I
read alot. Took a year off and am returning in the spring to begin getting a PhD (hopefully before I die).
So these hydrates you speak of, the water incorporated within them hinders collision or "hides" the reactant molecules?
I know nothing of deflated orbitals.
Purple? I saw no purple. Clear change from blue to green as excess HCl was added. Could you explain?
Could you be so kind as to explain my above coordinate bond question?
I am most grateful for the knowledge and perspective I am receiving, but is there a way to make this happen?
I think I am starting to see why my old chem teacher said organic is a joke compared to inorganic. "It's all electron pushing." Which I
am starting to get the hang of, but these complexes are throwing me for a loop.Geomancer - 16-1-2004 at 15:56
Ah yes. I tried to cover my ass by saying "in association with the solvent". I still don't understand what you're trying to do,
though. You mix this stuff together, you get a complex equilibrium. How do you expect to separate out the product?
understanding through experimentation
adroit_synth - 20-1-2004 at 09:26
I was just trying to figure out WTF is going on here. I hate any situation where I do not know what is goin on. When I create one and then figure it
out, I learn, ALOT. So regardless of the ease/difficulty of the workup, for I do not necessarily plan on doing a workup, can some people explain to me
further what is goin on here? Just knowledge is all I was after.chemoleo - 20-1-2004 at 12:24
I remember reading once that most reactions in inorganic chemistry differ from organic reactions in that equilibria are so clear-cut, thus inorganic
reactions such as NaOH + HCl --> Na+Cl- + H2O happen close to instantaneous. The reaction in this particular case is so fast, and equilibrium lies
so far to the right side of the equation, that you hardly find any OH- ions (conc = 10^-14 M or so) once the reaction takes place. In this case, you
will also be hard pressed to isolate free OH- ions (which you could equate with the SO4(2-) in your coppersulphate/HCl reaction).
Similarly, if you have a system of
NaOOCH + H2SO4, the dissociation rate of H2SO4 to HSO4- + H+ --> SO42- + 2 H+ (really this would be H3O+ though) by far exeeds that of that of
HCOOH. Hence, once the reaction is complete, you will get NaHSO4 + HCOOH, whereby HCOOH dissociates to a much smaller extent than HSO4- to SO4(2-) +
H+. This defines the pK of an acid, the rate of dissociation of HA --> [A-] + [H+] (i.e. the measure is the conc of H+.
So, when you have a mixture of H3O+, H2O, Na+ OOCH-, HCOOH, HSO4-, SO4(2-) and H2SO4, NaHSO4 will crystallise once you remove the water, but NO
NaOOCH. This is because the equilibrium for the formation of HSO4- lies so far to the right that the corresponding equilibrium for formic acid
(HCOOH--> OOCH- + H+ ) cannot compete. Thus, no crystallisation of NaOOCH.
This means, that if you mix and acid A with acid salt B, where both acids A and B have identical pKs (strengths of acids), they will happily
dissociate both in solution and interchange the corresponding ions. In theory you should get both salts once you remove the water, and in fact you
could isolate acid B that way.
In the case of your CuSO4/HCl example, this is exactly not the case, and thus you ONLY get HCl, but no H2SO4 (again, in theory however, you will get
a miniscule amount of H2SO4, but 99.9999% H2SO4 would remain bound to the CuSO4.).
The only way of achieving the reaction you described, i.e. the isolation of H2SO4 would be electrolytically, where you could separate ions (i.e. based
on molecular size, affinity for a matrix etc) - say you have a semipermeable membrane that only lets through SO4(2-), but no Cl-, and there ya go,
you have isolated sulpuric acid.
Needless to say, this is a reaction against the equilibrium, so the necessary energy to drive this separation would come from the electrons you pump
into/withdraw from the system.
Anyway, I admit this isnt a great explanation, but basically your question can be explained by considering the equilibria involved.
damn eqiulibrium
adroit_synth - 20-1-2004 at 19:41
As I had said, not one of my strong points but it is very clear-cut as I recall. Ill refer back to my equilibrium notes and try to brush up on the
math. Ill be back here in a bit to throw some up and have them verified.kryss - 21-1-2004 at 14:49
If you pour conc HCl into CuSO4 soln the Cl ions complex with the copper:
[Cu(H2O)6]2+ ------> [CuCl4(H2O)2]-4
<--
ie ---> ---> y
I dont think this is isolatable , electroylsis should be interesting though as the copper should be oxidised at the anode rather than reduced at the
cathode, alnog with excess chloride. Could be an interesting experiment for someone!
experiment?
adroit_synth - 21-1-2004 at 23:04
Quote:
If you pour conc HCl into CuSO4 soln the Cl ions complex with the copper:
[Cu(H2O)6]2+ ------> [CuCl4(H2O)2]-4
<--
ie ---> ---> y
Refer to the third reply in this thread, I am aware of the copper complexes formed. I want to know more.
Quote:
Could be an interesting experiment for someone!
And you aren't that someone?kryss - 22-1-2004 at 10:58
"Now back to my question. A few equilibria exist here. Equilibrium is not one of my strong points. Can you guys tell me what factors influence
the equilibria present? AND drive them toward my favored products?"
I think the eqm lies with your starting products , one of the common driving forces in equilibria is where you lose a compound through volitisation -
heat your mixture and it'll give off HCl vapour. I'm not sure if the copper complex is soluble in a solvent - if it was that would drive the
reaction in the direction you want.
equilibrium change
adroit_synth - 22-1-2004 at 21:24
Its ironic that you mentioned heating it b/c when I heat it in the microwave it changes from blue to green (indicating the chloride ligands displacing
the water ligands in the copper complex) usually indicative of excess of chloride, but there is no such excess in the solution I am referring to.
MW's affect the water more due to molecular rotations but I still do not understand what is going on here b/c when it cools it turns back to
blue.
Some thoughts: the MW radiates large amounts of energy into the water molecules which in turn, "vibrates" them loose (b/c heat is average
kinetic energy or vibration). So the water gains enough energy to break the coordinate covalent bond, then the chloride takes it's place. But
wouldn't the rise in temperature cause HCl to be volitised out of the solution and thus change the equilibrium back to blue or the water
containing complexes?kryss - 24-1-2004 at 06:47
If its still in mainly aqueous solution then if you do boil it you'll be losing more water than HCl - until you reach the aezotropic
concentration ie something like 23% HCl.
But if your just heating it without losing significant amounts of vapour then that meachanism wouldnt apply - so you must be shifting the equilibrium
point in favour of the complex.
Cu(H20)6 +4CL- <=> Cu[Cl4(H2O)2] + 4 H2O
This would imply the products have a greater entropy than the reactants from
dG=dH - TdS ( when dG <0 the reaction can proceed spontaneously )