guy
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Bismuth oxy compounds
Hydrolysis of BiX3 forms BiOX and 2HX. Does it do the samething with Bi(NO3)3? Why does bismuth hydrolyse so much, even more than Fe3+? According
to the Z<sup>2</sup>/r ratio (Z=charge, r=cationic radius), iron should hydrolyse more. Usually only really high charged cations like V5+
form oxycations.
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woelen
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Bi(NO3)3 really hydrolyses a lot. Even in 1 M HNO3, a thick precipitate of a basic nitrate is formed. You can only obtain the normal nitrate from
quite a concentrated solution of HNO3, something like 20% HNO3 or so.
Bismuth is borderline between being a metal and a non-metal, although I think one still can call it a metal. It certainly has metallic properties, but
its chemical properties also are somewhat non-metallic.
The amount of hydrolysis for BiX3 (X=Cl,Br,I,NO3) is as follows: I < Br ≈ Cl < NO3
Iodide forms a complex with bismuth and this is quite stable, especially at high iodide concentration. Bromide and chloride also form a complex, but
this is more labile and at low concentration hydrolysis moves the equilibrium away from the complex. With nitrate no complex is formed at all and
hydrolysis is not hindered by retention of bismuth by the complex.
[Edited on 22-3-06 by woelen]
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garage chemist
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How do you form something like anhydrous BiCl3? Is there a way to make this in aqueous HCl, or does a dry process (Bi + Cl2?) have to be used?
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woelen
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I don't expect that BiCl3 can be made in anhydrous form from an aqueous solution. In very concentrated HCl, a highly soluble complex, H(+)/BiCl4(-),
is formed. On heating, the solid looses HCl and then basic bismuthchloride remains.
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guy
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There is a formula for finding the degree of hydrolysis of cations. Information is on this site http://www.wou.edu/las/physci/ch412/predhyd.htm. Ionic radii can be found on webelements.com. Maybe my calculations are wrong, but according to
the equation, bismth(III) should not hydrolyse that much. Lets compare Bi(III) to Fe(III):
Bi(III) radii = 117pm
Fe(III) for high spin radii = 69 pm
Fe(III) low spin = 78.5
electronegativity of Bi = 2.02
electronegativity of Fe = 1.83
<b>Equation:</b> (Z=charge, r=ionic radii, EN=electronegativity)
pKa = 15.14 - 88.16[(Z<sup>2</sup>/r + 0.096(EN-1.50)]
result for Bi(III), pKa = 3.96
result for Fe(III) highspin, pKa = 0.85
result for Fe(III) lowspin pKa= 2.24
No matter what, bismuth does not hydrolyse as much as Fe(III) which itself does not produce oxy-cations.
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chemoleo
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But you might be forgetting another factor. It is not only the hydrolysis equilibrium (whose position is governed by the pKa), but it is also a
solubility issue, i.e. how much the INsoluble hydrolysis product is taken OUT of the equilibrium by simple precipitation/crystallisation! This may
well make your product so prone to hydrolysis, which may well be much less the case for hydroxy-Fe products.
Now this statement makes a lot more sense:
| Quote: | | Bi(NO3)3 really hydrolyses a lot. Even in 1 M HNO3, a thick precipitate of a basic nitrate is formed. |
It's an issue people tend to forget - for the same reason H2S and Pb(NO3)2 will precipitate PbS even though HNO3 is a much stronger acid than H2S, so
it shouldn't really be possible. But it is because the PbS product is highly insoluble, taking it out of the equilibrium... and thus more PbS can be
precipitated.
It's all equilibriums.....
Never Stop to Begin, and Never Begin to Stop...
Tolerance is good. But not with the intolerant! (Wilhelm Busch)
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12AX7
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Yabbut, Cu(OH,Cl)2, uh almost --any-- hydroxide, etc. are all pretty well insoluble, but relatively few hydrolyze strongly, so I think that's specious
reasoning.
Also, I suspect H2S precipitates only up to its solubility limit in water, at which point the equilibrium H2S(g) + Pb(NO3)2 <--> PbS + 2HNO3
ends. It would be downright nutty for CuS or PbS to be insoluble in strong HNO3, but to the extent that H2S stops bubbling or evaporating, I can see
it remaining insoluble.
Tim
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chemoleo
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| Quote: |
Yabbut, Cu(OH,Cl)2, uh almost --any-- hydroxide, etc. are all pretty well insoluble, but relatively few hydrolyze strongly, so I think that's specious
reasoning. |
They are insoluble. Having established that - it's a question then of at what pH this occurs, at what temp, and what the RATE of conversion is. The
RATE of conversion determines how fast insoluble product is accumulating. It could be very slow, and it is counteracted by the solubility of the
product.
Re PbS - I am actually not sure you understand what I am saying.
| Quote: | | I suspect H2S precipitates only up to its solubility limit in water |
No, because H2S is REMOVED from the water when it reacts with Pb2+, so H2S does never have to reach saturation in H2O. The limiting factor is the
solubility of PbS in an increasing concentration of HNO3, not the the solubility of H2S in H2o.
In other words, given high enough concentrations of HNO3, the PbS would redissolve in the HNO3, yes. THAT is when you have reached equilibrium. Until
then, you start accumulating lots of PbS. PbAc2 paper is used for H2S detection for that very reason, and HAc is a much stronger acid than H2S.
Never Stop to Begin, and Never Begin to Stop...
Tolerance is good. But not with the intolerant! (Wilhelm Busch)
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guy
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| Quote: |
But you might be forgetting another factor. It is not only the hydrolysis equilibrium (whose position is governed by the pKa), but it is also a
solubility issue, i.e. |
Solubility of BiOCl: 8.36e-5 M
Solubility of Fe(OH)3: 2.89e-10 M
Solubility of Bi(OH)3: 7.72e-11 M
Now you would be right if we were comparing the rxn
[Bi(H2O)6]<sup>3+</sup> <-> Bi(H2O)3(OH)3 + 3H+
with
[Fe(H2O)6]<sup>3+</sup> <-> Fe(H2O)3(OH)3 + 3H+
But then now, a new problem, why would it hydrolyse to BiOCl instead of Bi(OH)3? Unless the formula for BiOCl was actually BiCl3*Bi2O3...
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chemoleo
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But you aren't getting straightforward Fe(OH)3, you can easily get intermediate products, where it's only partially hydrolysed. Lots of products are
possible, because i.e. O=FeOH with better solubilities may be formed, or even [Fe2+ - Fe4+] species, which exist in complex together.
As to BiOCl vs Bi(OH)3 - does your theoretical calculation account for the fact of reactivities due to orbitals? O=Bi-Cl, is that your BiOCl? It could
well be that the O=Bi bond is much more stable due to orbital positioning than BiR3. Similar to carbon monoxide, CO exists but not CH2, or CCl2. Sorry
I can't come up with a better example, but I hope I get the point across.
[Edited on 23-3-2006 by chemoleo]
Never Stop to Begin, and Never Begin to Stop...
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guy
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I have seen the formulas for "sub" compounds as either an oxy-cation or as a double salt, where one is the hydroxide. Perhaps the latter is a more
correct form of an aqueous "sub" compound. For example, lead subacetate is Pb(CH3COO)2*2Pb(OH)2 (Source: http://www.jtbaker.com/msds/englishhtml/l2521.htm). If this is true then it would make more sense, since part of the Bi3+ would hydrolyse and
precipiate as the double salt.
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chemoleo
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Well yes. Lots of possibilities are possible  . Orbitals and double salts aside -
it all boils down to solubilities. So in essence - it's not the fact that Bi3+ hydrolyses more due to some unforeseen mechanism, it's the lack of
solubility of the *product* that causes more and more hydrolysis, regardless of the hydrolysis equilibriums!
Never Stop to Begin, and Never Begin to Stop...
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neutrino
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I don't follow. The hydrolysis would leave the solution more acidic, so wouldn't hydrolysis impede itself?
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chemoleo
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Oh it does. Once the H3O+ concentration is strong enough to impede it, you've reached equilibrium.
That's what I tried to say above
| Quote: | The limiting factor is the solubility of PbS in an increasing concentration of HNO3, not the the solubility of H2S in H2o.
In other words, given high enough concentrations of HNO3, the PbS would redissolve in the HNO3, yes. THAT is when you have reached equilibrium. Until
then, you start accumulating lots of PbS. PbAc2 paper is used for H2S detection for that very reason, and HAc is a much stronger acid than H2S.
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Never Stop to Begin, and Never Begin to Stop...
Tolerance is good. But not with the intolerant! (Wilhelm Busch)
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guy
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Is it possible to guess the formula by doing this experiment?:
<i>Theory: Find the Ka of the Bi3+ in one experiment. The find the pH of a solution made by dissolving Bi(NO3)3 in 1.0L water. Use the pH to
find the molarity of H+, plug in the Ka expression and see if the mol of Bi3+ is equal to amount dissolved. If it is close, then the formula is in
fact BiONO3.</i> <b>BiONO3 and Bi(NO3)3 can be replaced by BiOX and BiX3; X=halogen)</b>
1)BiONO3 is added to water(record amount). HCl is added until all dissolves(record amount used). Record pH. Amount of Bi3+ should be equal to
amount of BiONO3 added. Find equilibrium expression for BiONO3 + 2H+ <--> Bi3+ + H2O + NO3-.
2) In another experiment, a measured amount of Bi(NO3)3 is added to 1.0L water. Record pH. Convert to molarity and use to find [Bi3+]. Moles of
Bi3+ and moles of H+ together should equal initial amount of solid BiONO3.
[Edited on 3/23/2006 by guy]
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woelen
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From this discussion we can conclude that the simple formulas do not take into account sufficient phenomena. Now we have had:
Solubility of hydrolysed product;
Formation of complexes;
Radii of ions and charge of ions;
Equilibria driven to one side;
etc.
etc.
This matter is so complex that in practice there is only one answer. Let Nature decide itself and perform the experiments. I did with bismuth(III)
salts and these DO hydrolyse massively, even in 1 M HNO3! I also did with iron(III) salts and they also hydrolyse, but their hydrolysis is only faint,
compared with bismuth(III). The same experiments I also did with CdS. Dissolve this in conc. HCl and you get a clear colorless solution (and a
terrible smell of H2S). Dilute the liquid a few times and it becomes totally deep yellow and turbid, due to hydrolysis of Cd(2+), even in the very
strong acid conditions of the still quite strong HCl.
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