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Nicodem
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What do you mean with your claim No. 1?
Assuming volume additivity, in the final solution you would have got approximately [Na<sup>+</sup>] = 0,5 mol/L and
[Cl<sup>-</sup>] = 10,3 mol/L, concentrations which do not exceed your Ksp = 36 (mol/L)^2.
I might have calculated it wrong as I was so upset from your terribly unscientific claim No. 2 that thinking straight was not possible. We are not
talking about dilute solutions here! You can not use Ksp equations just like that! You seem to have no clue of how ions behave in solutions. There are
no free ions in aq. solutions. All ions are solvated. And since they are solvated the [H2O] does not equal c(H2O). The Ksp calculations, which neglect
[H2O] by definition, can only be done for salts of low solubility where the part of water molecules occupied with solvation can be neglected. It is
this in addition to the common ion effect that causes oversaturation. (In organic chemistry we call this phenomenon "salting out" and by using it, it
is possible to severely decrease the solubility of water soluble compounds. For example, water/acetone mixture will become biphasic when you add some
brine – acetone is unable to solvatize Cl<sup>-</sup> ions, so it rather forms a separate layer instead of interfering with the higher
energy process of Cl<sup>-</sup> solvation by H2O molecules.)
Besides, the recrystallization of crude NaCl out of conc. HCl standard way of purifying it (the solubility of NaCl at room temperature is low, while
it is much higher at boiling point). It is also used to grow large NaCl monocrystals.
…there is a human touch of the cultist “believer” in every theorist that he must struggle against as being
unworthy of the scientist. Some of the greatest men of science have publicly repudiated a theory which earlier they hotly defended. In this lies their
scientific temper, not in the scientific defense of the theory. - Weston La Barre (Ghost Dance, 1972)
Read the The ScienceMadness Guidelines!
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woelen
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| Quote: | | 1) What we are looking at here is the common ion effect |
You take 5 ml of a concentrated solution of NaCl and mix this with 55 ml of conc. HCl.
Let's do a rough computation, according to the standard theory about common ion effect:
Solubility of NaCl in water is 300 to 350 g/l, which is 5 to 6 mol/l (I do rough calculations, just to get the gist of what I want to say and do not
want to go into detail about numbers). Hence, the solubility product of NaCl would be appr. 30 mol2/l2.
Now, if we take 55 ml of conc. HCl, 34%, which is around 11 mol/l.
Now we add 5 ml of your concentrated solution of NaCl, let's say 5.5 mol/l NaCl. The new solution, after thorough mixing, will have appr. 10 mol/l of
chloride ion, and appr. 0.5 mol/l of sodium ion (your 5 ml of solution is diluted by mixing it with 55 ml of acid). So, the product of [Na(+)]*[Cl(-)]
is appr. 0.5*10 = 5 mol2/l2, which is well below the number of 30 mol2/l2.
So, there really is more to say about this than simple common ion effect. If that were the only effect, then there would be no precipitate of NaCl at
all. You have done the measurement of the precipitated NaCl and according to your measurement, only 0.5 grams of NaCl remain in solution at the 60 ml
of 31% HCl. Your 0.5 grams of NaCl in 60 ml of liquid is only roughly 0.15 mol/l. With a chloride ion concentration of appr. 10 mol/l, your theory
would predict a solubility of around 3 mol/l of NaCl, which is 20 times as much as what one find in a real situation.
[Edited on 6-1-08 by woelen]
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len1
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1) Woelen, you are right. I made a mistake, the solubility product if one assumes activity coefficients of 1 is well below the 36mol/L^2.
2) Thats still not too bad because activity coefficients in very concentrated solutions can differ substantially from 1. So if the activity coeff is
about 16 for 10mol/L Cl- solutions the solubility product holds, using molalities. The coefficient is very large though - suggesting non-linear
effects [Cl]^2 etc cant be neglected, its a useful first approx though.
3) My point about this not being a way to eliminate NaCl from solution also still holds - 1/3 of the salt remained in solution using liquid HCl - and
~ 0.14mol/L minimum NaCl remnant using HCl gas.
4) It has been shown that this is the common ion effect. I could get no precipitate with KNO3 is an example I gave. Woelen if you want anyone to
believe you, you must be able to precipitate other salts in HCl, that is the consequence of what you are saying. It would then run counter to what
garage chemist is claiming, that this is method is used to precipitate pure NaCl from solution. My experiments (irrespective of the numerical error)
show that he is right in this - though not as a way of clearing the soln of NaCl.
I emphasise that using activities to 'patch up' the mistake I made in the calculation is not some slight of hand way to maintain what I was saying.
That is because my assertion that this is the common ion effect is based entirely on 2) - not being able to get a precipitate with non-chloride salts.
The calculation is useful for judging the strength of the effect - a factor of six for the activity means that its actually quite strong, that is
the import of the numerical correction.
@Nicodem. With respect to the solubility product you are right. As for the rest, read up on activities. When a person shows hes prepared to cheat
using obfuscation as cover I dont argue with them - its a waste of time. If people want to read your posts in this thread each can draw their own
conclusions http://www.sciencemadness.org/talk/viewthread.php?tid=9717&a...
[Edited on 7-1-2008 by len1]
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woelen
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You could try to repeat exactly the same experiment as you did with NaCl, but now with 55 ml of 34% HCl and 5 ml of a 6M solution of NaBr instead of
6M NaCl. I'm quite sure that the precipitate will be near 100% of NaCl and only contains a small amount of bromide. I am inclined to think that the
same is true for 5ml of 6M NaNO3.
Most likely the solubility of KCl in conc. HCl is somewhat better than the solubility of NaCl in conc. HCl and that is why no precipitate is formed.
I never said that I believe that I can precipitate other salts in conc. HCl. If you have concluded that from my previous writings, then I just want to
say sorry, I might have been somewhat unclear  .
But I do believe that ANY Na-salt of sufficiently high solubility does lead to precipitation of NaCl (and not of that salt) under the conditions of
relatively small amount of salt, added to large excess of HCl (your 5 vs. 55 ml experiment is a good way of testing that).
I hope to find some time for this, tomorrow or Tuesday. Now it is too late for me, we have a baby crying in the night at least once and then there are
tasks with higher priority  (such as getting sufficient sleeping time).
[Edited on 6-1-08 by woelen]
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len1
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Yeah well, my experiment has shown that Na+ concentrations in excess of about 0.14mol/L in conc. HCl lead to NaCl precipitation, but thats just using
standard thermodynamics. So I dont need to repeat the exp. with NaBr, Im pretty sure of the result. BUT If there is some substantial change in the
nature of conc. HCl solutions, making the solvent non-polar or whatever, this should give a precipitate with salts such as KNO3. You can not make
exceptions willy nilly, else Ocams razor comes into play.
Good luck with the kid - ours have fortunately stopped crying now.
[Edited on 6-1-2008 by len1]
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garage chemist
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I once made some perchloric acid by mixing conc. NaClO4 solution and a large amount of HCl followed by filtration of the NaCl and workup via
distillation. It worked well, only a small amount of sodium salt remained as distillation residue.
That your KNO3 solution did not precipitate KCl surprises me a bit.
Perhaps KCl behaves different than NaCl here?
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len1
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Yes, KNO3 gives no precipitate under similar circumstances.
This is a very interesting phenomenon. For HCl up to about 20% what I got is roughlly in accord with the common ion effect (NaCl gives precipitate
NaNO3 doesnt). However above that there is definitely more than just the common ion effect, I have to take back my earlier statement. The activity
coefficient grows much faster than expected, so that at 37% the solution holds maximum sodium ion concentrations of the order of fractions of mol/L.
With such a small figure you have to say that sodium ion concentrations will be small no matter what the anion it comes with.
I disagree with woelen as to the nature of the effect however. I think the evidence is unequivocal. If there was a drastic change in the nature of
the solvent the solubility of salts such as KNO3 would be affected.
Interestingly KCl does precipitate from 32% HCl, but to a smaller extent than the sodium salt. It would appear the nitrate ion, despite its low
concentration has an effect in stabilizing the K+ in solutions of KNO3. NaNO3 does produce a precipitate - but less than NaCl. We have a clear
picture here. At very high [HCl] almost all water molecules are tied-up to the ions, and so the ion with greatest difference between solvation energy
and lattice energy stays in solution. Potassium solvates easier than sodium, while the nitrate anion acts the other way, increasing solubility for
both sodium and potassium. Based on this we expect KNO3 to be the most soluble salt of the four combinations.
I have repeated the exp as to the solubility of Na+ ions in 34% HCl, it is > 0.14mol/L as I earlier determined. I have determined this by adding
solutions of lower concentration to conc. HCL solutions and observing that it precipitates nothing.
PS There is a PS to this. You can get Na+ concentrations of 0.2 mol/L or less by gassing concentrated solutions of NaCl and perhaps NaBr, NaI. Salt
of Na with other anions, such as SO4- or NO3- PO43- will not be precipitated to nearly this degree from conc solutions - the common ion effect if you
like.
[Edited on 8-1-2008 by len1]
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woelen
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Another interesting experiment could be to take a precipitate of KCl from conc. HCl and add a few drops of conc. HNO3 and mix. It would be
interesting, if indeed the precipitate of KCl redissolves in this case. Of course, the total volume also increases, so one must be sure that the total
amount of dissolved KCl is (much) larger than the amount one would expect because of the increase of the volume.
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