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bismuthate
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Thanks. Also I have few thoughts.
1st Could a Ga Al alloy allow Al to react with boric acid?
2nd I am sorry if the equation was incorrect there seem to be many types of aluminium borate.
3rd I know that using a mobile device is a pain I typed this with a kindle with no keypad. It took forever.
[Edited on 14-10-2013 by bismuthate]
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Upsilon
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Ah, I see where we had a miscommunication here. You very well could be right, but I was not referring to reacting boric acid with aluminum. I intended
to heat the boric acid to decompose it into water vapor and boric oxide, THEN reacting it in a solution with aluminum. Sorry if that wasn't clear.
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WGTR
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You might like to consider, that if boric acid is heated to dehydrate it to boron trioxide; then by putting it back into a water solution, you are
rehydrating boron trioxide to boric acid again. Boron trioxide doesn't exist in an aqueous solution. The Wiki page for boron trioxide is a bit
deceptive, because it gives a solubility figure for it. In actuality, it is reacting with water to form boric acid.
If you look at the solubilities listed for boric acid and boron trioxide, the solubilities for boron are almost the same in both cases.
B2O3 solubility: 22 g/L (at unknown temperature).
(1 mol B2O3/69.6182 g B2O3) x (22 g B2O3/1 L) x (2 mol Boron/1 mol B2O3) = 0.63 mol Boron/L
H3BO3 solubility: 4.72 g/100 mL (at 20°C).
(1 mol H3BO3/61.83 g H3BO3) x (4.72 g H3BO3/100 mL) x (100 mL/ 0.1 L) x (1 mol Boron/1 mol H3BO3) = 0.763 mol Boron/L
If the B2O3 solubility was measured at a lower temperature than 20°C, then that could account for the small difference in calculated solubilities.
Anyway, those simple algebraic equations shown above were almost the first things learned in college chemistry. If you read up on Stoichiometry,
you'll get an idea of how to set up these types of equations.
[Edited on 14-10-2013 by WGTR]
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bismuthate
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Sorry for the confusion before. Maybe if you wanted to do the experiment you could find a solvent B2O3 is soluble in or if there is none dissolve the
aluminum in mercury. (quite an odd idea). The main rule is you need to keep the water away from your B2O3 because that will couneract all the time
spent dehydrating your boric acid.
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bfesser
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Threads Merged
14-10-2013 at 10:54 |
Upsilon
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I guess I see why the aqueous single replacement won't work; I had been wondering that myself. If I reduce the boric oxide with no solvent, then the
boron sample would be significantly contaminated. I assume it is possible to dissolve boric acid in propylene carbonate? This would make it easier to
obtain a pure sample.
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bismuthate
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In the begining of this thread I was told that borax does not react with Cl. This raise the question "how does it react with fluorine"? Under
anhydrous conditions of course.
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Upsilon
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Quote: Originally posted by bismuthate  | | In the begining of this thread I was told that borax does not react with Cl. This raise the question "how does it react with fluorine"? Under
anhydrous conditions of course. |
I'm going to guess there is some reaction with fluorine, but it depends on the reduction potential of tetraborate. For some reason it is exceedingly
difficult to find information on the reduction potential of polyatomic ions; in fact, I have yet to get any information on the subject. Regardless it
is much more practical to produce boric acid from borax + an acid.
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bismuthate
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This was not meant to be practical it was only scientific curiosity. I would not want to deal with fluorine.
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Upsilon
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Well anyway like I said you'd have to find a source telling you about the reduction potentials of polyatomic ions, and I wish you luck in your search.
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AJKOER
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An extract from Atomistry.com (link: http://boron.atomistry.com/ ) on Boron and its preparation:
"Amorphous boron is obtained by passing the vapours of the chlorine compound over heated sodium, or, quite similarly to silicon, by igniting the
oxygen compound with magnesium. After the removal of the admixtures, it forms a black powder of the density 2.5, which in many respects behaves
similarly to charcoal, but is more easily oxidised; this occurs more especially by means of strongly oxidising solutions even at the room temperature.
By the fusion of boron trioxide (vide infra) with aluminium, crystallised boron is obtained, which, on account of its hardness, has been called "
adamantine boron." It is not obtained quite pure in this way, but contains aluminium derived from its preparation. Since this metal is the element
most nearly related to boron, the product is not to be looked upon as a compound, but as a mixture (possibly with a diamond-like form of aluminium
isomorphous with boron, and not known by itself).
Boron containing carbon, and obtained from the two elements at a very high temperature, is of a similar character, and also possesses an adamantine
hardness. This also ought most probably to be regarded as a solid solution, and not as a chemical compound."
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bismuthate
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Thanks! The boron chlorides method gave me an idea. Could trimethlyborate (boron timethoxide) be reduced by sodium (or lithium) to yield sodium
methoxie and boron or am I missing the obvious?
[Edited on 19-10-2013 by bismuthate]
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Upsilon
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Just out of curiosity, what does borax decompose into when electrolyzed in an aqueous solution?
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bismuthate
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In solution NaOH and H3BO3 I would believe. However the two will quikly react.
P.S. it isn't decomposition.
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Upsilon
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If you were to use a Hoffman apparatus, however, wouldn't you be able to keep the solutions of NaOH and H3BO3 separate?
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bismuthate
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Yes, but it is more expensive than using HCl. Cool way to make NaOH though.
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Upsilon
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I already have a Hoffman apparatus I made out of PVC. Are you saying that the electricity will be more expensive than the HCl?
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bismuthate
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I'm not sure about you but I have a liter of acid for $2.
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WGTR
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You can isolate boric acid by electrolysis, but you need a suitable diaphragm to keep the anode and cathode compartments separate, and you need a
suitable non-soluble anode.
If you use a cathode compartment that is much larger than the anode compartment, this will make it easier to select a diaphragm. This is because it
limits the concentration of the NaOH in the cathode compartment to a low level. Industrially, obtaining concentrated NaOH with a diaphragm cell is
difficult, because the caustic effects of the concentrated hydroxide attack many materials that would otherwise be good diaphragms. At the same
time, borax is not very soluble, so your final concentration of NaOH probably wouldn't be too high even if the two compartments had equal volume.
Real cellophane might be a suitable cheap option for you.
If you start with a saturated solution of borax, as electrolysis proceeds your precipitate in the anode compartment should consist more and more of
boric acid.
The anode will have to be something that will not react under the cell conditions. Platinum would work great. You could experiment with other
materials to see what works for you.
Reacting HCl with borax would probably be much easier (and faster) than separating it out by electrolysis. Or just buying the boric acid from the
grocery store.
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Upsilon
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For my electrolysis experiments, I use graphite rods that I extract from common pencils. They hold up quite well, and it is no big deal of they need
replaced since they're so inexpensive.
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Upsilon
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Well I went to Lowes and picked up 2 gallons of 31.45% HCl for around $10. Will be experimenting soon.
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avrpunk
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Upsilon, graphite from pencils is quite expensive compared to a welding supply store which will have "gouging rods" in varying diameters. The only
preparation needed is to remove the copper coating from the part that will be immersed, which is quite easy given that you're doing electrolysis in
the first place, and the rest is graphite.
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Upsilon
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Alright, so I started with 100g of borax and added excess HCl and stirred. Afterward, after about 5 minutes the solution settled into a prominent
green-yellow layer floating above an opaque, thick white layer (about 3 times larger than the green layer) that I'm assuming is an extremely saturated
NaCl solution with excess NaCl floating around. I'm going to let it sit for a while yet longer and see what happens. Then I'll remove the HCl layer
and heavily dilute the other layer to hopefully yield some amount of boric acid.
UPDATE: After removing the excess HCl, I added more borax to get rid of any residual HCl, periodically throwing in a pinch of sodium carbonate to test
the presence of HCl. I also added some nom-distilled water to dilute the solution in hopes to get a better view of my product. What I see now is a
large top layer of frothy white solid above a layer of water. Underneath the water is another layer of white solid. Neither boric acid nor NaCl is
less dense than water, so what is this solid that is floating to the top?
[Edited on 23-10-2013 by Upsilon]
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violet sin
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Did you check solubility data on NaCl and B(OH)3? I don't think the thick opaque stuff is NaCl. You should let the ppt. settle out, decant off
HCl/NaCl liquid and wash the ppt. W/ ice cold Distilled water.
Diluting it will dissolve more of your product into the waste solution.
NaCl sol. data from wiki...... 359g/100ml ( no temp given )
B(OH)3 sol..."...."....."......... 2.52g/100ml @ 0'C
........................................ 27.53g/100ml @ 100'C
Borax sol....."....."....."....... No numbers give but stated easily sol. in water
The equations given in wiki for conversion products is 4mol B(OH)3 and 2 mol NaCl. It stands to reason 2 mol of highly soluble NaCl will stay in
sol. But the 4 moles of low sol. boric acid won't. ie your white ppt is boric acid no?
*************************************
But CO2 bubbles on small boric acid crystals is lighter than water, probably your floaters. If you started with just enough water to dissolve the
borax( add stirr, repeat. not super critical ) then calculated your HCl to be a slight excess needed to convert the borax. Add and wait for rxn.
Calculate NaCl moles produced, and reduce volume of sol. by heat till the salt is nearly saturated. And then chill to ppt. max B(OH)3, filter off,
wash w/ ice cold water, you should have fairly clean boric acid. You could then dissolve it in just enough boiling water and wrap you beaker in a
towel for insulation so cooling is slow. And get larger, easier to filter/wash nice boric acid.
[Edited on 23-10-2013 by violet sin]
[Edited on 23-10-2013 by violet sin]
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Upsilon
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I skimmed the surface of the settled solution with a spoon and gathered the floating substance. I put it in a new beaker and added a few hundred ml of
distilled water. Now only a fraction of the solid is floating above the water while the majority settled at the bottom. I'm going to repeat this
process and see if any still floats.
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violet sin
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You will be able to get boric acid the way you did, but I think it would be more rewarding to crunch the numbers and all next time. I did a quick run
through and if you started with 100g borax fully dissolved you would aprox 52ml of the 31.45% HCl, + throw in a slight excess. After rxn that would
make aprox 30.34g NaCl, 61.83g B(OH)3. You could calculate aprox yield off theoretical in the end. No guessing on neutralization HCl, baking soda
not needed. Heat to reduce sol. volume and bonus drive off excess HCl. Though it would be hard to reduce it till NaCl saturation as I had said,
would be near nothing left liquid wise with the boric ppt at bottom and only 30.34g NaCl. So just reduce it enough to not spend all day gravity
filtering for first isolation of boric acid. Then try the boiling water/towel on beaker thing( worked for me in the past quite well just haven't
tried on boric acid) for larger crystals and less drain time. Wash with ice cold water a few times and should be nice and clean.
There might be a slight mistake or two on the exact numbers, I calculated the needed HCl from weight/weight % of 1000ml @ 1.16g/ml. Wasn't labeled
strictly w/w % on MSDS but I don't think they would do w/v or volume/volume % with a gas dissolved in liquid. But I don't have a bottle handy to
check( its back home with my pool).
Man it sucks typing this up on an iTouch ): hope it helps
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