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Author: Subject: BariumCarbonate from BariumSulfate + Soda
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[*] posted on 12-9-2007 at 04:23
BariumCarbonate from BariumSulfate + Soda


I want to cook BaSO4 with Na2CO3 to get BaCO3 (+ Na2SO4)

This is said to work to 100 % (no BaSO4 left in the BaCO3), when cooked with concentrated soda-solution over quite a while.
My experience is: Some 20-30% BaCO3, rest BaSO4, even with quite concentrated solutions, 3-5 times the stoechiometric amount of soda and 12+ hours of cooking.
How can I get clean BaCO3 (in an time-economic manner) by this cooking-procedure (several cookings, washing etc.) ? (Have tried the melt, but it has to be _milled_, quite hard stuff, don't want that)
My BaSO4-powder is quite fine (average grain-size 6 micrometers).
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[*] posted on 12-9-2007 at 04:44


I think you answered your own question. You need to remove sulfate from the reaction mix to force the reaction, so repeated ( cooking, filter+wash ) cycles is what needs to be done. The wash doesn't need to be really thorough, just remove most of the Na2SO4.

A really large excess of Na2CO3 over remaining BaSO4 helps, say 10 to 15 times excess; industrially the used solution would be fractionally crystallised to recover much of the unreacted soda for further use.

If you are after the barium carbonate to make other barium compounds from, then you just live with some sulfate and filter it out of the solution of the barium salt.

In analytic chemistry, the melt method also used a many times excess of soda, treatment with hot water left a fairly fine powder of BaCO3 after the sodium salts were dissolved away.
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[*] posted on 12-9-2007 at 05:00


Interesting: Crystallize the soda; I have thought about that, but the solubilities at 20 and 100 [Celsius] (Na2CO3, Na2SO4) are so similar that it seemed to not be very applicable. At least one would want to remove the Na2SO4, so the soda could be used again for the same process ? May be partially recycle the soda, but at low efficiencies ? (Always to use about twice the stoechiometric soda-amount per mol BaCO3 ?)
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[*] posted on 12-9-2007 at 05:04


100% may be a stretch. Maybe 99.5% only. You should look into "what is said" for improvement to your experience:

http://dx.doi.org/10.1021/ac50065a014
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[*] posted on 12-9-2007 at 05:17


Great link ! Only that I have no access there; but the first (free) page contains enough info ... .
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[*] posted on 12-9-2007 at 13:25


Here ya are chief.

Attachment: ac50065a014.pdf (456kB)
This file has been downloaded 840 times

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[*] posted on 13-9-2007 at 04:31


Barium sulfate and sodium carbonate are both very cheap, and so (on a small scale) it's perfectly acceptable to sacrifice efficiency for quality.

Take your crude BaSO4, BaCO3, Na2SO4, Na2CO3 and H2O mixture, and collect the insoluble components by filtration.

Add a nominal excess of hydrochloric acid to the resulting BaSO4/BaCO3 mixture, and filter to give a solution of BaCl2 and HCl.

Next, add a nominal excess of Na2CO3 solution, and the resulting precipitate will be pure BaCO3.




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[*] posted on 14-9-2007 at 06:25


Great Idea !!! I have a lot of Ba(NO3)2, instead the BaCL2 . Thank you !!!
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[*] posted on 6-10-2012 at 09:43


ok i did the math on this, trying to figure out how efficient it is in theory. we have two equilibria1:

1.
BaCO3 <---> Ba2+ + CO32-
K1 = 2.58 * 10-9

2.
BaSO4 <---> Ba2+ + SO42-
K2 = 1.08 * 10-10

at the beginning we have a suspension of BaSO4 in a Na2CO3 solution of molarity a. in the absence of other species the BaSO4 dissociates slightly and causes for Ba2+ and SO4 to reach molarity b each. together, a new equilibrium is formed; CO32- steals some Ba2+ (let's say c mol/L) and precipitates (equation 1, from right to left); as a consequence, further BaSO4 goes into solution (let's say d mol/L) (as in equation 2). at equilibrium, the solubility products hold:

K1 = [Ca2+] * [CO32-] = (b - c + d) * (a - c)

K2 = [Ca2+] * [SO42-] = (b - c + d) * (b + d)

---> (b + d)/(a - c) = K2 / K1

and

(K1 / K2 + 1) * (b + d)2 - a * (b + d) - K2 = 0

which resolves to

b + d = a * (1 + sqrt(1 + 4 * (K1 + K2) / a2)) / (2 * (K1 / K2 + 1))

the ratio inside the square root is small, so this approximates to

a / (K1 / K2 + 1) = 4.02% a (*)

i.e. the final molarity of sulfate ions (which roughly indicates how much the metathesis progressed) is about 4% of the initial molarity of sodium carbonate dissolved. of course this says nothing about how fast the equilibrium is reached.

i also tried to work out the problem with CaF2 instead of BaSO4, but this seems to involve more than a simple quadratic equation, due to fluoride being monovalent :(

[1]
solubility products @ 25°C:
http://www.ktf-split.hr/periodni/en/abc/kpt.html

edit. it's clear from (*) that the stoichiometry of the undissolved substances plays no role. you wanna use an 'excess' Na2CO3 in the sense that you want it well saturated (i.e. a = 4.29 M @ 100°C), while at the same time you shouldn't mix in more BaSO4 than can be 'digested' (i.e. about 4% of 4.29 M, or 0.171 M) because the equilibrium represents a barrier that can't be crossed.

running the computation with calcium sulfate (more soluble) instead of barium sulfate, the value in (*) is over 99% of a.

using an approximation, it turns out that the fluoride in CaF2 can be leached to the extent of about 0.174 M for the same 4.29 M of Na2CO3.

[Edited on 6-10-2012 by tetrahedron]
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[*] posted on 7-10-2012 at 02:40


Quote: Originally posted by tetrahedron  
the fluoride in CaF2 can be leached to the extent of about 0.174 M for the same 4.29 M of Na2CO3


here goes:

a = [Ca2+]start

b = [F-]start

c Ca2+ + c CO32- ---> c CaCO3 (s)

d CaF2 ---> d Ca2+ + 2d F-

K1 = [Ca2+]end * [CO32-]end = (a - c + d) * (a - x)

K2 = [Ca2+]end * [F-]2end = (a - c + d) * (2 * (b + d))2

this system of equations gives (i'll spare you the details)

16 K1 x4 + 4 K2 x3 - 4 K2 a x2 - K22 = 0

where x = b + d. this intractable quartic can be reduced to

4 K1 x4 + K2 x3 - K2 a x2 ~~ 0

using the approximation K22 = 0. the only plausible solution is

x ~~ (- K2 + sqrt(K22 + 16 K1 K2 a)) / (8 K1)

here's where you can plug in a = 4.29 M and get x = 8.69 * 10-2 M, i.e. about 0.174 M of fluoride, as in my previous post. note that the ratio K2 / K1 is so small that (for a not too small) we can further approximate

x ~~ sqrt(K2 * a / (4 * K1))

i.e. the efficiency of the methathesis only increases with the square root of the initial concentration of Na2CO3 for the monovalent anion F-, as opposed to linearly in the case of the divalent anion SO42-.
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