decarbonate alkaline electrolyte

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robeyw
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decarbonate alkaline electrolyte

Post by robeyw »

I have a very strange problem restoring old flooded pocket plate NiCd cells. As I understand it, the new electrolyte would be a solution of 12 grams LiOH and 264 grams KOH in 1 liter water. With exposure to air over time it becomes carbonated. I tried to decarbonate it by reacting it with Calcium hydroxide. This is a slow process because the solubility of the Calcium hydroxide in electrolyte of about 200 ppm but it can be done. I judge it to be complete when injecting a bit of Ca(OH)2 saturated water into the middle of the solution produces no precipitate. (a precipitate is formed briefly because the solution is locally super saturated with calcium hydroxide but it redisolves. I inject it into the middle because it is much less dense than the solution under test.). I thought I would remove trace sulfate by reacting it with barium hydroxide but found a very heavy precipitate. Ultimately 80 grams of barium hydroxide per liter of electrolyte is required. The precipitate is white, dissolves quickly almost completely with HCl with evolution of a lot of gas. After heating the precipitate to dull red then cooling, the reaction with HCl is unaltered. All this points to the precipitate being mostly barium carbonate, but all the carbonate was supposed to have been removed using calcium hydroxide. So what happened?
robeyw
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followup

Post by robeyw »

Having had no response at all, here are some calculations and observations:
Starting with solubility constants from the CRC HB of Chemistry & Physics 84th ed :
BaCO2 2.58 E-9
Ba(OH)2 2.55 E-4
BaSO4 1.08 E-10
CaCO3 3.66 E-9
Ca(OH)2 5.02 E-6
The starting electrolyte had a specific gravity of 1.14. If this was the desired mix of hydroxides, it would be 167 g/l KOH + 8 g/l LiOH so it would be 2.98 M in KOH + .33 M in LiOH. Using these values which should approximate the finished solution,
Dissolving unlimited Ca(OH)2 with [OH-] known
[Ca++][OH-]^2= 5.02 E-6 so [Ca++]=.457 E-6 M
Then ionic carbonate can be calculated [Ca++][CO3--]=3.66 E-9 so [CO3--]=.008 M or .48 g/l

For barium [Ba++][OH-]2= 2.55 E-4 so [BA++]=2.6 E-5 M
[Ba++][CO3--]= 2.58 E-9 so [CO3--]= 9.9 E-5 M which is 80.7 times better than was achieved using calcium.
I observed using approximately 80 g/l Ba(OH)2 or .35 moles/liter rather than the calculated .008 to precipitate the additional carbonate, so the first problem is to explain this. It occurs to me that considering the very small second ionization constant for carbonate, a large amount of the carbonate might be bound as KCO3- +LiCO3-. Is that reasonable? Contrary indication is seen by preparing fresh electrolyte then introducing a few drops of Ca(OH)2 saturated in water. A fine precipitate forms and gradually settles. But from the previous experiment, Ca(OH)2 left atleast .35 moles/liter carbonate in solution. That is more than the new electrolyte should contain. So this thought of kCO3-+LiCO3- is not supported.

The second problem is to remove trace SO4--. The origional intent was to precipitate it with barium, but now this seems to be a wrong idea.
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