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I balanced the chemical reaction which is as follows:

NaCl + KHCO3 = NaHCO3 + KCl

I've also gathered solubility values which are the following:

  • KHCO3 22.4 g/100 mL (20 °C)
  • NaCl 35,9 g/100 mL (20°)
  • KCl 34.2 g/100ml (20°C)
  • NaHCO3 9.6 g/100ml (20 °C)

Since NaHCO3 is less soluble than the other compounds, I could separate it by boiling off some of the water up to a certain point, but I'm not sure how to predict which way the reaction will be going and if it will proceed completely.

How do I find out which way the reaction will proceed, and to which extent it will be complete?

I'd be glad for any help.

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    $\begingroup$ You show the bicarbonates, i.e., the hydrogen carbonates. Is that correct? $\endgroup$ – Ed V Jul 19 at 1:14
  • $\begingroup$ Sorry, I corrected the title! Thank you Ed. That being said and done I'd also be interested about the case of the carbonates. But I think I could extrapolate myself for them from the information I'll receive when this question is answered. $\endgroup$ – Hans Jul 19 at 12:38
  • $\begingroup$ Thanks a lot Ed!! $\endgroup$ – Hans Jul 19 at 12:45
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    $\begingroup$ You could run it over an kation exchanger column that is loaded with Na⁺. If we're talking "household" chemistry, such ion exchanger may be harvested from an old dishwasher, and it can be loaded with Na⁺ using NaCl (that's the salt the dishwasher asks for). $\endgroup$ – cbeleites supports Monica Jul 19 at 21:10
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Boiling off some water is a decent plan. Perhaps it should be done at reduced pressure so as to keep the temperature low. Magnesium and calcium bicarbonates decompose at temperatures above 60C in water to yield insoluble carbonates, and several data sheets for sodium bicarbonate suggest the same. One said decomposition was complete at 100C. I suppose you could consider that any H2CO3 in equilibrium with NaHCO3 could decompose to CO2 + H2O. Heat would just drive the equilibrium to completion. I suppose that's why I could find no solubility data for NaHCO3 or KHCO3 at temperatures higher than 60C.

One way to go about the preparation is to dissolve 58.5 grams (1 mole) of NaCl in 260 grams H2O, and 100 grams of KHCO3 (1 mole) in 260 grams H2O. Heat both to 60C; they will completely dissolve. Then pour one into the other (it may make a difference which way you do it) and stir. The amount of water is sufficient to dissolve 1 mole (84 grams) of NaHCO3 at 60C (but the other salts could and probably will affect this). So the mixed solution will be clear; all salts will be soluble.

Cooling to 20C will (should!) ppt 34.1 grams of NaHCO3, leaving 49.9 grams in solution (again, ignoring the effect of the KCl). Cooling to OC should ppt another 14 grams of NaHCO3, leaving 35.9 grams in solution. You could then boil off about half of the water before the solubility limit of the KCl is reached, but because of the possible loss of CO2, you may not get much more NaHCO3.

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  • $\begingroup$ James: Thank you!! $\endgroup$ – Hans Jul 19 at 17:48

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