When mixing HCl with K2CO3, the reaction that is usually given is the following:

K2CO3 + 2HCl → 2KCl + CO2 + H2O

I am wondering, however, if I could produce KHCO3 from K2CO3, by using half as many moles HCl. The reaction I imagine could happen would be expressed as follows:

K2CO3 + HCl → KHCO3 + KCl

I've gathered the solubility data for this reaction hereafter:

  • K2CO3: 112 g/100 mL (20°C)
  • HCl: 72,5 g/100ml (20°C)
  • KHCO3: 22.4 g/100 mL (20°C)
  • KCl: 34.2 g/100ml (20°C)

From solubility data I see that it wouldn't be extremely practical to conduct as both KHCO3 and KCl are fairly soluble, but I'm wondering if the reaction would occur, or if it would just produce half the amount of KCl + CO2 + H2O through the first reaction (which is generally attributed to these reagents).

If the reaction occurs, can I assume that the first reaction is strongly dominant until I get close to have added as many moles of HCl as the initial moles of K2CO3?

And if it doesn't occur, why does it not?

Note that this question had been asked already, but had received no satisfying answer.

  • $\begingroup$ You obviously have to dissolve the carbonate in water first. Look at the ions you have in solution then, and imagine slowing adding HCl, and remember the fastest reaction (very good rule of thumb) is always $\ce{H3O+ + OH- -> 2 H2O}$ $\endgroup$ – Karl Jul 21 '19 at 13:50

It is interesting to consider that either outcome is possible, depending on the order of addition.

If a solution containing 0.5 moles of HCl is added to a solution of 1 mole of K2CO3, slowly, with mild stirring, cold, the products will be 0.5 KCl + 0.5 KHCO3. Nothing is lost to the atmosphere.

However, if a solution of 1 mole of K2CO3 is added to a solution of 0.5 moles of HCl, it would be possible to get 0.5 KCl + 0.5 K2CO3 with slow addition, rapid stirring, warm/hot. The difference here is that the initial K2CO3 is completely overwhelmed by the HCl and produces KCl + H2CO3, which can be degraded to H2O + CO2. By the time half of the K2CO3 is added, the products are 0.5 KCl + CO2, and the CO2 could be mostly evolved as gas and disappears. The last half of the K2CO3 just comes in and sits there.

  • $\begingroup$ Thanks for your answer! You meant to write 2KCl + CO2 + H2O at beginning of the last paragraph right? $\endgroup$ – Hans Jul 21 '19 at 17:31
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    $\begingroup$ Oops! What I meant was that as you add the K2CO3 (1 mole in solution) to the HCl solution, at the halfway point, you can get reaction to KCl + 0.5 H2CO3 (which goes to H2O + CO2 and has the CO2 forced out of solution before it gets reacted with K2CO3 to form KHCO3). Thanks for noticing that! $\endgroup$ – James Gaidis Jul 22 '19 at 13:42

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