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.