I would like to make Na2CO3 react with HCl in a closed fixed volume vessel, and accordingly I'd like to understand how ensuing pressure will affect the equilibrium of my reaction (and to which extent my reaction will affect the level of pressure when equilibrium is reached).

I guess that the link between pressure and the reaction mostly has to do with CO2 retention into the solution, which leads me to want to understand how I can calculate CO2 solubility in water vs pressure.

Would anyone care to explain me how I could calculate CO2 solubility vs pressure? And tell me if there's anything big I am missing other than CO2 solubility?


1 Answer 1


The acidity of HCl is so great compared to H2CO3 that you have no equilibrium between Na2CO3 and HCl; the equilibrium is only between CO2 and water and the vapor space. H2CO3 is one of the forms of CO2 in water, but it is unnecessary to delve into this.

After your reaction, you will have a solution of NaCl in which some of your CO2 is dissolved; the rest of the CO2 will be in the vapor space. We will assume that the solubility of CO2 in this solution is the same as in pure water. Data from CRC Handbook, per 100 cc H2O at 1 atm pressure: 171 cm^3 at 0C, 90 cm^3 at 20C. In grams: 0.348 at 0C, 0.145 at 25C, 0.097 at 40C, 0.058 at 60C.

That's the chemistry. Now the mathematics is more complicated, and I'll leave that up to you. But a few points to help you get started: if you have a container with a volume of 100 cc and put 100 mL of water into it, you could also dissolve 171 cm^3 (0.348 grams) of CO2 into that same volume at 1 atm pressure (OK, maybe a slight volume increase - ignore it for simplicity). If you put only 50 mL H2O in that 100 cc volume and added 171 cm^3 of CO2, the pressure in the vessel would be greater than 1 atm (but less than 1.71 atm, which is the pressure you would have with no water in it at all).

  • $\begingroup$ This gives me clear structure of how to think about the issue and what I will need to learn and calculate. That's an awesome start for me as it is exactly what I need! Thank you James! $\endgroup$
    – Hans
    Jul 21, 2019 at 17:51

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