Re: CO2 and KH

> From: Michael Irlbeck <u7211aa at sunmail_lrz-muenchen.de>
> thank you for the discussion on the issue.

You're welcome.  But I'm afraid this will quickly get beyond my 
"practical" understanding of the CO2 reactions.  Buffering systems 
are difficult to understand and I'm not a chemist. 

> The question is:
> given that 
> CO2+H2O<-->[H2CO3]<-->HCO3-+H+

I'm not sure that is a "given".  There are various forms of CO2 in the
water and the proportion of these forms is determined by pH.  One form
is a combination of CO2+H2CO3 (dissolved carbon dioxide and carbonic
acid).  A second form is HCO3- (bicarbonate).  At a pH of about 6.37
(dependent on temperature), 50% is CO2+H2CO3 and 50% is HCO3-.  At a
pH of 8.37, all the CO2+H2CO3 becomes HCO3-.  At a pH of 10.25, 50% is
HCO3- and 50% is CO3--.  A graph of this can be found in the Baencsh
Atlas, Vol. 1 (pg 32).

As you add CO2, a small percentage of it becomes carbonic acid and 
reduces the pH, altering the proportions of the various forms.
Exactly how the chemical reactions occur is beyond me.  So we
have the following equilibiria

     CO2 + H2CO3 <-->  HCO3-  <--> CO3-- 

that is dependent on pH. 

When you use a test kit to measure CO2 concentrations, you are simply 
titrating the water sample with sodium hydroxide until all the
carbonic acid is converted to bicarbonate at a pH of 8.37.  Thus, the 
"free CO2" you are talking about is the amount of CO2 in both
dissolved form and in the form of carbonic acid. 

In a practical sense, the amount of "free CO2" in the water in a
typical, non-injected aquarium is very small (1-2 ppm).

> How much CO2 do you have to add to get a certain level of free CO2. How 
> is this a function of HCO3-? 

If you inject 15 mg/l (for example, you add 1.5 grams to a 100 l tank
*and keep it there*), you will have 15 mg/l of CO2 that plants can
use, free or otherwise.  The amount of HCO3- in the water will
determine the pH.  

I'm afraid that last part has a "hole" in it somewhere because I can't
convice myself that it is true.  I also can't answer the last part of
your question in a sensible way.  Time for a real chemist to carry on
from here.