CO2 and KH

> > 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).

I'm not sure your eqation expresses the whole reaction. To my knowledge,
CO2 combines with water to form H2CO3. But this H2CO3 is very short living
in H2O and quickly goes to either side of the equation (at 'normal' pH
values). Actually, to be more correct, it's a constant. That's why it's
usually put in brackets. The real concentration of it is neglectable for
our purpose. What makes the 'acid' is not the H2CO3, but the H+
dissociated from it, by definition. So we could simplify to: 

CO2+H2O <-> H+ + HCO3-

As you pointed out, at high pH values, we will get a significant ammount 
of CO3-- as well. I think the H+ ions should show up in the equation.
'Dependent on pH' is correct, of course as far as the ratio between H+ 
and HCO3- goes, but the balance of the equation will still be the same. 
If you add H+, some of the HCO3- will be used up to form CO2+H2O. (I'm 
just rambling on here to make it clear to myself mainly). Thus, the pH 
will NOT change (if you allow enough time for the extra CO2 to leave the 
water), as long as there is still enough HCO3- (--> buffering). 
So by adding acid (H+), carbonate hardness can be reduced effectively.

With free CO2 I meant the CO2 part of the equation or dissolved CO2 
useful to the plants. As I said, the [H2CO3] is neglectable.

> > 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 think I can understand it now a little better ( I consulted one of our 
chemists :-) ). If you add CO2 to the water, some of it will dissolve to 
H+ and HCO3-. The H+ ions will be caught away by HCO3- already in the 
water, if you have alkaline water with a high KH, for example. The HCO3- 
from the dissolved CO2 will be left over and keeping the HCO3- 
concentration stable. Now it's very clear why the KH does not change when 
you add CO2 to the water! It's also clear now why less CO2 will be needed 
to get a  certain concentration of dissolved CO2 when the KH is low. The 
ammount can be calculated with the Henderson Hasselbalch equation.