『1-4 A SECOND EXAMPLE: CARBON DIOXIDE IN WATER

　The hydrogen chloride reaction has only minor interest in geology but serves as a beautifully simple illustration of the principles of equilibrium because all the substances involved are gases. More technically, we say that all mixtures of H2, Cl2, and HCl constitute a single phase (unless, of course, the temperature is low enough for chlorine or hydrogen chloride to liquefy), the word phase meaning any part of a system which is homogeneous and which is separated from other parts of the system by sharp boundaries. The equilibria of greatest concern to geologists are those involving more than one phase -- liquid and gas, liquid and solid, or more complicated systems containing both liquid and gas together with several solid phases. Such equilibria are described as heterogeneous, in contrast to homogeneous equilibria like the HCl reaction.
　A simple two-phase system of everyday experience and of great importance geologically is the equilibrium between carbon dioxide and water. For a preliminary study of this system, one need only put a stopper in a bottle half full of water. Gaseous CO2 from the air in the top of the bottle dissolves in the water, and dissolved CO2 in the water escapes into the air. Soon a balance is established between these two processes:
　　　CO2 in air ⇔ CO2 dissolved in water　　　　　(1-8)
The importance of this reaction is due to the fact that the dissolved CO2 reacts with water to form an acid, H2CO3, carbonic acid. Thus we can rewrite the equation for the equilibrium,
　　　CO2 (in air) ＋ H2O ⇔ H2CO3　　　　　(1-9)
(We assume here that all the dissolved CO2 forms H2CO3. This is not strictly correct; to be more accurate, we should consider also the equilibrium CO2 (dissolved) ＋ H2O ⇔ H2CO3, which actually is displaced far to the left. To simplify this and future discussions, we omit this step. As long as the assumption is used consistently, the omission does not affect geologic arguments.)
　The equilibrium responds predictably to changes in conditions: if we increase the amount of CO2 by adding some from a tank, we speed up the forward reaction and more CO2 dissolves; if we decrease the pressure of CO2 by attaching a vacuum pump, the forward reaction is slowed and the equilibrium shifts to the left; heating the bottle decreases the solubility of the gas (or the stability of H2CO3), and again the equilibrium shifts to the left. Or we could disturb the reaction chemically by adding a base to neutralize the acid: if a little NaOH solution is poured into the bottle, some of the H2CO3 is destroyed, thereby slowing the reverse reaction and permitting more CO2 to dissolve.
　A numerical value for the equilibrium constant,
　　　K＝[H2CO3]／[CO2][H2O]　　　　　(1-10)
can be found by looking up the solubility of CO2. This is given in tables as 0.76 liter/liter of water at 25℃ when the pressure of CO2 is maintained at 1 atm. To express concentrations in solution, a common unit is moles per liter; the 0.76 liter of CO2 under these conditions would represent 0.76/24.5 or 0.031 mole, so that the concentration of H2CO3 in Eq.(1-10) may be written 0.031M. (The symbol M means moles of solute per liter of solution, and the number 24.5 is the volume in liters occupied by 1 mole of any gas at 25℃ and 1 atm pressure.) For the concentration of CO2 gas we use atmospheres of pressures, as we did for the gases in the HCl equilibrium; in this case its concentration will be 1 atm. For the concentration of H2O we could find moles per liter (1,000/18.016＝55.5), but in all dilute solutions the concentration of H2O is so nearly the same that we can treat it as constant and include it in the value of K. Hence we wrote
　　　K＝[H2CO3]／[CO2][H2O]＝0.031／1＝0.031＝10^-1.5　　　　　(1-11)
Note that the K in Eq.(1-11) is 55.5 times larger than the K in Eq.(1-10), since the unchanging concentration of water is included in the former.
　Having found the constant for the CO2 reaction, we can now use it to calculate how much H2CO3 is present in water exposed to ordinary air. Air contains 0.03％ CO2 by volume; this means a volume fraction of 0.0003 and therefore a partial pressure of 0.0003 atm (since partial pressure is approximately proportional to mole fraction, and this in turn to volume fraction). Hence we substitute in Eq.(1-11):
　　　K＝0.031＝[H2CO3]／0.0003
from which
　　　[H2CO3]＝0.031×0.0003＝10^-1.5×10^-3.5＝10^-5M
This seems a very small concentration of acid, but it is sufficient to make natural waters much better weathering agents than they could be without it.』