Al, I notice some are asking "how do you know that the ppm of CO2 in the bulb is the same as in the water?" The answer is: If there is a sealed quantity of air in contact with water which has CO2 in it, remembering that CO2 goes into and out of solution with water very easily, the air will soon reach an equilibrium with the water, with as much CO2 entering the air as is leaving it. Now, if that air is in contact with a sealed bit of water, that water and the air will also reach an equilibrium, with as much CO2 entering the water as is leaving it. No matter what the actual amount of CO2 in the air gap is, the equilibrium will be such that both batches of water will reach the same concentration of CO2. I tested this and reported the test on APC, at:
DIY Drop Checker - Page 5 - Aquatic Plant Central- aquascaping...a living art. You might want to add this explanation to your discussion.
Edit: You can do a mind experiment to prove this too: Assume a large body of water "A", with CO2 concentration C1, connected to an air passage "B", with CO2 concentration C2 in equilibrium with C1, connected to another small body of water "D", with CO2 concentration C3. Now assume that C3 is less than C1, therefore not in equilibrium with C2. That means more CO2 molecules will move from B to D than from D to B, which raises C3 until C3 is in equilibrium with C2. But, at equilibrium, the concentration in air is C2, so the concentration that is at equilibrium with C2 must be C1, since we already said that concentration was in equilibrium with C2. So, C1 and C3 must be the same. You can use the same logic for the case where you assume C3 is greater than C1.