This is from an old thread (http
/www.fishforums.net/content/forum/72100/Does-Carbon-Leak-toxins-back-into-water-/):
The chemicals that get adsorbed onto the actived carbon are in equilibrium with the tank water. The activated carbon will take up chemcials until that equilibrium is reached. What that equilibrium point is depends upon the concentration of the pollutant and how used up the carbon is already -- what percentage of the active spots on the carbon are still available.
Normally, when using carbon to take out medications for example, it will take up an overwhelming percentage of the chemicals. Say from 100% to 0.1%. What I mean is that 0.1% is the equilibrium. Now, you do a 25% water change. So, concentration of chemicals is now 0.075% in your tank. If you do not change the carbon at this time, the carbon will release some of the chemicals, again going towards equilibrium. Since, most of the chemicals are on the carbon now (99.9% of the original amount) given enough time (and this depends on temp, pH, etc.) the sysytem will tend toward equilibrium again. Meaning the carbon realeases 0.025% back into your water, to bring the concentration of chemcials back to 0.1% -- equilibrium.
Not a huge amount, but you see how a little bit can be released back. And long-term exposure, even at very low concetrations can lead to ill health.
Here is even a little more:
Lets call c the concentration of any chemical in the water that will be adsorbed.
If the adsorbtion at equilibrium follows the Langmuir isotherm (probably the most common -- there are others) the concentration of the chemical on the carbon, which I will call n is equal to:
n = c/(K+c)
K is the called the adsorption or equilibrium constant. It will vary depending upon what chemical we are talking about, and what quality the carbon is.
The total concentration of chemical is T
T=n+c
or using the definition of the adsorbed amount
T = c/(K+c) + c
Now, if we lower c a little (by doing a partial water change, for example) T does get lowered to T2, but so does n -- so call the new n2.
But again, T2=n2+c2 where n2=c2/(K+c2)
where does the difference adsorbed (n2-n) go? back into the water.
The carbon will uptake and release chemicals until equilibrium is reached, every time. So, yes, it is usually a small amount that will get released back -- carbon is very efficient uptaking chemicals when active sites are available -- but a small amount does get released.