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Hi everyone! I figured out how to convert CEC into something I understand...namely, how many grams of a cation (positively charged ion like ammonium, calcium, sodium, etc) the substrate can hold. Here is the equation;
Milligrams[mg] of cation substrate can hold = (Mass of substrate/100 grams) * Meqs/100 grams * molar mass of cation
Let us say, for example, that we have a 1 inch (2.53 centimeters) thick substrate of turface MVP (with a density of .593 grams per ml or cubic centimeter and a CEC of 30 meq/100 grams) in a 10 gallon (which has a floorplan of about 50.6 by 25.3 centimeters). To find the mass, multiply the length, width, height and density together like so;
2.53 * 25.3 * 50.6 * .593 = 1,920.64 grams
Divide the mass in grams by 100;
1,920.64/100 = 19.20 (we'll leave the decimals out for simplicity)
Then plug this into the equation above with the molar mass of your desired cation (we'll use ammonium...ammonium has a molar mass of 18.039, but we'll just use 18 here for simplicity) and turface's CEC of about 30 meq/100 grams;
19.2 * 30 * 18 = 10,368 milligrams of ammonium
So, this one inch thick layer of turface mvp in an aquarium could hold about 10,368 milligrams, or 10.368 grams, of ammonium (a milligram is 1/1,000th of a gram), assuming ammonium was the only cation being retained. Ammonium is a relatively small cation; the larger the cation, the more of it the substrate can hold (calcium, for example, has a molar mass of just over 40; this same turface substrate could hold over 23 grams of calcium).
Hope you guys find this useful
Milligrams[mg] of cation substrate can hold = (Mass of substrate/100 grams) * Meqs/100 grams * molar mass of cation
Let us say, for example, that we have a 1 inch (2.53 centimeters) thick substrate of turface MVP (with a density of .593 grams per ml or cubic centimeter and a CEC of 30 meq/100 grams) in a 10 gallon (which has a floorplan of about 50.6 by 25.3 centimeters). To find the mass, multiply the length, width, height and density together like so;
2.53 * 25.3 * 50.6 * .593 = 1,920.64 grams
Divide the mass in grams by 100;
1,920.64/100 = 19.20 (we'll leave the decimals out for simplicity)
Then plug this into the equation above with the molar mass of your desired cation (we'll use ammonium...ammonium has a molar mass of 18.039, but we'll just use 18 here for simplicity) and turface's CEC of about 30 meq/100 grams;
19.2 * 30 * 18 = 10,368 milligrams of ammonium
So, this one inch thick layer of turface mvp in an aquarium could hold about 10,368 milligrams, or 10.368 grams, of ammonium (a milligram is 1/1,000th of a gram), assuming ammonium was the only cation being retained. Ammonium is a relatively small cation; the larger the cation, the more of it the substrate can hold (calcium, for example, has a molar mass of just over 40; this same turface substrate could hold over 23 grams of calcium).
Hope you guys find this useful