Hello everyone
I'm working on treatment of a synthetic wastewater and I want to set the COD of the wastewater on 500 mg/lit. Does anyone know that how much Ethanol should I add to 50cc wastewater to reach the desired COD ?
One gram of ethanol has a theoretical Oxygen Demand of 192/180 gm of Oxygen or about 1.05 gm/ gm.
2.09 g COD/g Ethanol
Ethanol (C2H5OH) has a degree of reduction of 12 (mol electrons/mol ethanol).
Oxygen has a degree of reduction of -4 (mol electrons/mol oxygen) and a molecular weight of 32 g/mol, which corresponds to 8 g oxygen (=COD)/mol electrons. Multiplying 12 by 8, this results in 96 g COD/mol ethanol, or, 96/46 = 2,09 g COD/g ethanol.
You may get help through this page, also for the future needs of calculations.
https://mimoza.marmara.edu.tr/~kyapsakli/enve202/Lecture5_Chemical%20Oxygen%20Demand.pdf
I really appreciate your answers, is the ThOD (Theritical Oxygen Demand) of Ethanol equal to it's COD (Chemical Oxygen Demand) ? does anyone have any experience about it ?
Dear Sajjad,
ThOD will always be slightly more than COD (In reality, compounds are not 100 % oxidized).
You can do few trials by adding varying amount of Ethanol and checking actual COD. You will certainly get the relationship between ThOD and COD. The same can be used for your future experiments.
Brother, add sucrose. 1 gm sucrose is generally equivalent to 980-1000 mg/L of total COD. Meanwhile, you can also prepare your own recipe by adding chemical available in your lab. Ethanol is a potencial inhibitor to the microorganisms. Therefore, if you are going to do biological (activated sludge) or bio-chemical treatment, do not use ethanol for synthetic wastewater preparation. But, if you are trying to prepare pharmaceutical wastewater than adding ethanol would be benficial. For other types of wastewater, add sucrose, glucose or fructose.
2.09 g COD/g Ethanol is the correct answer as stated by Shahad Abdulkareem
The rest is basic math.
C2H5OH + 3 O2 -> 2 CO2 + 3 H2O
You need 3 Molecules of Oxygen (3 mol = 3 * 32,00 g = 96,0 g) in order to oxidise 1 Molecule of ethanol (1 mol = 2*12,01+16+6*1,01 = 46,1 g), thus the COD is 96/46,1 = 2,08 g O2/g Ethanol.
But if you analyse it and you don't use the official dichromate method, it will maybe not fully oxidised and the measured value will become much too low. I made this experience years ago with a COD analyser which worked with electrochemical oxidising. This analyser detected most COD but failed with ethanol.
Currently, you have high performance COD meters that give you direct and reliable measurement
I always keep this old printout of a WE&T Got Carbon? handy when designing reactors and biological treatment needing carbon dosing.
According to this source, roughly 1.65 g COD/mL 100% EtoH.
Hope this helps
Ethanol is an organic chemical compound. Measurement of COD is depend on the method (Open reflux or closed reflux). If COD is greater then 40 mg/L, the dilution of sample is necessary.
Sajjad Ashtari Larki yes the COD of the ethanol can easily be calculated or given as the degree of reduction of ethanol (C2H5OH) is 12 (mol electrons/mol ethanol).... When you divide 12 by 8, you get 96 g COD/mol ethanol, or 96/46 = 2,09 g COD/g ethanol.
Ethanol is an organic chemical compound. Measurement of COD is depend on the method
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