Thread: Problem regarding COD determination

Hello,I have a problem regarding COD determination using titration. Fyi my sample (palm oil mill effluent) volume is equal to 2.5 ml,potassium digestion, 1.5ml and H2SO4 reagent, 3.5 ml. Same as blank but just substitute the sample with distilled water. Molarity of potassium digestion is equal to 0.01667 M. For titration with 0.1 M FAS,only 2 ml of FAS required for blank while for sample i cannot proceed titrating since after I add ferroin indicator into the sample solution it instantaneously turn into red. As far as I know, the solution has to be blue - green in color and change into reddish-brown after titration with FAS solution. Why is that so? can anyone explain it to me? How to solve this problem? Do i have to manipulate the volume of each reagent and sample by increase the amount of potassium digestion solution or decrease the amount of sample? THANKS

The indicator is turning red because your sample solution is already reduced, which means you either have to add less of your sample, or add more of your oxidizing agent, because not all of your sample is being oxidized. I would ask a teacher (are you doing this for school?) for which is more appropriate, I don't have first hand experience with this experiment.

Thanks for the reply. I would really appreciate if u can ask your teacher for an accurate answer. I ‘m not sure if we can change the amount of the oxidizing agent and sample cause what I got from internet and books , mostly the amount of the reagent and sample are same as above. If I have to increase the amount of oxidizing agent is it okay if the amount of both oxidizing agent and H2SO4 are the same? But today I did prepare 6 dilutions for each sample as below

Sample+dH2O
2.0 + 0.5
1.5 + 1.0
1.0 + 1.5
0.5 + 2.0
0.2 + 2.3
0.1 + 2.4

Only last and the second last can be proceed with titration. Sadly the results obtained after calculated using equation gave huge number of COD (after multiply with dilution factor) which is 5700 mg/l (usually from the previous study for my sample is around 2500 mg/ l). I really need your teacher’s answers right now. ..Thanks again