solve for BOD5 of river water

A sample of 200mL of water was collected from a river. From this sample, 5mL was diluted to 1L. A BOD bottle (which is 300mL ) was filled with this diluted water and aerated. The dissolved oxygen content of the diluted sample was 7.8mg/L initially. After 5 days, the dissolved oxygen content had dropped to 4.8mg/L. What is BOD5 of the river water? (5.2), the oxygen consumed in the first five days is BOD5=PDOiDOf=0.0309.03.0=200mg/L total amount of oxygen needed to decompose the carbona tion of the waste can be found by rearranging (5.11): L0=(1ekt)BODs=(1e0.225)200=300mgL five days, 200mg/L of oxygen demand out of the total 30 ld have already been used. The remaining oxygen demand woul be (300200)mg/L=100mg/L. Reaction Rate Constant k eaction rate constant k is a factor that indicates the rate of bi tes. As k increases, the rate at which dissolved oxygen is used ultimate amount required, L0, does not change. The reactio number of factors, including the nature of the waste itself (fo rs and starches degrade easily while cellulose does not), th microorganisms to degrade the wastes in question (it may healthy population of organisms to be able to thrive on th estion), and the temperature (as temperatures increase, so d lation). ypical values of the BOD reaction rate constant, at 20C, otice that raw sewage has a higher rate constant than either lluted river water. This is because raw sewage contains a la y degradable organics that exert their oxygen demand ainder that decays more slowly. e of biodegradation of wastes increases with increasing ter hese changes, the reaction rate constant k is often modi uation: kT=k20(T20)