Answer to Question #94510 in Inorganic Chemistry for mayuresh

Question

1. A 50 mL sample of water contains 500 ppm of dissolved oxygen (DO). The water sample is diluted to 100 mL. After 5days of incubation the DO value of the water sample reduces to 400 ppm. Calculate the BOD of the water sample.
\

Answer 1

BOD₅ or Biological Oxygen Demand is a value of oxygen consumed by microorganisms in 5 days and its calculation follows the expression:

$BOD_5=\frac{D_0-D_5}{P}$

D₀ and D₅ are the concentrations of diluted oxygen before and after incubation in mg/L or nearly the same ppm and P is a dilution factor or $\frac{V_{final}}{V_{initial}}$

$P=\frac{100mL}{50mL}=2$

$BOD_5=\frac{500mg/L-400mg/L}{2}=50mg/L$

PS: ppm is actually not correct unit, which is not accepted by SI system. It should be at least expressed as ppmw or ppmv.