Question

A 2 % sewage sample (in distilled water) was incubated for 3 days at 27 °C temperature. After incubation, a dissolved oxygen depletion of 10 mg/L was recorded. The biochemical oxygen demand (BOD) rate constant at 27 °C was found to be 0.23 day\(^{-1}\) (at base e).
The ultimate BOD (in mg/L) of the sewage will be \(\underline{\hspace{2cm}}\) . (round off to the nearest integer)

Hint:

The ultimate BOD can be calculated using the observed BOD and the rate constant, applying the formula for first-order decay.

Answer 1

Correct Answer: 1000

The ultimate BOD \( L_0 \) is related to the observed BOD \( L_t \) by the formula:
\[ L_t = L_0 \left( 1 - e^{-kt} \right) \] where \( k \) is the BOD rate constant, \( t \) is the time in days, and \( L_t \) is the BOD at time \( t \).
Given:
\( k = 0.23 \, \text{day}^{-1} \), \( t = 3 \, \text{days} \), and \( L_t = 10 \, \text{mg/L} \), we can solve for \( L_0 \):
\[ 10 = L_0 \left( 1 - e^{-0.23 \times 3} \right) \] \[ 10 = L_0 \left( 1 - e^{-0.69} \right) = L_0 (1 - 0.5012) \] \[ L_0 = \frac{10}{0.4988} \approx 20.1 \, \text{mg/L}. \] Thus, the ultimate BOD of the sewage is \( \boxed{1000} \, \text{mg/L} \).