WE&T Magazine - Water Environment Federation

BROWSE THIS ISSUE

Resources

Certification Quiz

Test your knowledge of biochemical oxygen demand

True or False Questions:

Nitrogenous demand is considered an interferent in the determination of 5-day biochemical oxygen demand (BOD).
If nitrification is inhibited, results are reported as CBOD.
Any final dissolved oxygen (DO) reading that is less than 1 mg/L may not be used in the BOD calculation.
Any final DO reading that does not yield a total depletion of 2 mg/L is fine to use to calculate BOD.

Multiple Choice Questions:

What are the components of a GGA standard solution?

A. 198 mg of glucose and 198 mg of glutamic acid per 1 L of water
B. 150 mg of glucose and 150 mg of glutamic acid per 1 L of water
C. 200 mg of glucose and 200 mg of glutamic acid per 1 L of water
D. 99 mg of glucose and 99 mg of glutamic acid per 1 L of water

What is the true value of the GGA standard?

A. 198 mg/L
B. 300 mg/L
C. 150 mg/L
D. 500 mg/L

Which of the following is the proper range for seed correction factor (SCF)?

A. 0.1 – 1.1 mg/L
B. 1.0 – 2.0 mg/L
C. 0.5 – 1.5 mg/L
D. 0.6 – 1.0 mg/L

All samples for BOD are incubated for 5 days at what temperature?

A. 20°C ± 1°C
B. 25°C ± 1°C
C. 20°C ± 5°C
D. 25°C ± 5°C

Which of the following pH ranges should samples be within prior to analysis?

A. 6.0 to 8.0
B. 7.0 to 9.0
C. 6.5 to 7.5
D. 5.0 to 7.0

Using the following data, calculate the seed correction factor.

Initial DO concentration	9.00 mg/L
Final DO concentration	5.25 mg/L
Volume of seed used in control bottle	20 mL
Volume of seed used in samples and GGA	4 mL
Volume of bottle used	300 mL

A. 0.75
B. 1.50
C. 0.05
D. 0.60

Using the following data, calculate the GGA standard result.

Initial DO concentration	9.00 mg/L
Final DO concentration	4.50 mg/L
Seed correction factor	0.6
Volume of seed added to GGA	4 mL
Volume of GGA added to 300-mL bottle	6 mL

A. 198 mg/L
B. 195 mg/L
C. 293 mg/L
D. 150 mg/L

Using the following data, calculate the 5-day BOD of a sample.

Initial DO concentration	9.00 mg/L
Final DO concentration	4.00 mg/L
Seed correction factor	0.6
Volume of seed added to sample	4 mL
Volume of sample added to 300-mL bottle	100 mL

A. 1.47 mg/L
B. 25.3 mg/L
C. 13.2 mg/L
D. 15.2 mg/L

Questions developed by Joe Boyd, technical specialist at Environmental Express (Charleston, S.C.), and reviewed by the Operations forum Editorial Advisory Board and the Association of Boards of Certification (Ankeny, Iowa) Wastewater Treatment Validation and Examination Committee.

Answer Key:
Click here to show or hide the answer key

True. If an inhibitory chemical is used, results are reported as CBOD. If not, the BOD is the sum of carbonaceous and nitrogenous demands.
True. CBOD stands for carbonaceous BOD.
True. According to Standard Methods for the Examination of Water and Wastewater, final DO readings must be at least 1 mg/L. However, in practice, sometimes terms, such as less than or greater than, are used to qualify BOD results that do not meet the criteria. For example, a lab analyzes several dilutions of an influent sample and all the results over-deplete — that is, leave less than 1 mg/L DO. Technically speaking, the analyst must report a non-result for that sample because it doesn’t meet quality control requirements; in reality, a lab will report the lowest dilution result with a qualifier: “greater than 300 mg/L.”
False. According to Standard Methods for the Examination of Water and Wastewater, a DO depletion of at least 2.0 mg/L is required. However, industry practice is to use terms such as less than or greater than to qualify BOD results that do not meet the criteria. For example, if a lab analyzes several dilutions of an effluent sample, including a 100% volume sample and all the results under-deplete — that is, do not deplete by at least 2 mg/L DO — technically speaking, the analyst must report a non-result for this sample because it doesn’t meet quality control requirements. In reality, a lab will report the highest dilution result with a qualifier: “less than 2 mg/L.”
b.
a.
d.
a.
c.
a. Seed correction factor = (initial DO – final DO) × (bottle volume/control seed volume) × (sample seed volume/bottle volume). So, (9.00 mg/L – 5.25 mg/L) × (300 mL/20 mL) × (4 mL/300 mL) = 0.75.
b. GGA result = (initial DO – final DO – seed correction factor) × (bottle volume/GGA volume). So, (9.00 mg/L – 4.50 mg/L – 0.6) × (300 mL/6mL) = 195 mg/L.
BOD = (initial DO – final DO – seed correction factor) ÷ (sample volume/bottle volume). So, (9.00 mg/L – 4.00 mg/L – 0.6) ÷ (100 mL/300 mL) = 13.2 mg/L.

References:

ARCHIVES

Certification Quiz

Test your knowledge of biochemical oxygen demand