BOD-AIMAN

1.0 INTRODUCTION

Biochemical oxygen demand (BOD) is used to measure the amount of dissolved

oxygen needed by microorganism (ex: aerobic biological organisms) in order to

break down organic matter present in water sample at certain temperature over a

specific time period.

Organic waste normally is decomposed by microorganism. Exist of the organic

matter will caused the aerobic bacteria decomposed the waste by consuming the

dissolved oxygen in water. Therefore, the demand of oxygen will be increased so

that the BOD level also will be increased. The higher the BOD level, the lower the

dissolved oxygen (DO) level in water. This is because the available oxygen in the

water is being used by bacteria to decompose the waste. As a result, BOD test

commonly used as the indicator to show the cleanliness of the waste water. The

higher BOD value is an indication of poor water quality.

Normally, when carried out the BOD test, the water sample is incubation at 20°C

over a period of 5 days and it is called as BOD5. However, in this experiment BOD3

is used instead of BOD5. Thus, the water sample is incubation at 30°C over a period

of 3 days. By doing so, the time used to conduct the experiment can be shorten.

2.0 OBJECTIVE

To measure the strength of the water sample ( water, wastewater, etc) based on the

amount of oxygen needed to stabilize the organic matter in the sample.

3.0 LEARNING OUTCOME

3.1 Student able to describe the importance of BOD in the environmental studies.

3.2 Students able to measure the BOD of samples with the right sample size.

4.0 THEORY

Biochemical Oxygen Demand is a common, environmental procedure for

determining the extent to which oxygen within a sample can support microbial life.

This method is popular in many environmental laboratories analyzing waste water,

compost, sludge, and soil samples.

BOD directly affects the amount of DO in water bodies. The greater the BOD, the

more rapidly oxygen is depleted in the water body, leaving less oxygen available to

higher forms of aquatic life. The following formula can be used to calculate the

BOD,

BOD=(DO¿¿initial−DO final)×300

sample¿¿¿¿

Once a general range for the BOD of a sample has been determined, the dilution

can be established which will ensure that at least once dilution will meet the criteria

for valid BOD results. The following formulas are used to calculate the minimum

and maximum estimated dilution:

Min. mL sample added to BOD bottle =

min allowable depletion ×Volumeof BOD bottleEstimated BOD

Max. mL sample added to BOD bottle =

maz allowable depletion×V olime of BOD bottleEstimated BOD

When a measurement is made of all oxygen consuming materials in a sample, the

result is termed “Total Biochemical Oxygen Demand” ( TBOD ) , or often just

simply “Biochemical Oxygen Demand” (BOD). Because the test is performed over

a five day period, it is often referred to as a “five Day BOD”, or a BOD5.

In addition, this procedure is only suitable for samples void of serious matrix

interferences. To gain a broader appreciation of oxygen demand, additional avenues

of interest may be explored including CBOD (carbonaceous oxygen demand), COD

(chemical oxygen demand), and TOC (total organic carbon).

Because of complications measuring this ultimate BOD (BODu), BODu is usually

extrapolated from laboratory 5-day BOD bottle tests

BODt = BODu ( 1 – e-kt ).

In many biological treatment plants, the facility effluent large numbers of nitrifying

organisms which are developed during the treatment process. These organisms can

exert an oxygen demand as they convert nitrogenous compounds (ammonia and

organic nitrogen) to more stable forms (nitrites and nitrates). At least part of this

oxygen demand is normally measured in a five day BOD.

Sometimes it is advantageous to measure just the oxygen demand exerted by

organic (carbonaceous) compounds, excluding the oxygen demand exeried by the

nitrogenous compounds. To accomplish this, the nitrifying organisms can be

inhibited from using oxygen by the addition of a nitrification inhibitor to the

samples.The result is termed “Carbonaceous Biochemical Oxygen Demand “ or

BOD.

5.0 EQUIPMENTS AND REAGENTS

5.1 3 units 300 mL BOD Bottles

5.2 Incubator , capable of maintaining 30 ± 1 °C

5.3 2 units 100 mL beaker

5.4 100 mL graduated cylinder

5.5 2 units 25 ml meaning pipettes

5.6 DO meter

5.7 pH meter

5.8 Phosphate Buffer

Dissolve 8.5g KH2PO4, 21.7g K2HPO4, 33.4g Na2HPO4, and 1.7g NH4Cl in

deionized water. Adjust pH to 7.2, if necessary, with either 1 N H2SO4 or

NaOH. Dilute to one liter.

5.9 Magnesium Sulfate

Dissolve 22.5g MgSO4.7H2O and dilute to one liter.

5.10 Calsium Chloride

Dissolve 27.5g CaCl2 and dilute to one liter.

5.11 Ferric Choloride

Dissolve 0.25g FeCl3.6H2O and dilute to one liter.

6.0 PROCEDURE

1. The waste water was collected from specified source.

2. Three units of 300ml BOD bottles were labelled separately; one BOD bottle for

blank of dilution water, another two BOD bottles for minimum and maximum

volume of water sample respectively.

3. The minimum and maximum volumes of water samples were calculated by

using formula.

4. The pH value and temperature of the water sample were determined. The

sample must be neutralize before test is performed.

5. Each BOD bottle was first filled with the proper volume of sample size by using

pipettes and then followed by dilution water that completely filled up the 300 ml

bottle. For the blank BOD bottle, the bottle only filled with dilution water.

6. The initial DO for each bottle was determined by DO meter and the data was

recorded on the table.

7. After that, the bottles stoppers were inserted to the bottles and small amount of

dilution water was pour on top of the stopper.

8. The bottles were placed in the incubator at 30°C and incubated for 5 days.

9. The final DO was recorded after 5 days of the experiment.

10. Finally, the BOD for each dilution was calculated.

7.0 RESULT AND CALCULATION

BIOCHEMICAL OXYGEN DEMAND ( BOD )

Data : 29 March 2015 Analyst: Group5 members

Time: 2.00 pm – 4.00 pm

Sample Details:

Waste water from water plant behind Kolej Kediaman Tun Fatimah was used as

sample for this BOD3 test. No pretreatment is done as the water sample is obeys the

criteria and exists as a neutral water sample.

pH 6.41

Temperature, ºC 22.35°C

Sample

Type

Sample

ID

Volume

Sample

(mL)

Dilution

Factor

Initial

DO

(mg/L)

Final

DO

(mg/L)

DO

Depletion

(mg/L)

BOD

(mg/L)

BOD3 Blank 0 0 8.30 7.29 1.01 -

BOD3 1 4 0.01 8.33 6.30 2.03 203.00

BOD3 2 10 0.03 8.29 5.87 2.42 80.67

Average BOD3 = 141.83 mg/L

8.0 DATA ANALYSIS

8.1 Calculate for the average BOD.

Average BOD=Sample1+Sample 22

¿203+80.67

2

= 141.83 mg/L

8.2 Show all the calculation and state if any of the data needs to thrown out.

Sample Blank

Dilution factor , P=Volume of sample ,V s

Total volume , V T

¿ 0300

= 0

DO Depletion = DOi - DOf

= 8.30 – 7.29

= 1.01 mg/L

BOD = None

Sample 1

Dilution factor , P=Volume of sample ,V s

Total volume , V T

¿ 4300

= 0.01

DO Depletion = DOi - DOf

= 8.33 – 6.30

= 2.03 mg/L

BOD= DO i−DO fDilution factor , P

= 2.030.01

= 203..00 mg/L

Sample 2

Dilution factor , P=Volume of sample ,V s

Total volume , V T

¿ 10300

= 0.03

DO Depletion = DOi - DOf

= 8.29 – 5.87

= 2.42 mg/L

BOD= DO i−DO fDilution factor , P

= 2.420.03

= 80.67 mg/L

8.3 The dillution water blank cannot deplete more than 0.2 mg/L. Was this

criteria met?

From the above calculation of the dilution water blank, it is shown that the

criterion is not valid. This is because the DO depletion of dilution water blank

obtained from the experiment is 1.01 mg/L. This value is more than 0.2 mg/L

may due to poor quality control of dilution water when preparing the dilution

water. If the dilution water is kept in poor environment this also will caused

growth of microorganism in the dilution water. In the end, the DO depletion

value sure will deviate from 0.2 mg/L.

8.4 Does you sample shows‘a toxic effect’?

Toxicity in BOD testing means the characteristic of a sample that causes it to

interfere with biochemical oxidation of organic materials during incubation.

Toxic materials present in the sample might spoil the biochemical process.

Thus, proper test to identify whether the waste water sample contains toxicity

is needed in order to take proper action to eliminate those toxicants.

According to Standard Methods 5210B, if the average of all BOD bottles that

meet the criteria of 2.0 mg/L or more of dissolved oxygen (DO) depleted with

at least 1 mg/L DO retained, toxicity will not be occurred. Since the sample

(Sample 1 + Sample 2) has average value of DO depleted more than 2, thus

the toxic effect does not occur.

8.5 Could you rely on your BOD results? Why?

The BOD result in this experiment cannot rely on because of some errors were

occurred. Although the sample is free from toxicity, the dilution water blank

does not meet the requirement. The dilution water was prepared and stored for

some times without proper quality control before the experiment. This caused

growth of some biological that cannot be seen by naked eyes. Existence of

biological in the dilution water caused inaccuracy of the result. Moreover, the

waste water used in this experiment does not undergo pretreatment process. It

may contain dissolved heavy metal. The presence of heavy metal make the

result obtained may be different.

8.6 By referring to Sewage/Effluent Standard (DOE, Malaysia ) could your

sample be discharge to river untreated? If not, suggest the associated

treatment for BOD removal.

According to Third Schedule of Environmental Quality Act, 1974 under the

Environmental Quality (Sewerage and Industrial Effluent) Regulation, 1979,

regulation 8(1), 8(2) and 8(3), the parameter limits of effluent of Standard A

and B are as follow:

Parameter Unit Standard A Standard B

(i) Temperature °C 40 40

(ii) pH Value - 6.0 - 9.0 5.5 - 9.0

(iii) BOD5 at 20°C

(Almost the same with BOD3)mg/L 20 50

(iv) COD mg/L 50 100

Although the temperature = 22.35°C and pH value = 6.41 of the waste water

are satisfy the requirement, however, the average BOD for the waste water is

141.43 mg/L and it is exceed the BOD requirement for both Standard A and

Standard B. Thus, the waste water sample cannot be discharged to river

untreated. The process which can used to treat waste water is aeration.

Aeration is the process by which air is circulated through, mixed with or

dissolved in a liquid or substance. Normally, this treatment used in the

secondary treatment of wastewater through aerating mixers. By doing so, high

dissolved oxygen content of waste water will be produced. In the end, BOD

level of wastewater will reduced.

9.0 DISCUSSION

Q1. a) Calculate the min / max volume of sample to be added to prepare as

estimated BOD 150mg/L.

Estimated BOD=(min .¿max . allowabledepletion x total volumesample water )

Suppose that estimated BOD of an influent sample is 150 mg/L.

Water sample ( min ) = (2 mg/L X 300 mL) / 150 mg/L

= 4 m/L

Water sample ( max ) = (5 mg/L X 300 mL) / 150 mg/L

= 10 m/L

b) Why must samples containing cautic alkalinity or acidity be adjusted

before preparing BOD dilution?

This is because if the sample too alkalinity or acidic will prevent the growth

of bacteria. Thus, in order to conduct BOD experiment, the sample must be

neutral which is in the range of 6.0 to 8.0. In this pH range, the microbes are

most active.

c) The completed composite sample arrive in the lab at eg:,___) What is the

latest day and time the sample can be started for BOD.

The sample must be tested in time less than 24 hours. If it is begin within 2

hours of collection, cold storage is unnecessary. However, if the test will

begin within more than 2 hours, the sample must keep at or below 4°C.

d) Describe the function of BOD bottle’s cap and seal water.

BOD bottle’s cap and seal water are used to ensure that no air bubble is

trapped in the bottle that will affect the characteristics of sample. Moreover,

it is also to prevent further oxygen dissolving in the sample.

Q2. a) Why must samples containing residual chlorine be dechlorinated before

preparation of BOD dilutions?

This is because chlorine can kill the microbes and bacteria in samples. If the

sample is destroyed, the result of BOD test will be affected.

b) What reagents are required to chemically dechlorinate a BOD sample?

The reagents are sulphuric acid solution, acetic acid solution, potassium

iodide solution or sodium sulfite ( Na2SO3 ).

c) What must be done to samples which have been dechlorinated or adjusted

for pH variations?

The sample must be seeded. In the seeding process, the healthy active

organism is added. Lastly, a seed correction is used in the BOD calculation.

Q3. a) State the formula to calculate

i. Seed correction

Seed correction = BOD of seed material x volume of seed material

300

ii. BOD5 (seeded)

BOD5 (mg/L) = ( D1−D2 )−(B 1−B 2) f

P

where;

D1 = DO of diluted sample immediately after preparation

D2 = DO of diluted sample after 5 days incubation at 20°C

B1 = initial seed DO

B2 = final seed DO

f = seed fraction

P = decimal volumetric fraction of sample used

b) Calculate the seed correction and BOD5 (seeded) for the data given as

below

   BOD5 of Seed Material   95 mg/L

  Dilution #1 mL of seed material 2 mL

mL of sample 100 mL

Start D.O. 7.8 mg/L

Final D.O. 2.9 mg/L

Seed correction = (95 x 2 ) / 300

= 0.633 mg/L

BOD5 (seeded) = [(7.8 – 2.9) - 0.633] x 300 / 100

= 12.801 mg/L

Q4. a) 30 mL of wastewater are placed in a 300 mL BOD bottle. The sample is

diluted to fill the bottle. The DO concentrations at the beginning and the

end of 5-day incubation period are 7.3 mg/L and 1.8 mg/L respectively.

What is the BOD?

BOD5=DO i−DO f

Dilution factor ,P

¿(7.3−1.8 ) x300

30

= 55 mg/L

b) The BOD5 of a wastewater was determined to be 250 mg/L. If the reaction

coefficient was 0.23 l/d, calculate

i) ultimate BOD,

BODt =Lo(1−e−kt)

Lo=250

1−e−(0.23)(5 )

Lo = 365.84 mg/L

ii) BOD3

BODt =Lo (1−e−kt )

BOD3 = 365.84 ( 1- e−(0.23)(3))

= 182.34 mg/L

iii) BOD remaining at 3 days

BOD remaining at 3 days = BODu – BOD3

= 365.84 – 182.34

= 183.50 mg/L

Q5. a) What is seeding process in BOD measurement?

Seeding is a process of adding live bacteria and microorganism to a sample.

If the samples tested contain materials which could kill or injure the

microorganisms, the condition must be corrected and healthy active

organisms added.

b) Explain preparation of seed material.

Select a material to be used for seeding which will have a BOD of at

least 180 mg/L. This will help ensure that the seed correction meets the

0.6 mg/L minimum specified in ³StandardMethods´.

Place the material in a suitable container and incubate at 20°C for 3

days. Usually, settle draw domestic sewage prepared in the manner

above will have sufficient BOD for use as a seed material.

As an alternative, commercially available seed material may be used.

The seed correction should not exceed 1.0 mg/L BOD; therefore, care

should be taken not to use too strong a seed material for the test.

Make sure that seed material is a relatively stable which produces a good

seed correction in every test situation.

c) What materials can be used to seed a BOD sample?

Settled raw sewage or commercially prepared seed material are the most

common sources. However, any source of water which can provide a

suitable population of organisms can be used.

Q6. What is the significance of dissolved oxygen?

Dissolved oxygen allows animals to breathe in water and it provides a suitable

habitat for the other animals. Bacteria in the water also use this oxygen to

break down animals and plants. If the oxygen level is reduced, the animals

begin to die. It is also to maintaining the aquatic life and aesthetic quality of

streams and lakes.

Q7. a) With regard to precision, ten percent duplicate or replicate samples should

be run. This would result in one duplicate sample or one replicate sample

being run every ten samples. Differentiate replicate and duplicate sample.

Replicates are two or more separate water samples collected in the field

from the same site and depth. It is used to determine the errors involved

in sample collection. If there are no errors in the collection and analysis,

and then the difference between two replicate analysis indicates the

natural variability in the water at that location.

Duplicates are two or more lab analyses on the same water sample. It is

used to determine the percentage difference between two samples in

order to estimate the error involved in the analysis.

b) When are DO levels at their highest and their lowest and why?

DO level is highest if the water is free from polluted materials like toxic

and bacteria. When the numbers of bacteria is decreased, the oxygen

demand also decreased and resulted in highest DO level. On the other hand,

water will have lowest DO level if water is highly polluted.

Microorganisms in water will used up the oxygen so will cause DO level

decreases. Moreover, DO level is often highest on the daytime as aquatic

plant will undergoes photosynthesis to produce oxygen. The DO level will

decreases during night as photosynthesis cannot occur without sunlight.

10.0 CONCLUSION

In conclusion, the objective of the experiment is achieved. The experiment was

carried out to measure the strength of the waste water sample based on the amount

of oxygen needed to stabilize the organic matter in the sample.

Normally, BOD5 will be used to determine the BOD level in water. However, in

this experiment, BOD3 is used instead of BOD5. By using BOD3, the time of

experiment can be shorten and the effect of experiment almost the same with BOD5.

Since the average BOD3 obtained from experiment is 141.83 mg/L so that it can be

concluded that the waste water from water plant behind Kolej Kediaman Tun

Fatimah has been polluted. The higher BOD3 reading will result in lower DO level.

When the DO level is lower, this indicated that the water is unsuitable for living

organisms. Besides, this waste water cannot simply discharge to the river untreated.

It must be treating first by any possible treatment to make sure that water is safe and

will not to be polluted to the river.