1.0 INTRODUCTION

Microorganisms such as bacteria are responsible for decomposing organic waste.

When organic matter such as dead plants, leaves, grass clippings, manure, sewage, or even

food waste is present in a water supply, the bacteria will begin the process of breaking down

this waste. When this happens, much of the available dissolved oxygen is consumed by

aerobic bacteria, robbing other aquatic organisms of the oxygen they need to live.

Biochemical oxygen demand or BOD is a procedure to determine the amount of oxygen

consumed by the microorganisms in the waste water chemically. It is commonly used as the

indicator to show the cleanliness of the waste water.

2.0 OBJECTIVE

To measure the amount of dissolved oxygen in the water for a specified period of time

and temperature.

1

3.0 APPARATUS AND MATERIAL

1. 5 units 300 ml BOD Bottles

The ideal bottle for incubating diluted sample of sewage, sewage effluents,

polluted water and industrial wastes to determine the amount of oxygen.

Figure 3.1

2. Incubator, capable of maintaining 20 ± 1 °C

To maintains the optimal temperature, humidity and other conditions.

Figure 3.2

3. 100 ml beaker

The simple container for stirring, mixing, and heating the liquid.

Figure 3.3

2

4. 100 ml graduated cylinder

For measuring the volumes (amounts) of liquids.

Figure 3.4

5. 2 units 25 ml meaning pipettes.

For transferring or measuring out small quantities of liquid

Figure 3.5

6. DO meter

To measure the amount of dissolve oxygen in the liquid.

Figure 3.6

3

7. pH meter

to measure the hydrogen-ion concentration or pH value in a solution,

indicating its acidity or alkalinity.

Figure 3.7

8. Phosphate Buffer

Dissolve 8.5 g KH2PO4, 21.7 g K2HPO4, 33.4g Na2HPO4, and 1.7g NH4Clin

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

NaOH. Dilute to one liter.

9. Magnesium Sulfate

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

10. Calcium Chloride

Dissolve 27.5g CaCl2 and dilute to one liter.

11. Ferric Choloride

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

4

4.0 PROCEDURE

1. 500mL of waste water collected from Tasik Kemajuan (Tasik G3) using plastic

containers.

2. The waste water was then tested by using pH meter. It was tested to be neutral. The

reading for the both pH and temperature were recorded in the Bench Sheet.

3. The sample was divided by using the meaning pipette. 2mL of sample poured into the

first, second and the third BOD Bottle. Then, BOD Bottle filled up with dilution water

with 300mL. The forth BOD Bottle filled up with dilution water and labeled as

‘Blank’.

4. The sample in the bottle is stirred with DO meter and placed into the bottle to get the

DO reading. 1 – 4 steps repeated for the second and the third BOD Bottle.

Figure 4.1

5. All the three BOD Bottle reading recorded in the Bench Sheet as initial DO.

6. The BOD Bottle placed into BOD incubator with temperature around 20 °C.

Figure 4.2

7. The final DO taken after 5 days of the experiment for three bottles.

5

5.0 DATA RESULT

BIOCHEMICAL OXYGEN DEMAND ( BOD)

Date:02 / 03 / 16

Time : 11 A.M

Analyst:

Sample Details:

Source: Tasik Kemajuan (Tasik G3), UTHM

7.1pH 250CPretreatment:

Alkalinity/Acidity

Sample Volume: 500 mL

I N NaoH : 0 mLI N N2SO4: 0 mL

Comments:

Neutral. This is because the sample water pH is 7.1 which is

considered as neutral. (6.5 – 7.5) = neutral

Sample Type

Sample ID

Volume Sample( mL)

DilutionFactor

Initial DO

(mg/L)

Final DO

( mg/L)

DO Depletion( mg/L)

BOD(mg/L)

BOD1

Blank 2 0.83/207.5 = 0.004

4.76 3.93 4.76 – 3.93 = 0.83

(0.83 x 500) = 207.5 2

BOD2Blank 2 0.004 4.98 4.25 0.73 182.5

BOD3Blank 2 0.004 5.56 5.10 0.46 115

Average BOD( show the calculation)

Cancelled Data/ Result:

BOD1 = 207.5/3 = 69.17 mg/L

BOD2= 182.5/3 = 60.83 mg/L

BOD3= 115/3 = 38.33 mg/L

6

6.0 DATA ANALYSIS

1. Calculate for the average BOD

( 207.5 + 182.5 + 115 ) = 168.33 mg/L

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

BOD1, DO depletion = DOi – DOf = 4.76 – 3.93 = 0.83 mg/LBOD5 = (0.83 x 500)/2 = 207.5 mg/L

BOD2, DO depletion = DOi – DOf = 4.98 – 4.25 = 0.73 mg/LBOD5 = (0.73 x 500)/2 = 182.5 mg/L

BOD3, DO depletion = DOi – DOf = 5.56 – 5.10 = 0.46 mg/LBOD5 = (0.46 x 500)/2 = 115 mg/L

3. The dillution water blank can not deplete more than 0.2 mg/L. Was this criteria met?

From the calculation, it is shown that the criterion is not valid. This is because

the DO depletion of dilution water blank obtained from the experiment is

more than 0.2mg/L. This 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.

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 less than 2,

thus the toxic effect is occur.

7

5. Could you rely on your BOD results? Why?

No. Because some errors were occurred. Although the sample is occur 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.

6. By referring to Sewage/Effluent Standard (DOE, Malaysia ) could your sample be to

discharge 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, the parameter limits of effluent of Standard A and B are as follow:

Parameter Unit Standard A Standard B

(i) Temperature 40oC

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

(iii) BOD5 at 20°C (Almost the same with BOD3), 20 – 50mg/L

(iv) COD 50 – 100 mg/L

Although the temperature = 25°C and pH value = 7.1 of the waste water are

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

168.33 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.

8

7.0 DISCUSSION

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

BOD 400 mg/L.

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

BOD dilutions?

Caustic alkalinity and acidity can prevent the growth of bacteria during the test

which prevents the use of oxygen.

c)The completed composite sample arrive in the lab at

e.g : 5 days , ±3 hours ) What is the latest day and time the

sample can be started for BOD.

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

BOD bottle’s cap is needed to reduce evaporation of the water seal during

Incubation and to avoid trapping air bubbles inside the bottles.

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

BOD dilutions?

The presence of chlorine in a sample will inhibit the growth of bacteria during

the BOD test.

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

Sodium sulfite solution, 0.0250 N; Potassium iodide solution, 10%; acetic acid (1+1) or sulfuric acid (1+50); Starch indicator.

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

variations?

They must be seeded and a seed correction used in the calculation of the BOD.

9

Q3. a) State the formula to calculate

i) seed correction

o Seed Correction = (Seed BOD x mL seed in sample dilution)/300mL

ii) BOD5 (seeded)

o BOD mg/L = (DO depletion – Seed correction) x 300/mL ofsample

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 = (Seed BOD x mL seed in sample dilution)/sample volume

= 95 mg/L x 2 mL

300 mL

= 0.633 mg/L

BOD mg/L = (DO depletion – seed correction) x sample volume/mL of sample

= ( 7.8 – 2.9 )mg/L – 0.633mg/L

(100/300) mL

= 12.801 mg/L

10

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 = 7.3 mg/L – 1.8 mg/L

( 30/300 )mL

= 55 mg/L

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

was 0.23l/d, calculate i) ultimate BOD, ii) BOD3 and iii) BOD remaining at 3 days

i. BOD5= 250 mg/L

yt= Lo (1 – 10 – kt)

where yt = 250 mg/L, t = 5 days, k = 0.231/day

250 = Lo (1 – e– 0.23 × 5)

Lo = 365.84 mg/L

ultimate BOD,Lo = 365.84mg/L

ii. BOD3

BOD3= 250 = 365.84 (1 – e– 0.23 × 3)

= 182.34 mg/L

iii. BOD remaining at 3 days

BOD, Lo – BOD3 = 365.84 – 182.34 = 183.50 mg/L

11

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 “Standard Methods”, current Edition. Place the material

in a suitable container and incubate at 20°C for 24 - 36 hours. Usually, settled

raw domestic sewage prepared in the manner above will have sufficient BOD

for use as a seed material. If not, small quantities of digester supernatant,

return activated sludge, or an acclimated seed material can be used to increase

the potency of the seed material used for the test. 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. The key to a good seed correction is a relatively

stable seed material 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.

12

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

wouldresult in one duplicate sample or one replicate sample being run every ten

sample. 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 analysis 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.

13

8.0 CONCLUSION

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

out to measure the amount of dissolved oxygen in the water for a specified period of time and

temperature. 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 168.33 mg/L so that it can be concluded that the waste water

from water plant behind Tasik Kemajuan (Tasik G3), UTHM 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.

14

9.0 APPENDIX

Parameter Unit Standard  A B

(1) (2) (3) (4)(i) Temperature oC 40 40(ii) pH Value - 6.0-9.0 5.5-9.0(iii) BOD at 20oC mg/L 20 50(iv) Suspended Solids mg/L 50 100(v) Mercury mg/L 0.005 0.05(vi) Cadmium mg/L 0.01 0.02(vii) Chromium, Hexavalent mg/L 0.05 0.05(viii) Chromium, Trivalent mg/L 0.20 1.0(ix) Arsenic mg/L 0.05 0.10(x) Cyanide mg/L 0.05 0.10(xi) Lead mg/L 0.10 0.5(xii) Copper mg/L 0.20 1.0(xiii) Manganese mg/L 0.20 1.0(xiv) Nickel mg/L 0.20 1.0(xv) Tin mg/L 0.20 1.0(xvi) Zinc mg/L 2.0 2.0(xvii) Boron mg/L 1.0 4.0(xviii) Iron (Fe) mg/L 1.0 5.0(xix) Silver mg/L 0.1 1.0(xx) Aluminium mg/L 10 15(xxi) Selenium mg/L 0.02 0.5(xxii) Barium mg/L 1.0 2.0(xxiii) Fluoride mg/L 2.0 5.0(xxiv) Formaldehyde mg/L 1.0 2.0(xxv) Phenol mg/L 0.001 1.0(xxvi) Free Chlorine mg/L 1.0 2.0(xxvii) Sulphide mg/L 0.50 0.50(xxviii) Oil and Grease mg/L 1.0 10(xxix) Ammoniacal Nitrogen mg/L 10 20(xxx) Colour ADMI* 100 200*ADMI-American Dye Manufacturers Institute

15

9.0 REFERENCES

Websites :

http://www.lenntech.com/why_the_oxygen_dissolved_is_important.htm

Lenntech B.V.1998-2016.Why oxygen dissolved in water is important.

https://en.wikipedia.org/wiki/Biochemical_oxygen_demand

Wikipedia, 2016.Biochemical oxygen demand.

http://www.apgqa.com/newsletters/6_2003_demand.asp

Jessica Raney, 2016.To Seed or Not To Seed - BOD Explained, Dilution, pH

Adjustment, Seeding and BOD Calculations.

16