- BOD Lab Report

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.

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

and temperature.

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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

4. 100 ml graduated cylinder  For measuring the volumes (amounts) of liquids. 2

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 7. pH meter  to measure the hydrogen-ion concentration or pH value in a solution, indicating its acidity or alkalinity. 3

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 H 2SO4 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.

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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.

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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.

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DATA RESULT

BIOCHEMICAL OXYGEN DEMAND ( BOD) Analyst: Date:02 / 03 / 16 Time : 11 A.M

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Sample Details: Source: Tasik Kemajuan (Tasik G3), UTHM 250C

7.1pH Pretreatment:

Comments:

Alkalinity/Acidity Sample Volume: I N NaoH : I N N2SO4:

Sampl e Type

0 0

Sample ID

500

BOD3

considered as neutral. (6.5 – 7.5) = neutral

mL mL

Volume Sampl e ( mL)

Dilution Factor

Initial DO (mg/L)

Final DO ( mg/L)

0.83/207.5 = 0.004

4.76

3.93

4.76 – 3.93 = 0.83

(0.83 x 500) = 207.5 2

Blank

2

0.004

4.98

4.25

0.73

182.5

Blank

2

0.004

5.56

5.10

0.46

115

Average BOD Cancelled Data/ Result:

= 207.5/3 = 69.17 mg/L

BOD2= 182.5/3 = 60.83 mg/L BOD3= 115/3 = 38.33 mg/L

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BOD (mg/L)

2

( show the calculation)

BOD1

DO Depletion ( mg/L)

Blank BOD1 BOD2

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

mL

DATA ANALYSIS 1. Calculate for the average BOD 

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

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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/L BOD5 = (0.83 x 500)/2 = 207.5 mg/L  BOD2, DO depletion = DOi – DOf = 4.98 – 4.25 = 0.73 mg/L BOD5 = (0.73 x 500)/2 = 182.5 mg/L  BOD3, DO depletion = DOi – DOf = 5.56 – 5.10 = 0.46 mg/L BOD5 = (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.

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 7

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.

7.0

DISCUSSION

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

400

mg/L.

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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.

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) 9

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

Dilution #1

BOD5 of Seed Material

95 mg/L

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

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 10

( 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

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. 11

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.

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 12

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.

8.0

CONCLUSION

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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 BOD 5. 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.

9.0 APPENDIX

Parameter

Unit

Standard A

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B

(1)

(2)

(3)

(4)

(i)

Temperature

o

C

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

15

(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

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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.

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