OBJECTIVE
For biochemical oxygen demand (BOD) experiment, the objective is to measure the amount of dissolved oxygen in the wastewater for a specific period of time and temperature.
INTRODUCTION Biochemical Oxygen Demand is a common, environmental procedure for determining the extent to which oxygen within a sample can support microbial life. The microorganisms use organic matter as a food source through oxidation which oxygen is consumed. According to Linda (1994), the test for Biochemical Oxygen Demand is especially important in waste water treatment, food manufacturing, and filtration facilities where the concentration of oxygen is crucial to the overall process and end products. . The study orientation was on the beneficial uses of water which was focused on, water for domestic water supply, fisheries and aquatic propagation, livestock drinking, recreation and agricultural use (DOE Malaysia, 1985). The higher the BOD, the more oxygen will be demanded from the waste to break down the organics. The BOD test is performed by incubating a sealed wastewater sample for the standard 5-day period, then determining the change in dissolved oxygen content. The bottle size, incubation temperature, and incubation period are all specified. Most wastewaters contain more oxygen demanding materials than the amount of DO available in air-saturated water. Therefore, it is necessary to dilute the sample before incubation to bring the oxygen demand and supply into appropriate balance. Because bacterial growth requires nutrients such as nitrogen, phosphorous, and trace metals, these are added to the dilution water, which is buffered to ensure that the pH of the incubated sample remains in a range suitable for bacterial growth. Complete stabilization of a sample may require a period of incubation too long for practical purposes, therefore, 5-day period has been accepted as the standard incubation period. Quality controls, standards and dilutions are also run to test for accuracy and precision (Linda, 1994). There are some the environmental impacts of BOD which is the sewage containing high BOD create environmental and health problems. Moreover, it interferes with the aquatic life. Organic pollution is harmful to fish as it tends to reduce the amount of dissolved oxygen. Last but not least, it defines the strength of domestic wastes and industrial wastewaters APPARATUS & MATERIALS
6 units of 300mL BOD bottles with glass stoppers Volumetric pipette Incubator at 25°C
REAGENTS
Dilution water containing nutrient buffer and seed
Water sample
PROCEDURE 1. The dilution water was prepared using a BOD nutrient buffer pillow. 2. The experiment was divided into seeded and unseeded dilution water. 3. For seeded dilution water, the BOD bottle was filled up with the dilution water and was labeled as blank seeded sample. For other two BOD bottles were filled up with dilution water as well as inserted the 3mL and 6mL of water sample to each of the BOD bottles. 4. For each of BOD bottles were added nutrients and seed. 5. Step 3 was followed for unseeded dilution water. 6. The bottles were closed with stopper. The stopper was tightly twisted into place and pressed down and inverted the bottles several times to mix. 7. The initial dissolved oxygen concentration was measured in each of bottle both for unseeded and seeded sample. 8. The dilution water was added to the lip of the BOD bottles to make water seal. 9. The bottles were incubated in the incubator for 7 days at 20°C. 10. After 7 days, the remaining dissolved oxygen concentration was measured in each bottle. 11. The BOD value was calculated. RESULT To calculate the BOD5 value, For unseeded dilution water
BOD 5=
( DO initial− DO final ) × bottle volume s ample volume
For seeded dilution water
BOD 5=
( D 1−D 2 ) −(B 1−B2) f P
Where; D1=initial DO concentration of sample, mg/L D2= final DO concentration of sample, mg/L
B1= initial DO concentration of blank, mg/L B2= final DO concentration of blank, mg/L f= fraction seeded dilution water volume in sample to blank P= decimal fraction of sample in 300mL bottle
Sample, mL Seeded
Blank
DO initial, mg/L 7.28
DO final, mg/L 7.15
BOD7, mg/L
300 300−0/¿ ¿ mL mg mg ¿ 7.49−7.23 −(7.28−7.15 )¿ L L ¿
(
)
¿0 3mL sample
7.33
7.22
300−3 ¿ 300 ¿¿ ¿ ¿ mg mg ¿ 7.33−7.22 −(7.28−7.15 )¿ L L ¿
(
)
¿−1.87 6mL sample
7.28
7.37
300 300−6/¿ ¿ 300 mL ¿ 300 6/ ¿ ¿ mg mg ¿ 7.28−7.37 −(7.28−7.15 )¿ L L ¿
(
)
¿−10.87 Unseede d
Blank
7.49
7.23
mg )× 300 mL L 0 mL
¿ (7.49−7.23
¿0
3mL sample
7.34
mg )×300 mL L 3 mL
7.09
¿ (7.34−7.09
¿ 25 .0 6mL sample
7.33
mg )×300 mL L 6 mL
7.06
¿ (7.33−7.06
¿ 13.5
Calculation for BOD5 The formulae to calculate BOD5 is;
Ultimate BOD=BOD 5/ ( 1−e−(k ) (T ) ) BOD 5=Ultimate BOD × ( 1−e−( k ) (T ) ) where k= reaction constant per day T= time
Unseede d
Sample, mL Blank
BOD7, mg/L 0
Ultimate BOD, mg/L
¿ 0/(1−e−(0.23) (7 )) ¿0
3mL sample
25
13.5
¿ 25.0/(1−e−(0.23) (7 ))
Blank
0
¿ 13.5/(1−e−( 0.23) (7 ))
−e−(0.23)(5) ¿ 16.875× ¿ ¿ 11.53
− ( 0.23) ( 7 )
¿ 0/(1−e =0
1−e−(0.23 )(5) ¿ 31.25× ¿ ¿ 21.36
¿ 16.875 Seeded
1−e−(0.23 )(5) ¿ 0× ¿ ¿0
¿ 31.25 6mL sample
BOD5, mg/L
)
1−e−(0.23 )(5) ¿ 0× ¿ ¿0
3mL sample
-1.87
¿−1.87 /(1−e−( 0.23) (7 )) ¿−2.34
6mL sample
-10.87
¿−10.87 /(1−e−( 0.23) (7 )) ¿−13.6
1−e−(0.23 )(5) ¿−2.34 × ¿ ¿−1.60
1−e−(0.23 )(5) ¿−13.6 × ¿ ¿−9.29
DISCUSSION On the BOD experiment, we have gathering the BOD results. Theoretically, the BOD value is calculate in 5 days but in this experiment the BOD value is indicated until day 7. So, we are using the ultimate BOD formulae to find the exact value of BOD5. According to INQWS (Index National Quality Water Standards), the good BOD5 value is less than 5ppm, which is suitable for human daily consumption. High concentrations of dissolved oxygen (DO) predict that oxygen uptake by microorganisms is low along with the required break down of nutrient sources in the sample. On the other hand, low DO readings signify high oxygen demand from microorganisms, and can lead to possible sources of contamination depending on the process. The result showed that for unseeded sample, is in higher numbers, indicate that the water sample is highly contaminated. A high BOD indicates a high content of easily degradable, organic material in the sample. Apparently, the sample water for 3mL is in Class V since the BOD 5 value is more than 12ppm,whereas for 6mL sample water is in Class IV, based on DOE Water Quality Index Classification. This showed the dissolved oxygen is depleted since there are many microbiological organisms that have in the water sample. Thus, it can be harmful to human and environment condition. Theoretically, for BOD5, the number of seeded must be higher than seeded because in seed packet contain nutrient consists of phosphorus and nitrogen which is the major metal for growth of microbiology organisms. Hence, it should consume a lot of oxygen than unseeded sample. But in this case, the value for seeded sample for both 3mL and 6mL of water sample showed the negative results. This may due to some errors during the lab experiment. The problem may come from diluting the BOD sample. During the time to fill up the BOD bottles, the student should take turns interspersed the sample and dilution water, by the same time shake the bottle regularly in order for the mixture to be mix together before taking the DO value.
CONCLUSION
From the data collected above, the value of BOD 5 for unseeded 3mL and 6mL of water sample are 21.36mg/L and 11.53mg/L respectively. For seeded 3mL and 6mL of water samples are -1.60mg/L and -9.29mg/L respectively. Therefore, as the conclusion, only value of BOD5 for unseeded 3mL and 6mL of water sample has polluted and contaminated sample, whereas for seeded showed an error during the experiment.
REFERENCES Department of Environment Malaysia, (1985). Development of Water Quality and Standards for Malaysia Linda S. M. (1994). Biochemical Oxygen Demand. Retrieved from http://www.rpi.edu/dept/chem-eng/Biotech-Environ/Environmental/BOD/coda.htm
