Determine BOD of Bagmati River Water Sample: Calculation

REQUIREMENTS:

APPARATUS REQUIRED:

1. Bod bottle
2. Conical flask
3. Burette
4. Beaker
5. Pipette
6. Measuring Cylinder

Chemicals required:

1. Sodium thiosulphate
2. Alkaline potassium Iodide
3. Magnese  sulphate
4. Starch
5. Conc. H₂SO₄

THEORY:

   Biological oxygen demand is a measurement of amount of dissolved oxygen(DO) that is used by aerobic microorganisms when decomposing organic matter in water. Nitrates and phosphates are plant nutrients and can cause plant life and algae to grow they die quickly. This contributes to organic waste in water.

Generally, BOD level in water indicate.

BOD level in mgll Indicators

         1 – 2 Very Good

         3 – 5 fair: moderate polluted

         6 – 9 poor: some what polluted

        10 or >10                          very poor: very polluted

PROCEDURE:

                 First of all, the dilution process was done. 1l of distilled water was taken in the volumetric flask. 1ml of phosphate buffer was added along with 1ml of Mg SO₄. 7H₂O, 1ml of CaCL₂  was added along with 1ml of MgSO₄,7H₂O and then 1ml of FeCl₃. 6H₂O was added. The solution was mixed well for the complete preparation, 50ml of sample was taken from the sample bottle of 2 litre bottle.

Now finally for BOD test, 2 BOD bottle were taken. The bottle was wrapped with carbon paper and was kept in BOD in cupboard at 20⁰C for 5days. 

In second bottle DO content of sample water was determined by using winkerls Podsmetric method.

CALCULATION FORMULA:

       BODs= (DO initial DO )-(Do five day (DO₅)*D.F.

OBSERVATION TABLE:

1. For Sundarijal

Initial observation:

S.N.	Volume of sample	Initial buretle reading	Final buretle reading	Different buretle reading	Concurrent reading
1	50ml	0	1.7	1.7
2	50ml	1.7	3.6	1.9	1.9
3	50ml	3.6	5.5	1.9

Final observation:

S.N.	Volume of sample taken	Initial buretle reading	Final buretle reading	Different buretle reading	Concurrent reading
1	50 ml	0	1.6	1.6
2	50 ml	1.6	3.1	1.5	1.5
3	50 ml	3.1	4.6	1.5

2. For Pashupati

Initial observation:

S.N.	Volume of sample taken	Initial buretle reading	Final buretle reading	Different buretle reading	Concurrent reading
1	50 ml	0	1.6	1.6
2	50 ml	2.4	4.1	1.7	1.7
3	50 ml	4.1	5.8	1.7

Final observation:

S.N.	Volume of sample taken	Initial buretle reading	Final buretle reading	Different buretle reading	Concurrent reading
1	50 ml	0	1.6	1.6
2	50 ml	1.6	3.1	1.5	1.5
3	50 ml	3.1	4.6	1.5

3. For Buddhanagar

Initial observation:

S.N.	Volume of sample taken	Initial buretle reading	Final buretle reading	Different buretle reading	Concurrent reading
1	50 ml	0	2.2	2.2
2	50 ml	2.2	3.5	1.3	1.3
3	50 ml	3.5	4.8	1.3

Final observation:

S.N.	Volume of sample taken	Initial buretle reading	Final buretle reading	Different buretle reading	Concurrent reading
1	50 ml	18.9	19.7	0.8
2	50 ml	19.7	20.6	0.9	0.9
3	50 ml	20.6	21.5	0.9

4. For chovar:

Initial observation:

S.N.	Volume of sample taken	Initial buretle reading	Final buretle reading	Different buretle reading	Concurrent reading
	50 ml	0	1.2	1.2
	50 ml	1.2	2.3	1.1	1.1
	50 ml	2.3	3.4	1.1

Final observation:

S.N.	Volume of sample taken	Initial buretle reading	Final buretle reading	Different buretle reading	Concurrent reading
	50 ml	0	1.7	1.7
	50 ml	1.7	3.0	1.3	1.3
	50 ml	3.0	4.3	1.3

CALCULATION:

1. For sundarijal

Do initial = (ml*g) of Na₂S₂O3*8*100/ V₂(V₁-V/V₁)

= 1.9*0.025*8*1000/50(300-4/300)

= 7.7032 mg/l

Do (5 days) = 1.5*0.025*8*1000/50(300-4/300)

= 6.081 mg/l

BOD of Sundarijal = ( Do initial- Do₅day) *d.f.

= (7.7032-6.081)*10

= 16.222 mg/l

2. Calculation of Pashupati

Do inintial = 1.7*0.025*8*1000/50(300-4/300)

= 6.89 mg/l

Do (5 days) = 1.5*0.025*8*1000/50(300-4/300)

= 6.081 mg/l

BOD = (6.89-6.081)*10

= 8.09 mg/l

3. Calculation of Buddhanagar

Do initial = 1.3*0.025*8*1000/50(300-4/300)

= 5.27 mg/l

Do (5 days) = 0.9*0.025*8*1000/50(300-4/300)

= 3.648895 mg/l

BOD= (5.27-3.64)*10

= 16.3 mg/l

4. Calculation of Chovar

Do initial = 1.3*0.025*8*1000/ 50(300-4/300)

= 5.27 mg/l

Do (5 days)= 1.7*0.025*8*1000/50(300-4/300)

= 4.45976 mg/l

BOD = (5.27-4.45)*10

= 8.2 mg/l

CONCLUSION:

               From the above experiment we can conclude that the comparative study of biological oxygen demand (BOD) at different location of Bagmati river was done.