BIOCHEMICAL
OXYGEN DEMAND (B.O.D.) TEST
AIM:- To
determine the Biochemical oxygen demand of the sewage sample.
APPARATUS:-
i)
Glassware
ii)
B.O.D. bottle or incubation bottle
iii)
Incubator.
THEORY:-
The B.O.D.
of sewage (domestic or industrial) or any polluted waste water is defined as
the amount of oxygen required for the biochemical decomposition of dissolve
organic solids to occur under aerobic conditions & under standardized time
& temperature. The laboratory test for B.O.D is an empirical test conduced
at a temperature of 200C & an incubation period of 5 days (sometimes 7 days). This test is based on the principle of determination of
dissolved oxygen before & often incubation for a specified period (5 days)
at standard temperature (200C). the difference in D.O value gives oxygen utilized by the bacteria for the
oxidation of organic matter, which is expressed as mg/lit. the rate of
oxidation depends upon the amount & type of organic matter, dissolved
oxygen, temperature, types of organisms & environmental conditions. Under
natural conditions, complete oxidation of organic matter takes about 2-3 months
but with in 10 days nearly 90% biological demand is satisfied after which the
rate of oxygen consumption becomes very slow. In the laboratory, 5-days BOD is
tested which gives 60-70% of total demand.
The
decomposition or decay of organic substance take place by the help of enzymes,
which are produced by the secretion of protein molecules from the living
organisms. The organic matter & some of the complex mineral portions of
those substances are utilized as a source of energy by a series of living
organisms. A series of biochemical reaction is thereby set in motion, &
polluted waters are returned to a normal state or purity.
RELEVANCE:-
The BOD has
tremendous applications in the field of sanitary engineering. Some of these
applications are as follows.
i)
To measure the strength of sewage & wastes.
ii)
To determine the amount of pollution in a stream.
iii)
In the design of water treatment facilities.
iv)
To determine the efficiency of the treatment units.
v)
In the establishment of stream & effluent standards
for stream pollution regulatory agencies.
THE NORMAL VALUES OF B.O.D FOR RAW
& TREATED SEWAGE (DOMESTI) ARE GIVEN BELOW.
Nature of sewage
|
BOD5 at 200C
in mg/l
|
1, Strong sewage
2. Average sewage
3. Weak sewage
4. Std. filter sewage effluent
5. Very good filter sewage effluent
|
450
– 500
350
250
20
5 – 10
|
REAGENTS FOR
DILUTION WATER:-
i)
Distilled water for dilution.
ii)
Phosphate buffer solution consisting of 8.5g KH2
PO4, 21.75g K2 HPO4, 33.4g NA2HPO4.7H2O
and 1.7g of NH4CL dissolved in 1 lit. PH of this solution is 7.2.
iii)
Magnesium sulphate solution consisting of 22.5g MgSo4
7H2O dissolved in distilled water & diluted water diluted to 1
lit.
iv)
Calcium chloride solution consisting of 27.5g cacl2
dissolved in distilled water & diluted to 1 lit.
v)
Ferric chloride solution, consisting of 0.25g Fecl3.
6H2o dissolved in distilled water & diluted to 1 lit.
REAGENT FOR D.O
MEASUREMENT: (By AZIDE MODIFICATION METHOD)
i)
Manganese
sulphate solution: Dissolve 480g mnso4.4H2O in
distilled water and dilute it to 1 lit.
ii)
Alkali –
Iodide – Azide reagent: Dissolve 500g Na oH and 150g KI in distilled
water and dilute it to 1 lit. add 10g sodium Azide(NaN3) dissolved
in 40ml distilled water
iii)
Concentrated
H2So4.
iv)
Starch:-
Dissolve 5g starch in 1 lit boiling distilled water.
v)
Sodium
tiosulphate titrant.:- Dissolve 6.205g Na2 S2 O3.5H2O
in distilled water and dilute it to 1 lit. Standardize it using standard
potassium dichromate solution.
vi)
Standard
potassium dichromate solution (0.025N):- Dissolve 1.22g K2CR2
O7 in distilled water and dilute it to 1 lit.
PROCEDURE:-
Standardization of
sodium thiosulphate titrant:-
a) Take
100ml distilled water in a flask and add 10ml KI solution. Add 10ml 1+9 H2SO4.
b) Add
20ml K2CR2O7 solution and store in dark place
for five minute.
c) Titrate
with Na2S2O3 titrant, adding starch towards
the end when a pale straw colour develops. Continue titration to the colourless
end point. Note down the volume of titrant as V2 ml.
d) Determine
the normality of the titrant ‘N2’ using the relation N1V1
= N2V2 where
N1 = normality K2
CR2 O7 solution = 0.02
V1 = volume of K2
CR2 O7 = 20ml
Preparation of
dilution water and placing it in BOD bottle
a) Aerate
the distilled water saturation with D.O. Take 1 lit of this dilution water and
1ml of each of phosphate buffer, MgSo4, CaCl2 and FeCl2
solutions. Stir to mix the contents.
b) Preparing
samples of different dilution. The following dilutions are recommended
depending upon the expected BOD of the waste water sample.
25%, 40%, 60%,
75% and 100%
To make dilution of say 25%, take 1
lit measuring cylinder and fill it about half with dilution water by siphoning.
Add 250ml sample and make it 1 lit with dilution water stir well to mix
properly. Transfer this by siphon into two BOD bottles.
c) Repeat
the procedure for other dilutions.
d) Place
one bottle in the incubator and test the second bottle immediately for D.O.
using Azide – Modification method discussed later. Let this value be “D1”.
e) Incubate
the sample for 5 days at 200C and test for D.O. again. Let this
value be “D2”.
Alternate method
of diluting the sample:-
In this
method, the sample sewage is diluted to the required extent by direct pipetting
a given volume of sewage into a 300ml BOD bottle and filling the remaining
volume of bottle with dilution water and calculating the dilution factor. For
example, if 5ml of waste water is placed in the BOD bottle and mixed with
dilution water to make 300ml diluted sample, the dilution ratio will be equal
to 300/5=60. The following table gives the range of BOD that can be measured
with various dilutions expressed either percentage mixtures or as ml directly
pipetted into 300ml BOD bottles.
By using %
mixture
|
By direct pipetting into 300ml BOD bottle
|
||
% mixing
|
Range of BOD in
mg/lit
|
ml
|
Range of BOD in mg/lit
|
0.01
0.02
0.05
0.10
0.20
0.50
1.0
2.0
5.0
10.0
20.0
50.0
100.0
|
20000-70000
10000-35000
4000-14000
2000-7000
1000-3500
400-1400
200-700
100-35
40-140
20-70
10-35
4-14
0-7
|
0.03
0.05
0.10
0.20
0.50
1.0
2.0
5.0
10.0
20.0
50.0
100.0
300.0
|
30000-105000
12000-42000
6000-21000
3000-10500
1200-4200
600-2100
300-1050
120-240
60-210
30-105
12-42
6-21
0-7
|
Estimation of D.O.:-
a) To
the sample in the BOD bottle, add 2ml manganese sulphate solution, 2ml Alkali –
iodide – Azide reagent, well below the surface.
b) Stopper
and mix by inverting bottle at least 15 times. Let the brown precipitate settle
for 5 minutes.
c) Add
2ml conc. H2SO4 and stopper. Gently invert to dissolve
the precipitate totally.
d) Take
203ml of bottle water in a flask and titrate with 0.025N sodium thiosulphate
solution.
e) Add
1 to 2ml starch and continue titration till the first disappearance of pale
blue colour.
f) Note
down the volume of titrant used upto colourless end point.
Note:- 200ml of sample have to be used in the titration. The
volume corresponding to
200ml of original sample after correction for a total of 4ml
(2ml manganese solution and 2ml alkali-azide reagent) in a 300 ml bottle is
200*300
= 202.7ml = 203ml.
(300-4)
OBSERVATIONS:-
S.No
|
Burette
|
Vol. of Titrant used
|
|
Initial
|
Final
|
||
CALCULATIONS:
Initial D.O. of dilute sample (before incubation)
In mg/l = D1 = ml of 0.025N sodium thiosulphate
used.
Final D.O. of dilute sample (after incubation)
In mg/l = D2 = ml of 0.025N sodium thiosulphate
used.
BOD of the sample in mg/l = D1 – D2
(if dilution is expressed in percentage)
P
Where P = Decimal fraction of sample used.
BOD of the sample in mg/l = (D1 – D2)*
dilution ratio (if sample is directly pipetted into BOD bottle)
Where dilution ratio = Volume of diluted sample
Volume of
undiluted sewage sample
RESULTS:-
INTERPRETATION OF
RESULTS:
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