1.
Cairo University - Faculty of Engineering
Petrochemicals from Oil & Gas
Exp. 4 & 5
COD & BOD

2.
Experiment # 4:Chemical Oxygen
Demand [COD] Chemical Oxygen Demand
[COD]

3.
• Chemical oxygen demand (mg O2 / lit.) is the amount of
oxygen required for reacting with the organic (harmful)
matter present in waste water, both soluble or insoluble
(suspended) matters, producing CO2 and H2O. In this
experiment, organic compounds are oxidized to carbon
dioxide and water by a boiling acid dichromate solution:
𝐶𝑎𝐻𝑏𝑂𝑐 + 𝑑𝐶𝑟2𝑂7
−2
+ 8𝑑𝐻+1
ℎ𝑒𝑎𝑡
𝑎𝐶𝑂2 + 𝑒𝐻2𝑂 + 2𝑐𝐶𝑟+3
Definition

4.
 Waste water sample.
 Distilled water.
 Potassium dichromate (K2Cr2O7).
 Sulfuric acid (H2SO4).
 Silver sulfate (Ag2SO4).
 Mercuric sulfate (HgSO4).
Materials

5.
Procedure
1. Take a sample of waste water (2.5 ml) in a standard test tube.
2. Add K2Cr2O7 (1.5 ml) to the above sample.
3. Add 3.5 ml of a solution containing H2SO4, Ag2SO4 and HgSO4 to the above
mixture. This solution is known as "digestion solution" which is prepared by
adding Ag2SO4 and HgSO4 to 1 kg of H2SO4.
4. Repeat the above procedure with a sample of distilled water (2.5 ml) in another
test tube.
5. Heat the two test tubes in the reactor for 2 hrs. at a temperature of 150 ºC and
after that leave them to cool.
6. Use the spectrophotometer to detect the COD (in mg/lit.) value for the waste
water sample.

6.
Notes
 K2Cr2O7 is used as an oxidizing agent (source of oxygen needed to react with organic
matters).
 H2SO4 is a digesting agent which helps in decomposing the organic matters to be easily
reacted with oxygen.
 Ag2SO4 is used to reduce the volatility of the organic matters exist in the waste water
sample and keep them in liquid phase. If those matters vaporized, the measured value of
COD will be incorrect.
 HgSO4 is used to avoid oxidation of 𝐶𝑙− if it exists in wastewater as salt. This will lead
to high misleading value of COD since 𝐶𝑙−
is oxidized by K2Cr2O7 into Cl2.
 The distilled water sample is used as a blank sample which allows the calibration of the
spectrophotometer. The COD value for this sample is zero.

7.
Experiment # 5:
Biological Oxygen Demand
[BOD5]

8.
Experiment # 5: Biological Oxygen
Demand [BOD5]
• Biological oxygen demand (mg O2 / lit.) is the amount of
oxygen required to be used up by bacteria so as to
decompose the waste matters in a liter of wastewater. This
test may need at least 3 months to be finished: the
standard test defines it as BOD5 as it is performed within
5 days only. During those 5 days, about 70 – 80% of
degradation is achieved.

9.
Difference between COD and BOD
 In the COD test we completely oxidize the wastes,
whether biodegradable (i.e. can be decomposed by
bacteria) or non – biodegradable.
 In the BOD test we oxidize the biodegradable wastes
only.
∴ 𝑪𝑶𝑫 > 𝐵𝑂𝐷

10.
Respiratory Method
(Respirometric determination of the BOD5
using the OxiTop® measuring system)
• Determination of the BOD5 of undiluted samples of sewage
containing high levels of industrial pollutants may be
considerably impaired(damaged) by the presence of inhibitors or
toxic substances. Measurements can only be carried out after the
sample has been diluted with dilution water that contains a
sufficient amount of nutrients and microorganisms in order to
reduce the interfering substances to an acceptable level.

11.
Respiratory Method
(Respirometric determination of the BOD5
using the OxiTop® measuring system)
Equipment:
 OxiTop®control.
 Sample bottle.
 Stirrer rods.
 Rubber stopper.
 Overflow flask.

12.
Respiratory Method
(Respirometric determination of the BOD5
using the OxiTop® measuring system)
Reagents and chemicals used
• Phosphate buffer solution.
• Magnesium sulfate solution.
• Ferric chloride solution.
• Aerated water for dilution which contains a certain amount of sewage
water.
• Sodium hydroxide (NaOH) tablets.

13.
Procedure
1) Take 108 ml of the wastewater (or any other volume according to the expected
levels of BOD5).
2) Measure its pH using the pH meter. Its pH must be in the range 6.5 – 8.5. If the
value of the pH <6.5 (acidic), put one droplet of NaOH (1N) and if the value of
the pH >8.5 (alkaline), put one droplet of HCl (1N).
3) Pour the above sample in the 432 ml overflow flask.
4) Fill the 432 ml overflow flask with dilution water.
5) Pour the mixture into a sample bottle.
6) Place a magnetic stirrer rod into the sample bottle.
7) Insert a rubber stopper into the neck of each bottle.

14.
Procedure
8) Use tweezers to add two tablets of sodium hydroxide through the stopper.
9) Tightly screw the OxiTop® Controller / OxiTop® measuring sensor onto the measuring bottle.
10) Start the measurements using the OxiTop® Controller or OxiTop®. This is done by pressing and
hold the two buttons of the controller together until "zero" appear on the digital screen of the
controller.
11) Place the sample bottle on a stirring platform in the thermostat cabinet or thermostat box.
12) Make a blank sample by filling the 432 ml. overflow flask with dilution water only. Repeat the
steps 5 through 11 for this blank sample.
13) Switch on the motor drive of the stirring system after all of the bottles have been placed in the
cabinet or box.
14) Incubate the samples at 20°C ± 1K for 5 days.
15) Read the measurements after 5 days.

15.
Procedure
16) Calculate the BOD5 from the following relation:
Where:
o BOD5)sample: is the required value of the biological oxygen demand of the sample in mg /l.
o 𝐷𝑖𝑙𝑢𝑡𝑖𝑜𝑛 𝑓𝑎𝑐𝑡𝑜𝑟 =
𝑉𝑜𝑙𝑢𝑚𝑒 𝑜𝑓 𝑜𝑣𝑒𝑟𝑓𝑙𝑜𝑤 𝑓𝑙𝑎𝑠𝑘 (432 𝑚𝑙.)
𝑉𝑜𝑙𝑢𝑚𝑒 𝑜𝑓 𝑤𝑎𝑠𝑡𝑒𝑟 𝑤𝑎𝑡𝑒𝑟 𝑠𝑎𝑚𝑝𝑙𝑒 (108 𝑚𝑙.)
o BOD5: the reading of BOD taken in the fifth day.
o BOD5)blank: the value of BOD of the blank sample.
𝐵𝑂𝐷5)𝑠𝑎𝑚𝑝𝑙𝑒 = 𝐷𝑖𝑙𝑢𝑡𝑖𝑜𝑛 𝑓𝑎𝑐𝑡𝑜𝑟 ∗ 𝐵𝑂𝐷5 − 𝐵𝑂𝐷5)𝑏𝑙𝑎𝑛𝑘
∗
𝑣𝑜𝑙𝑢𝑚𝑒 𝑜𝑓 𝑤𝑎𝑡𝑒𝑟 𝑜𝑓 𝑑𝑖𝑙𝑢𝑡𝑖𝑜𝑛 (432 − 108)
108

16.
Notes
a) The waste water used here is industrial waste water which does not
contain bacteria. This because industrial waste water is not a suitable
environment for bacteria to live as its temperature is high, its pH is
either high (alkaline) or low (acidic) and contains toxic pollutants such
as heavy metals and toxic hydrocarbons. All these conditions will kill
bacteria in the waste water. So, if the sample contains <103 bacteria /
ml, 100 – 200 ml. of settled sewage should be added to each liter of
dilution water. This "seeding" procedure provides a sufficiently large
bacterial population to enable bio-oxidation to occur.

17.
Notes
b) As the amount of pollutants in the waste water (the concentration of
pollutants) is high, the waste water sample should be diluted with a
suitable amount of water of dilution to decrease the load on bacteria.
The factor of dilution (i.e. the ratio between the flask volume and the
waste water volume) is determined according to the expected range of
BOD which can be deduced by measuring the value of the COD to the
sample firstly. There are different volumes of overflow flask other than
the 432 ml. flask (smaller volumes are used for higher concentrations
of pollutants and hence higher BOD).

18.
Notes
c) You must put the waste water sample in the overflow flask and then pour the
water of dilution on it for not losing any droplet of waste water sample.
d) In this experiment, the BOD is measured by calculating the depletion of oxygen
(by measuring its partial pressure) in the gas space above the waste water in the
sample bottle. So, the sample bottle must not be completely filled with water.
e) Aerated water is used here to ensure that the depletion of oxygen is due to the
bacteria consumption and not as dissolved oxygen in water (i.e. if the water used
here is not saturated with oxygen before the experiment, oxygen from the space
inside the bottle will dissolve in water) and this will give error in calculating the
BOD value.

19.
Notes
f) Sodium hydroxide tablets are used to absorb the carbon dioxide formed from the
reaction. If CO2 left as it is, it will interfere with the measured value of oxygen
partial pressure.
g) The controller should be tightly screwed onto the measuring bottle to prevent any
excess amount of oxygen to enter to the bottle during the test.
h) The values of COD and BOD (after measuring each one) are compared together.
If the value of BOD is near that of COD (ex. COD = 12000 mg/l., BOD = 10000
mg/l.) then, the biological treatment will be useful as the majority of the pollutants
are biodegradable. On the other hand, if BOD value is far from that of COD (ex.
COD = 12000 mg/l., BOD = 1000 mg/l.) then, the biological treatment will be
useless because the majority of pollutants are non – biodegradable.