Chemical oxygen demand. mujahid hussain

Chemical oxygen Demand
(COD)
6/11/2017 1
Mujahid hussain (M.Phil Botany)
Department of Botany,
University of Sargodha,
Sargodha
(mujahid.hussain7877@gmail.co
m)

Chemical oxygen demand
 Chemical oxygen demand (COD) is a measure of the capacity of
water to consume oxygen during the decomposition of organic
matter and the oxidation of inorganic chemicals such as ammonia
and nitrite.
OR
 Chemical Oxygen Demand or COD is a measurement of the oxygen
required to oxidize soluble and particulate organic matter in water.
OR
 The COD value indicates the amount of oxygen which is needed for
the oxidation of all organic substances in water in mg/l or g/m3
 SI units is milligrams per litre (mg/L).
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Why is COD important?
 Chemical Oxygen Demand is an important water quality
parameter because provides an index to assess the effect
discharged wastewater will have on the receiving
environment.
 Higher COD levels mean a greater amount of oxidizable
organic material in the sample, which will reduce dissolved
oxygen (DO) levels.
 A reduction in DO can lead to anaerobic conditions, which is
deleterious to higher aquatic life forms.
 The COD test is often used as an alternate to BOD due to
shorter length of testing time.
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How is COD measured?
 A common method for Chemical Oxygen Demand analysis is
Method 410.4.
 The method involves using a strong oxidizing chemical,
potassium dichromate Cr2O72-, to oxidize the organic matter
in solution to carbon dioxide and water under acidic
conditions.
 The higher the chemical oxygen demand, the higher the
amount of pollution in the test sample.
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 Often, the test also involves a silver compound to encourage
oxidation of certain organic compounds and mercury to
reduce the interference from oxidation of chloride ions. The
sample is then digested for approximately 2 hours at 150°C.
The amount of oxygen required is calculated from the quantity
of chemical oxidant consumed.
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Procedure
 1)A 50 mL waste water sample is collected. 10 mL of 0.25 N
K2Cr2O7 is added to the water sample and to 50 mL of
distilled water.
 You add dichromate to distilled water AND to your water
sample.
 What does dichromate do in distilled water?
 Nothing There’s nothing for it to oxidize.
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2) Both samples are heated to 50°C for 30 minutes.
The dichromate is being allowed to react. It is
oxidizing organic material!!!
The samples are allowed to cool for 10 minutes and
then titrated with 0.1015 N iron (II) ammonium
sulfate ( FAS).
 The waste water sample requires 15.36 mL of titrant,
while the blank sample requires 23.65 mL to reach a
1,10 phenanthroline endpoint.
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 3) The samples are allowed to cool for 10 minutes and then
titrated with 0.1015 N iron (II) ammonium sulfate ( FAS).
 6 Fe2+ + Cr2O72- + 14 H+ →
6 Fe3+ + 2 Cr3+ + 7 H2O
 The titration reaction has 6:1 stoichiometry of the Fe2+ titrant
to the Cr2O72-.
 Titrating the solutions with Fe2+ is telling us how much
dichromate is left over!

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4)The waste water sample requires 15.36 mL of titrant,
while the blank sample requires 23.65 mL to reach a 1,10
phenanthroline endpoint
We have different amounts of dichromate in the 2
different samples, does this make sense?
Yes, we reduced some dichromate in the “dirty” sample
while the distilled water should have all the dichromate it
started with!
The difference between the two samples is the amount of
dichromate reduced and, therefore, the amount of organic
material oxidized
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express solution concentrations (like the Fe2+) in
“normality
Normality = equivalent moles of solute
 L solution
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 This is a redox titration – equivalence is about
electrons.
 6 Fe2+ + Cr2O72- + 14 H+ → 6 Fe3+ + 2 Cr3+
+ 7 H2O
 Each iron atom transfers 1 electron.
 Each dichromate molecule involves 6 electrons.
 This means that 1 M Fe2+ = 1 N Fe2+
 1 M Cr2O72- = 6 N Cr2O72-
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i2M1V1 = i1M2V2
iFeMCrVCr = iCrMFeVFe
6 * MCr* 50 mL = 1* 0.1015 M * 15.36 mL
MCr = 5.20x10-3 M Cr2O72-
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N1V1 = N2V2
NCrVCr = NFeVFe
NCr* 50 mL = 0.1015 N * 15.36 mL
NCr = 3.12x10-2 N Cr2O72-
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For reference water
N1V1 = N2V2
NCrVCr = NFeVFe
NCr* 50 mL = 0.1015 N * 23.65 mL
NCr = 4.80x10-2 N Cr2O72-
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Pure water:
NCr = 4.80x10-2 N Cr2O72-
“Dirty” water:
NCr = 3.12x10-2 N Cr2O72-
The difference between the two is the amount reduced! Since the volume is
the same, you can just subtract:
4.80x10-2 N - 3.12x10-2 N = 1.68x10-2 N Cr2O72- reduced
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We could express the impurity based strictly on the
dichromate used: the more dichromate required, the
more organic material that must have been there.
You could calculate the COD using the scheme
COD (mg/L) = 8000 (mL blank – mL sample) [Fe2+]
 mL sample
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COD (mg/L) = 8000 (mL blank – mL sample) [Fe2+]
mL sample
COD (mg/L) = 8000 (23.65 – 15.36 mL) [0.1015 M]
50 mL
COD = 135 mg/L
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Inorganic interference
 Some sample of water contain high level of
oxidizanle inorganic material which may
interfere with the determination of COD.
 Chloride is often the most serious source of
interference .Its reaction with potassium
dichromate follows the equation
 6 Fe2+ + Cr2O72- + 14 H+ →
 6 Fe3+ + 2 Cr3+ + 7
H2O
 Prior to the addition of other reagents,
Mercuric sulfate can be added to the sa mple
to elimate chloride interference.
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Where is COD measured?
Influent wastewater streams for process control
Effluent wastewater streams to sewer or environment
for regulatory compliance
Applications:
Industrial effluent pollution management
Municipal sewage monitoring
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Chemical oxygen demand. mujahid hussain

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