Environmental Engineering Reference
In-Depth Information
and the sample is transported to the lab for titration. The ''fixation'' reaction between
DO and Mn
2þ
is as follows:
Mn
2þ
þ2OH
þ½O
2
!MnO
2
ðsÞþH
2
O
ð6
:
9Þ
DO rapidly oxidizes an equivalent amount of the dispersed divalent manganous
hydroxide precipitate to oxides of higher valence state. The sample completely fills
the 300 mL bottle to ensure no further oxygen is introduced. After being transported
to the laboratory (within 6 h of sample collection), the sample is acidified with
concentrated H
2
SO
4
. In the presence of iodide ions (I
) in an acidic solution, the
oxidized manganese reverts to the divalent state, with the liberation of iodine (I
2
)
equivalent to the original DO content:
MnO
2
ðsÞþ2I
!Mn
2þ
þ I
2
þ2H
2
O
ð6
:
10Þ
The released I
2
can then be titrated with standard solution of sodium thiosulfate
(Na
2
S
2
O
3
) using a starch indicator:
I
2
þ2S
2
O
3
2
! S
4
O
6
2
þ2I
ð6
:
11Þ
The titration end point can be easily detected visually from blue to colorless. By
summing Eq. 6.9 to 6.11, the overall reaction of the Winkler method is:
2S
2
O
3
2
þ2H
þ
þ½O
2
! S
4
O
6
2
þH
2
O
ð6
:
12Þ
The above reaction indicates that 1 mol of O
2
is equivalent to 4 mol of thiosulphate
(S
2
O
2
3
) in the final titration. The DO can be calculated by the following equation:
NV8000
V
S
DOðmg
=
LÞ¼
ð6
:
13Þ
where N and Vare the normality and volume (mL) of Na
2
S
2
O
3
, respectively, V
s
is the
sample volume (typically 200 mL is withdrawn from 300 mL sample).
A thorough knowledge of the above chemical reactions helps analysts to
command the key step(s) for an accurate DO measurement. Understanding the
reaction stoichiometry is also required to fully comprehend the formula (Eq. 6.13)
used to calculate DO. For example, one may wonder how the conversion factor of
8,000 is included in Eq. 6.13. The example below is used to illustrate this point.
EXAMPLE 6.1.
(a) From the reaction stoichiometry shown in Eq. 6.12, derive the
conversion factor (8000) shown in Eq. 6.13. (b) The Standard Method (SM
20
) does not provide
any chemical reactions shown from Eq. 6.9 to 6.12, but calls for a equivalency of 1 mL
0.025 M Na
2
S
2
O
3
and 1 mg DO/L. Demonstrate this equivalency. (c) Explain whether a
positive or negative interference would it cause if oxidizing agents or reducing agents are
present in wastewater and effluent samples for DO measurement.
SOLUTION:
(a) From S
2
O
3
2
to S
4
O
6
2
, the oxidation number changes from þ2toþ2.5.
Since the net change isþ1 per molecule, meaning one mole of electron per mole S
2
O
2
3
is lost
to O
2
. This also means that 1 mol of thiosulfate (S
2
O
3
2
) is equal to 1 equivalent of the same
species. N V equivalents of S
2
O
3
2
will be equal to N V mol of S
2
O
3
2
or N V=4 mol
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