Dissolved Oxygen
Aquatic plants and phytoplankton (single
cell floating plants) release oxygen into the
water as a product of photosynthesis
CO2
O2
Oxygen is Water Soluble Gas
H2O
O2
H2O
H2O
H
O
2
O
O2
2
O2
O2 H O H O
H
O
2
2
2
O2
O2 H O
2
H
O
2
HO
2
What issues does that suggest?
Solubility is limited : in pure water
100% Saturation Lavel
-As temperature increases, the solubility decreases
20
18
16
14
12
10
8
6
4
2
0
0
100% DO Saturation
5
10
15
20
25
Temperature (C)
30
35
-As the atmospheric pressure increases, the
solubility increases.
Normal solubility of oxygen in pure water
at 1 atm and 25° C is 8 mg/L.
This is a modest value – oxygen is considered to be a
poorly soluble gas in water!
Weak intermolecular force: Which one?
Impure water will typically have a value
less than 8 mg/L
Solubility is about equilibrium
Keep in mind that “solubility” is an equilibrium value
representing the MAXIMUM amount that can be
dissolved.
Equilibrium is not achieved instantaneously – it takes
time for oxygen to be absorbed (or desorbed) from
water.
Oxygen can be lost to or gained from the air after
collection. (usually gained)
Collection of water samples
Special sample collection devices must be used
that seal with no air.
Bottle needs to be overfilled then capped.
“Fixing” the oxygen content
Immediately after collection, sometimes before
reaching the lab, the oxygen content of the samples
is “fixed” by conversion to another material that is
later titrated in the lab.
Even after fixing, you need to minimize biological
activity in the samples that could create new oxygen
by “chewing” on the chemicals.
How do you minimize biological activity?
Ice – if you aren’t warm blooded, you always slow
down in the cold.
Dark – many water species are photosynthetic and
can’t do anything in the dark.
Poison – add enough chemicals in the fixing process to
kill a lot of the normal biological species in the water
sample.
The Winkler Method
Fixing O2:
1) In a basic solution, the addition of MnSO4 fixes
the O2 in a precipitate ( MnO2).
2Mn+2(aq)+O2(g)+4OH-(aq)→2MnO2(s)+2H2O(l)
Oxidation number of Mn?
2) Acidified iodide ions, I-, are oxidized
MnO2(s)+2I-(aq)+4H+(aq)→Mn2+(aq)+I2(aq) +2H2O(l)
The Winkler Fixing
2 Mn2+ + O2 + 4 OH- → 2 MnO2 (s) + 2 H2O
MnO2 (s) + 2 I- + 4 H+ → Mn2+ + I2 + 2 H2O
Do you see the brilliance of this two-step sequence?
The first step converts O2 to MnO2 under basic conditions.
The second step converts MnO2 to I2 under acidic conditions.
When you acidify the solution – you prevent the first reaction!!!
Any oxygen that dissolves later can’t react!
The Winkler Method: Titration
1) 2Mn+2(aq)+O2(g)+4OH-(aq)→2MnO2(s)+2H2O(l)
2) MnO2(s)+2I-(aq)+4H+(aq)→Mn2+(aq)+I2(aq) +2H2O(l)
3) The I2 produced is then titrated with Na2S2O3 and
therefore the amount of O2 originally present is
determined.
2 S2O3(aq)+I2(aq) →2I-(aq) + S4O62- (aq)
So, there are 3 reactions:
2Mn2+ + O2 + 2 OH- → 2MnO2 (s) + 2H2O
MnO2 (s) + 2 I- + 4 H+ → Mn2+ + I2 + 2 H2O
2 S2O32- + I2 → 2 I- + S4O62-
For every one mole of O2 in the water,
2Mn2+ + O2 + 2 OH- → 2MnO2 (s) + 2H2O
2MnO2 (s) + 4 I- + 8 H+ → 2Mn2+ + 2 I2 + 4 H2O
4 S2O32- +2 I2 → 4 I- + 2 S4O624 mol of S2O32- are used
https://www.youtube.com/watch?v=u96zAOJACkc
( a)
2Mn2+ + O2 + 2 OH- → 2MnO2 (s) + 2H2O
The addition of excess manganese and hydroxide ions added to each water sample forms a
precipitate (solid). Mn+2 is oxidized by the dissolved oxygen in the water sample.
( b –c )
MnO2 (s) + 2 I- + 4 H+ → Mn2+ + I2 + 2 H2O
A strong acid acidifies the solution and converts the iodide ion (I-1) into an iodine molecule (I2),
causing the precipitate to dissolve (b) and the solution to turn brownish-orange (c).
(d)
2 S2O32- + I2 → 2 I- + S4O62-
The solution is put on top of a stir plate and titrated with a thiosulfate solution.
The titration is complete when the I2 has reacted: the solution is colorless.
A sample problem:
250.0 mL of waste water is collected and
fixed using the Winkler method. Titration of
the I2 produced requires the addition of
12.72 mL of a 0.0187 M Na2S2O3 solution.
What is the O2 content of the wastewater
expressed in mg/L?
Where would you start?
Moles! Moles! Moles!
Remember:
M= Molarity = mole = moldm-3
dm3
12.72 x10-3 dm3 Na2S2O3 * 0.0187 M Na2S2O3 =
0.238 x10-3mol Na2S2O3 = 0.238 x10-3 mol S2O32And so…
2Mn2+ + O2 + 2 OH- → 2MnO2 (s) + 2H2O
MnO2 (s) + 2 I- + 4 H+ → Mn2+ + I2 + 2 H2O
2 S2O32- + I2 → 2 I- + S4O620.2379x10-3mol S2O32- * 1 mol I2
*
2 mol S2O32-
1 mol MnO2
1 mol I2
= 0.1189 x10-3mol MnO2 * 1 mol O2 = 0.05947x 10-3mol O2
2 mol MnO2
2Mn2+ + O2 + 2 OH- → 2MnO2 (s) + 2H2O
MnO2 (s) + 2 I- + 4 H+ → Mn2+ + I2 + 2 H2O
2 S2O32- + I2 → 2 I- + S4O62You could go directly from a ratio of O2: S2O32- = 1:4
0.238 x10-3mol S2O32- = 0.0595x10-3mol of O2
4
Wait a minute….say it again……
2Mn2+ + O2 + 2 OH- →
2x ( MnO2 (s) + 2 I- + 4 H+ →
2x (
2 S2O32- + I2 →
2MnO2 (s) + 2H2O
Mn2+ + I2 + 2 H2O )
2 I- + S4O62- )
You could go directly from a ratio of O2: S2O32- = 1:4
0.238 x10-3mol S2O32- = 0.0595x10-3mol of O2
4
Finishing…
0.0595 x10-3mol of O2
0.2500 dm3
= 0.0595 x10-3mol of O2
0.2500 dm3
=2.38x10-4 mol O2 / L of waste water.
= 2.38 x10-4 mol O2 * 32.0 g O2 /mol =
= 7.62 x 10-3 g of O2 in 1 L of waste water =
= 7.62 mg/L
What does that answer mean?
= 7.62 mg/L = 7.62 x 10-3 g O2 / 1 kg water
= 7.62 x 10-6 kg O2 / kg water
= 7.62 ppm (part per million)
Is the water saturated in O2?
Saturated pure water at room T° is ~ 8 mg/L.
BOD=Initial DO-Final DO
(after 5 days)
What is the BOD for water A?
If all the DO is used up the test is invalid, as in B above
To get a valid test dilute the sample, as in C above. In this
case the sample was diluted by 1:10. The BOD can then be
calculated by:
BOD = (I – F) D
D = dilution as a fraction
D = vol. of diluted sample / vol. of initial sample)
BOD = (8 – 4) 10 = 40 mg/L