Dissolved Oxygen
Photosynthesis: Your one-stop shop for all of your oxygen needs!
Carbon Dioxide
(from air)
Water (from ground)
Oxygen (to
air)Carbohydrate (plant material)
Solar energy + 6CO2 + 6H2O → C6H12O6 + 6O2
CO2 O2
Aquatic plants and phytoplankton (single cell floating plants) release oxygen into the water as a product of photosynthesis
Oxygen is Water Soluble Gas
H2O
H2O
H2O
H2O
H2O
O2 H2OH2O
H2OO2
O2
O2
O2
O2
H2OO2
H2O
What issues does that suggest?
Solubility is limited : in pure water
-As temperature increases, the solubility decreases 100% DO Saturation
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0 5 10 15 20 25 30 35Temperature (C)
100
% S
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-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:
Solution A, B, C 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)
Solution D
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) + S4O6
2- (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- + S4O6
2-
For every one mole of O2 in the water,
4 mol of S2O32- are used
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 S4O6
2-
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!
12.72 mL Na2S2O3 * 0.0187 M Na2S2O3 =
0.238 mmol Na2S2O3 = 0.238 mmol S2O32-
And so… m = milli = 10-3m = milli = 10-3
m = milli = 10-3
Remember:
M= Molarity = mole = mmol L ml
Where would you start?
2Mn2+ + O2 + 2 OH- → 2MnO2 (s) + 2H2O
MnO2 (s) + 2 I- + 4 H+ → Mn2+ + I2 + 2 H2O
2 S2O32- + I2 → 2 I- + S4O6
2-
0.2379 mmol S2O32- * 1 mol I2 * 1 mol MnO2
2 mol S2O32- 1 mol I2
= 0.1189 mmol MnO2 * 1 mol O2 = 0.05947 mmol O2
2 mol MnO2
Where would you start?
2Mn2+ + O2 + 2 OH- → 2MnO2 (s) + 2H2O
MnO2 (s) + 2 I- + 4 H+ → Mn2+ + I2 + 2 H2O
2 S2O32- + I2 → 2 I- + S4O6
2-
You could go directly from a ratio of O2: S2O32- = 1:4
0.238 mmol S2O32- = 0.0595 mmol of O2
4
Finishing…
0.0595 mmol of O2 = 0.0595 x10-3mol of O2
0.2500 L 0.2500 L
=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
= 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.
What does that answer mean?
How would you determine the BOD ( biological oxygen demand) ?
BOD=Initial DO-Final DO (after 5 days)
If all the DO is used up the test is invalid, as in B aboveTo 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
What is the BOD for water A?
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