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

02468

101214161820

0 5 10 15 20 25 30 35Temperature (C)

100

% S

atur

atio

n La

vel

-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?