Dissolved Oxygen. CO 2 O2O2 Aquatic plants and phytoplankton (single cell floating plants) release oxygen into the water as a product of photosynthesis.

Dissolved Oxygen

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)

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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 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) + 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,

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-

4 mol of S2O3

2- 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- + S4O6

2-

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!

12.72 x10-3 dm3 Na2S2O3 * 0.0187 M Na2S2O3 =

0.238 x10-3mol Na2S2O3 = 0.238 x10-3 mol S2O32-

And so…

Remember:

M= Molarity = mole = moldm-3

dm3

2Mn2+ + O2 + 2 OH- → 2MnO2 (s) + 2H2O

MnO2 (s) + 2 I- + 4 H+ → Mn2+ + I2 + 2 H2O 2 S2O3

2- + I2 → 2 I- + S4O62-

0.2379x10-3mol S2O32- * 1 mol I2 * 1 mol MnO2

2 mol S2O32- 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- + S4O6

2-

You 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- → 2MnO2 (s) + 2H2O

2x ( MnO2 (s) + 2 I- + 4 H+ → Mn2+ + I2 + 2 H2O ) 2x ( 2 S2O3

2- + I2 → 2 I- + S4O62- )

You could go directly from a ratio of O2: S2O32- = 1:4

0.238 x10-3mol S2O3

2- = 0.0595x10-3mol of O2 4

Finishing…

0.0595 x10-3mol of O2 = 0.0595 x10-3mol of O2

0.2500 dm3 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

= 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?

What does that answer mean?

Saturated pure water at room T° is ~ 8 mg/L.

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?