1. Dissolved Oxygen

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

3. Oxygen is Water Soluble Gas H2OH2O H2OH2O H2OH2O H2OH2O H2OH2O O2O2 H2OH2O H2OH2O H2OH2O O2O2 O2O2 O2O2 O2O2 O2O2 H2OH2O O2O2 H2OH2O What issues does that suggest?

4. Solubility is limited : in pure water -As temperature increases, the solubility decreases 100% DO Saturation 0 2 4 6 8 10 12 14 16 18 20 05101520253035 Temperature (C) 100% Saturation Lavel - As the atmospheric pressure increases, the solubility increases.

5. 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

6. 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)

7. Collection of water samples Special sample collection devices must be used that seal with no air. Bottle needs to be overfilled then capped.

8. “ 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.

9. 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.

10. The Winkler Method Fixing O 2 : 1) In a basic solution, the addition of MnSO 4 fixes the O 2 in a precipitate ( MnO 2 ). 2Mn +2 (aq) +O 2(g) +4OH - (aq) →2MnO 2(s) +2H 2 O (l) Oxidation number of Mn? 2) Acidified iodide ions, I-, are oxidized MnO 2(s) +2I - (aq) +4H + (aq) →Mn 2+ (aq) +I 2(aq) +2H 2 O (l)

11. The Winkler Fixing 2 Mn 2+ + O 2 + 4 OH - → 2 MnO 2 (s) + 2 H 2 O MnO 2 (s) + 2 I - + 4 H + → Mn 2+ + I 2 + 2 H 2 O Do you see the brilliance of this two-step sequence? The first step converts O 2 to MnO 2 under basic conditions. The second step converts MnO 2 to I 2 under acidic conditions. When you acidify the solution – you prevent the first reaction!!! Any oxygen that dissolves later can’t react!

12. The Winkler Method: Titration 1) 2Mn +2 (aq) +O 2(g) +4OH - (aq) →2MnO 2(s) +2H 2 O (l) 2) MnO 2(s) +2I - (aq) +4H + (aq) →Mn 2+ (aq) +I 2(aq) +2H 2 O (l) 3) The I 2 produced is then titrated with Na 2 S 2 O 3 and therefore the amount of O 2 originally present is determined. 2 S 2 O 3(aq) +I 2(aq) →2I - (aq) + S 4 O 6 2- (aq)

13. So, there are 3 reactions: 2Mn 2+ + O 2 + 2 OH - → 2MnO 2 (s) + 2H 2 O MnO 2 (s) + 2 I - + 4 H + → Mn 2+ + I 2 + 2 H 2 O 2 S 2 O 3 2- + I 2 → 2 I - + S 4 O 6 2- For every one mole of O 2 in the water, 2Mn 2+ + O 2 + 2 OH - → 2MnO 2 (s) + 2H 2 O 2MnO 2 (s) + 4 I - + 8 H + → 2Mn 2+ + 2 I 2 + 4 H 2 O 4 S 2 O 3 2- +2 I 2 → 4 I - + 2 S 4 O 6 2- 4 mol of S 2 O 3 2- are used

14. https://www.youtube.com/watch?v=u96zAOJACkc ( a ) 2Mn 2+ + O 2 + 2 OH - → 2MnO 2 (s) + 2H 2 O 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 ) MnO 2 (s) + 2 I - + 4 H + → Mn 2+ + I 2 + 2 H 2 O A strong acid acidifies the solution and converts the iodide ion (I-1) into an iodine molecule (I 2 ), causing the precipitate to dissolve (b) and the solution to turn brownish-orange (c). (d) 2 S 2 O 3 2- + I 2 → 2 I - + S 4 O 6 2- The solution is put on top of a stir plate and titrated with a thiosulfate solution. The titration is complete when the I 2 has reacted: the solution is colorless.

15. A sample problem: 250.0 mL of waste water is collected and fixed using the Winkler method. Titration of the I 2 produced requires the addition of 12.72 mL of a 0.0187 M Na 2 S 2 O 3 solution. What is the O 2 content of the wastewater expressed in mg/L?

16. Where would you start? Moles! Moles! Moles! 12.72 x10 -3 dm 3 Na 2 S 2 O 3 * 0.0187 M Na 2 S 2 O 3 = 0.238 x10 -3 mol Na 2 S 2 O 3 = 0.238 x10 -3 mol S 2 O 3 2- And so… Remember: M= Molarity = mole = moldm -3 dm 3

17. 2Mn 2+ + O 2 + 2 OH - → 2MnO 2 (s) + 2H 2 O MnO 2 (s) + 2 I - + 4 H + → Mn 2+ + I 2 + 2 H 2 O 2 S 2 O 3 2- + I 2 → 2 I - + S 4 O 6 2- 0.2379x10 -3 mol S 2 O 3 2- * 1 mol I 2 * 1 mol MnO 2 2 mol S 2 O 3 2- 1 mol I 2 = 0.1189 x10 -3 mol MnO 2 * 1 mol O 2 = 0.05947x 10 -3 mol O 2 2 mol MnO 2

18. 2Mn 2+ + O 2 + 2 OH - → 2MnO 2 (s) + 2H 2 O MnO 2 (s) + 2 I - + 4 H + → Mn 2+ + I 2 + 2 H 2 O 2 S 2 O 3 2- + I 2 → 2 I - + S 4 O 6 2- You could go directly from a ratio of O 2 : S 2 O 3 2- = 1:4 0.238 x10 -3 mol S 2 O 3 2- = 0.0595x10 -3 mol of O 2 4 Wait a minute….say it again……

19. 2Mn 2+ + O 2 + 2 OH - → 2MnO 2 (s) + 2H 2 O 2x ( MnO 2 (s) + 2 I - + 4 H + → Mn 2+ + I 2 + 2 H 2 O ) 2x ( 2 S 2 O 3 2- + I 2 → 2 I - + S 4 O 6 2- ) You could go directly from a ratio of O 2 : S 2 O 3 2- = 1:4 0.238 x10 -3 mol S 2 O 3 2- = 0.0595x10 -3 mol of O 2 4

20. Finishing… 0.0595 x10 -3 mol of O 2 = 0.0595 x10 -3 mol of O 2 0.2500 dm 3 0.2500 dm 3 =2.38x10 -4 mol O 2 / L of waste water. = 2.38 x10 -4 mol O 2 * 32.0 g O 2 /mol = = 7.62 x 10 -3 g of O 2 in 1 L of waste water = = 7.62 mg/L

21. = 7.62 mg/L = 7.62 x 10 -3 g O 2 / 1 kg water = 7.62 x 10 -6 kg O 2 / kg water = 7.62 ppm (part per million) Is the water saturated in O 2 ? What does that answer mean? Saturated pure water at room T° is ~ 8 mg/L.

22. BOD=Initial DO-Final DO (after 5 days) 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 What is the BOD for water A?