Happy Days video (2:30)

 Consider the balanced redox reaction of potassium manganate(VII) with ammonium iron(II) sulfate. 5Fe 2+ + MnO H +  5Fe 3+ + Mn H 2 O notice spectator ions are left out of the equation and only focusing what was reduce and oxidized  In a titration to determine the concentration of a potassium manganate(VII) solution, 28.0 cm 3 of the potassium manganate(VII) reacted completely with 25.0 cm 3 of a mol/dm -3 solution of iron(II) sulfate. Determine the concentration, in g dm -3, of the potassium manganate(VII) solution.

from the periodic table

 aquatic life depends on CO 2 an O 2 dissolved in water  O 2 is non-polar, while H 2 0 is polar › therefore, solubility of oxygen in water is very low  it decreases with increase in temperature  0°C is 14.6 ppm (parts per million)  20°C is 7.6 ppm (parts per million)

 can be used to measure the degree of pollution in a water sample › saturate a sample of water with oxygen › incubate for 5 days to allow microorganisms to oxidize the organic matter › use an iodine/thiosulfate redox titration to measure amount of remaining dissolved oxygen › the measurement is biochemical oxygen demand (BOD)  defined as the amount of O 2 required by aerobic biological organisms to break down organic material present in a given water sample for 5 days  measured in ppm or mg/L (same thing)

 when organic matter is discharged into water, it provides a source of food for bacteria › when bacteria multiply and their uptake of oxygen is greater than what is replaced by photosynthesis, the body of water becomes depleted of oxygen  in these anaerobic conditions  methane (CH 4 ) is produced instead of CO 2  hydrogen sulfide (H 2 S) and ammonia (NH 3 ) instead of water, sulfates, and nitrates

 An iodine/thiosulfate redox titration is carried out to measure the dissolved oxygen present in a water sample: A 50 cm 3 water sample is taken and saturated with oxygen for five days › The following reactions (don’t need to know) are related to the Winkler method  2 Mn 2+ (aq) + O 2 (aq) + 4OH – → 2 MnO(OH) 2 (s)  MnO(OH) 2 (s) + 2 I – (aq) + 4H + → Mn 2+ (aq) + I 2 (aq) + 3H 2 O  2 S 2 O 3 2- (aq) + I 2 (aq) → S 4 O 6 2- (aq) + 2 I – (aq) › It was found that 5.25 cm 3 of a mol dm -3 solution of sodium thiosulfate (Na 2 S 2 O 3 ) was required to react with the iodine produced.

 Determine the concentration of dissolved oxygen in ppm in the sample of water › molarity of O 2 is 1.31 x mol dm -3 › ppm is 4.19  Deduce the BOD, in ppm, of the water sample assuming the maximum solubility of oxygen in water is 9.00 ppm at 293K. › BOD is 4.81 ppm  Comment on the BOD value obtained. › Results show reasonable water quality  untreated sewage has a BOD range of ppm.