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Happy Days video (2:30)
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Consider the balanced redox reaction of potassium manganate(VII) with ammonium iron(II) sulfate. 5Fe 2+ + MnO 4 - + 8H + 5Fe 3+ + Mn 2+ + 4H 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 0.0100 mol/dm -3 solution of iron(II) sulfate. Determine the concentration, in g dm -3, of the potassium manganate(VII) solution.
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from the periodic table
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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)
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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)
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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
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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 0.00500 mol dm -3 solution of sodium thiosulfate (Na 2 S 2 O 3 ) was required to react with the iodine produced.
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Determine the concentration of dissolved oxygen in ppm in the sample of water › molarity of O 2 is 1.31 x 10 -4 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 100-400 ppm.
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