Redox bod ppt
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This document discusses oxidation-reduction (redox) reactions and their environmental applications. It defines key terms like: 1) Biochemical oxygen demand (BOD) which measures the amount of oxygen used by microorganisms to break down organic waste in water, with higher BOD indicating more pollution. 2) Parts per million (ppm) which is used to measure very dilute solutions like dissolved oxygen in water, important for sustaining aquatic life. 3) The Winkler titration method for determining dissolved oxygen concentration in water samples, which involves a series of redox reactions and titration with sodium thiosulfate.
Redox bod ppt
- 1. 9.1 OXIDATION AND REDUCTION By: Merinda Sautel Alameda Int’l Jr/Sr High School Lakewood, CO msautel@jeffco.k12.co.us
- 2. Topic 9.1 Happy Days video (2:30)
- 3. Example Problem Consider the balanced redox reaction of potassium manganate(VII) with ammonium iron(II) sulfate. 5Fe2+ + MnO4 - + 8H+ 5Fe3+ + Mn2+ + 4H2O 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 cm3 of the potassium manganate(VII) reacted completely with 25.0 cm3 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.
- 4. on’t want to use the previous formula because I already know how to do this and don’t want to learn a new formula. 𝑴𝑨 = 𝒏 𝑽 … look at balanced equation…. 𝑴𝑩 = 𝒏 𝑽 𝟎. 𝟎𝟏𝟎𝟎𝑴 = 𝒏 .𝟎𝟐𝟓𝟎𝑳 …mole(n) ratio… 𝑴𝑩 = 𝒏 .𝟎𝟐𝟖𝟎𝑳 n = .000250 mole(n) ratio is 5 to 1 n = .0000500 𝑴𝑩 = .𝟎𝟎𝟎𝟎𝟓𝟎𝟎 .𝟎𝟐𝟖𝑳 𝑴𝑩 = 𝟎. 𝟎𝟎𝟏𝟕𝟗 need answer in g dm-3, not Molarity 𝟎. 𝟎𝟎𝟏𝟕𝟗 𝐦𝐨𝐥 𝐊𝐌𝐧𝐎𝟒 𝒙 𝟏𝟓𝟖. 𝟎𝟒 𝒈 𝐊𝐌𝐧𝐎𝟒 𝟏 𝒎𝒐𝒍 𝐊𝐌𝐧𝐎𝟒 0.283 g dm -3 from the periodic table
- 6. Enviromental application of redox chemistry aquatic life depends on CO2 an O2 dissolved in water O2 is non-polar, while H20 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)
- 7. ppm is used for very dilute solutions = 𝒎𝒂𝒔𝒔 𝒐𝒇 𝒄𝒐𝒎𝒑𝒐𝒏𝒆𝒏𝒕 𝒊𝒏 𝒔𝒐𝒍𝒖𝒕𝒊𝒐𝒏 𝒊𝒏 𝒈𝒓𝒂𝒎𝒔 𝒕𝒐𝒕𝒂𝒍 𝒎𝒂𝒔𝒔 𝒐𝒇 𝒔𝒐𝒍𝒖𝒕𝒊𝒐𝒏 𝒊𝒏 𝒈𝒓𝒂𝒎𝒔 x 106 .00419 g in 1 liter of solution (1 liter H2O = 1000 ml = 1000 g) .00419 g / 1000 g x 106 = 4.19 ppm to sustain a healthy aquatic environment, dissolved oxygen should be >6 ppm
- 8. BOD Biochemical oxygen demand (also called biological oxygen demand), or BOD for short, is used as a measure of the quality of water. It is a measure of the amount of oxygen used by microorganisms to oxidise the organic matter in the water.
- 9. Any organic pollutants in river water will be decomposed (oxidised) by microorganisms (aerobic bacteria) in the water and this process uses up dissolved oxygen. The higher the BOD, the more organic waste there is in water.
- 10. If, for instance, sewage is released into a river or lake this will greatly increase the BOD – the water is more polluted. BOD is defined as the amount of oxygen used by the aerobic microorganisms in water to decompose the organic matter in the water over a fixed period of time (usually 5 days) at a fixed temperature (usually 20 °C).
- 11. The basic principle in measuring BOD is to compare the initial amount of dissolved oxygen in a sample of water with the amount present when the sample has been incubated for 5 days at 20 °C. If a dissolved oxygen concentration of 9 ppm, which after incubation for 5 days falls to 4 ppm, the BOD is 9 − 4, or 5 ppm.
- 12. A typical method for determining the amount of dissolved oxygen is the Winkler titration method. 1. The basic chemistry behind the Winkler method is that manganese(II) sulfate is added to the water and the manganese(II) ions are oxidised 2. Manganese(II) sulfate is converted to manganese(II) hydroxide in the presence of hydroxide ions.
- 13. The sample is acidified with sulfuric acid to produce manganese(IV) sulfate: There is no change in oxidation number in this reaction. Iodide ions are oxidised to I2 by the manganese(IV) ions:
- 14. This iodine can then be titrated against a standard sodium thiosulfate solution: The outcome from these equations is that the number of moles of dissolved oxygen is 1/4 of the number of moles of sodium thiosulfate used in the titration – or the mass of oxygen is eight times the number of moles of sodium thiosulfate.