QUANTITATIVE DETERMINATION OF DISSOLVED OXYGEN CONTENT BY WINKLER REDOX TITRATIONR.Y.A. RIVERADEPARTMENT OF FOOD SCIENCE AND NUTRITION, COLLEGE OF HOME ECONOMICSUNIVERSITY OF THE PHILIPPINES, DILIMAN, QUEZON CITY 1101, PHILIPPINESDATE SUBMITTED: 21 APRIL 2015DATE PERFORMED: 7 APRIL 2015

ANSWERS TO QUESTIONS1. Give the pertinent chemical equations and stoichiometry in the standardization of NaSO.IO + 8I + 6H 3I + 3HO (1)2SO + I SO + 3I (2)2. Explain the purpose of the addition of HSO and excess KI during standardization and why the acid was added before KI.The production of I needed excess KI and sulfuric acid to be added in the solution. Yhe sulfuric acid provided H ions while KI provided I ions. If the acid was added first instead of KI crystals, HIO will be formed instead of I.3. Give the pertinent chemical equations and stoichiometry in the sample analysis.MnSO Mn + SOMn + 2OH Mn(OH)O + 4Mn(OH) + 2HO + 2HO 4Mn(OH) / O + 4Mn(OH) 4MnO(OH) + 2HO2Mn(OH) + 2I + 6H 2Mn + I + 6HO / 6H +2MnO(OH) + 2I 2Mn + I +4HOI + I I2SO + I SO + 3I4. Explain stepwise how I was produced from the dissolved O in the water sample. Explain briefly why the reagents are added in a definite sequence.MnSO4 Mn2+ + SO4-MnSO4 was added to provide Mn2+ ions in the solution. Mn2+ ions are necessary since Winkler Method is based in the ability of dissolved O2 to oxidize divalent manganese added to the solution. NH4HCO3 was added to convert Mn into oxygen-sensitive carbonates and eliminate the influence/interference of other dissolved organic compounds.

Mn2+ + 2OH- Mn(OH)2The Mn2+ ions will then bind to the free OH- ions in the solution or to the OH- from the added NaOH to form Mn(OH)2.

O2 + 4Mn(OH)2 + 2H2O 4Mn(OH)3 / O2 + 4Mn(OH)2 4MnO(OH)(s) + 2H2OThe produced Mn(OH)2 will then be oxidized if dissolved oxygen is present to form either Mn(OH)3 or MnO(OH) precipitate. 2Mn(OH)3 + 2I- + 6H+ 2Mn2+ + I2 + 6H2O / 6H+ + 2MnO(OH)(s) + 2I- 2Mn2+ + I2 + 4H2OAddition of KI introduces excess I- into the solution and the phosphoric acid provides H+ ions that acidifies the precipitate and oxidizes the excess iodide in the solution. Mn2+ and I2 will then be produced.I2 + I- I3-Produced I2 will react with excess I- to form an iodine cmplex, I3-.

2S2O32- + I3- S4O6- + 3I-Upon titration with thiosulfate, the iodine complex will be reduced to I-.

The reagents are added accordingly to ensure the formation of I3-. Changes in the sequence will yield unwanted products not needed in the titration.

5. Give the reason why starch is was used as the indicator in this analysis and why it was added towards the end of the titration.6. How is the analysis (an iodometric process) different form an iodimetric one?The analysis involved an iodometric process because the analyte which was an oxidizing agent was added to excess iodide to produce iodine. The iodine produced was determined by titration with sodium thiosulfate. An iodimetric process, unlike the previous one, the analyte is a reducing agent and is titrated directly with a standard iodine solution.7. Form the calculated ppm O, identify the degree of water pollution and ability of the water sample source to sustain aquatic life.The calculated dissolved oxygen in water sample obtained from the pond near the Math building was . This value lies between the range which means that the water quality is 8. Predict the effect, if any, of each on the following on the DO oxygen content obtained:a. The water sample is made to stand overnight before analysisCalculated DO can either increase or decrease. Since the water sample was obtained from a pond, oxygen consuming microorganisms could be present in the sample. These microorganisms undergo cellular respiration that require oxygen, thus the calculated DO will decrease. Moreover, photosynthetic phytoplanktons could also be present in the samples. These organisms on the other hand give off oxygen as a product of the light reaction phase of photosynthesis thus calculated DO could increase. However, if the sample was left to stand overnight, the planktons will concentrate on the dark reactions phase of photosynthesis, and the light reaction phase will not proceed since sunlight is needed in this phase. Dissolved oxygen will then stop increasing once the light supply is no longer available but will still decrease because of the presence of oxygen consuming microbes.b. MnSO is added and the solution made to stand for an hour before the alkaline KI was added.Mn will react with O to produce MnO, compound insoluble in water. Because of this, calculated DO will decrease in value.9. What are the possible sources of errors and their effect on the calculated parameters? Rationalize.A possible source of error can be committed during the collection and treatment of the water sample. Presence of air bubbles during collection will give a higher amount of calculated DO. Incorrect sequence of adding the reagents in the treatment of the water sample would result to a gross error. Different unwanted products whose effect on the calculated DO is indeterminate might be formed. Prolonged interval in the adding of reagents could also be a source of errors. For example, prolonged interval in the addition of MnSO and KI would lead to the production of insoluble MnO which will consume dissolved oxygen in the sample and thus decrease the calculated DO. Another possible source of error is during titration if the sulfuric acid is added first before the KI crystals. This will cause the formation of HIO and less formation of I. Consequently, the amount of I will decrease and the amount of the calculated DO will also decrease.