Quantitative Determination of Dissolved Oxygen Content by Winkler Redox Titration

QUANTITATIVE DETERMINATION OF DISSOLVED OXYGEN CONTENT BY WINKLER REDOX TITRATION R.Y.A. RIVERA DEPARTMENT OF FOOD SCIENCE AND NUTRITION, COLLEGE OF HOME ECONOMICS UNIVERSITY OF THE PHILIPPINES, DILIMAN, QUEZON CITY 1101, PHILIPPINES DATE SUBMITTED: 21 APRIL 2015 DATE PERFORMED: 7 APRIL 2015

ANSWERS TO QUESTIONS 1. Give the pertinent chemical equations and stoichiometry in the standardization of Na₂S₂O₃. IO₃⁻ + 8I⁻ + 6H⁺ →3I₃⁻ + 3H₂O (1) 2S₂O₃²⁻ + I₃⁻ → S₄O₆⁻ + 3I⁻ (2) 2. Explain the purpose of the addition of H₂SO₄ and excess KI during standardization and why the acid was added before KI. The production of I₃⁻ 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, HIO₃ will be formed instead of I₃⁻. 3. Give the pertinent chemical equations and stoichiometry in the sample analysis. MnSO₄ → Mn²⁺ + SO₄⁻ Mn²⁺ + 2OH⁻ → Mn(OH)₂ O₂ + 4Mn(OH)₂ + 2H₂O + 2H₂O ↔ 4Mn(OH)₃ / O₂ + 4Mn(OH)₂ ↔ 4MnO(OH) + 2H₂O 2Mn(OH)₃ + 2I⁻ + 6H⁺ → 2Mn²⁺ + I₂ + 6H₂O / 6H⁺ +2MnO(OH) + 2I⁻ → 2Mn²⁺ + I₂ +4H₂O I₂ + I⁻ → I₃⁻ 2S₂O₃²⁻ + I₃⁻ → S₄O₆⁻ + 3I⁻ 4. Explain stepwise how I₃ was produced from the dissolved O₂ in the water sample. Explain briefly why the reagents are added in a definite sequence. MnSO4 → Mn2+ + SO4MnSO¬4 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)2 The Mn2+ ions will then bind to the free OH- ions in the solution or to the OHfrom the added NaOH to form Mn(OH)2. O2 + 4Mn(OH)2 + 2H2O ↔ 4Mn(OH)3 / O2 + 4Mn(OH)2 ↔ 4MnO(OH)(s) + 2H2O The 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 + 4H2O Addition 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- → I3Produced I2 will react with excess I- to form an iodine cmplex, I3-. 2S2O32- + I3- → S4O6- + 3IUpon 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 O₂, 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 analysis Calculated 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 of 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. MnSO₄ is added and the solution made to stand for an hour before the alkaline KI was added. Mn²⁺ will react with O₂ 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. Diferent unwanted products whose efect 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 MnSO₄ 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 HIO₃ and less formation of I₃⁻. Consequently, the amount of I₂ will decrease and the amount of the calculated DO will also decrease.