Dissolved Oxygen in Water Sample Question 1. Calculate the dissolved oxygen for 300 ml sample of...
Dissolved Oxygen in Water Sample
Question 1. Calculate the dissolved oxygen for 300 ml sample of water that was analyzed using Winkler method with azide modification. The sample were titrated against standard 0.025 N of sodium thiosulphate, and the required volume of sodium thiosulphate to reach the colorless endpoint was 13.5ml.
Question 2. A student conducted an experiment to measure the
biochemical oxygen demand for 400 ml of waste water. She diluted
the sample by adding 600 ml of oxygen saturated distilled water.
Her dissolved oxygen results of the samples were as follow:
DO for waste water sample on day 0 = 8.24 mg/L
DO for waste water sample after 5 days = 7.01 mg/L
DO for blank sample on day 0 = 9.17 mg/L
DO for blank sample after 5 days = 9.09 mg/L
Calculate BOD5 20°C.
(Note dilution fraction = volume of waste water / (volume of waste
water + volume of dilution water).
Homework Answers
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Dissolved Oxygen in Water Sample
Question 1. Calculate the dissolved oxygen for 300 ml sample of...
QUESTION1 A 15-mL sample is added to à 300-ml 8OD bottle. Dilution water is added to the sample bottle and the initi...
QUESTION1 A 15-mL sample is added to à 300-ml 8OD bottle. Dilution water is added to the sample bottle and the initial dissoved axygen concentration is measured as 8.5 mg/L. After the sample is sealed, the laboratory incorrectly takes a measurement of dissolved oxygen reading on day 6 of 3 mg/L. If the BOD reaction rate coefticient for the sample is 0.30/day, estimate what the 5-day BOD should have been QUESTION 2 A 15-mL sample is added to a 300-mL...
Q 1(a) [13.33 Marks] Levels of dissolved oxygen in water can be determined by the Winkler...
Q 1(a) [13.33 Marks] Levels of dissolved oxygen in water can be determined by the Winkler method in the Leaving Certificate Chemistry course as follows: Under alkaline conditions manganese(II) sulphate produces a white precipitate of manganese(II) hydroxide: Mn2+(aq) + 2OH-(aq) → Mn(OH)2(s) This reacts with the dissolved oxygen to produce a brown precipitate. 2Mn(OH)2(s) + O2(aq) → 2MnO(OH)2(s) Addition of concentrated H2SO4 releases free iodine from KI. MnO(OH)2(s) + 4H+(aq) + 21-(aq) + Mn2+(aq) + 12(aq) + 3H2O(1) The free...
While interning at a wastewater treatment plant, you conduct a seeded BOD5 test of the lant's...
While interning at a wastewater treatment plant, you conduct a seeded BOD5 test of the lant's effluent. For the experimental bottle, you insert 30 mL of test sample into the 300-mL ottle and fill the rest with seeded dilution water. You fill a blank bottle with seeded dilution ater. You measure the following dissolved oxygen (DO) concentrations at the start and end f the 5-day test: Bottle Initial DO (mg/L) Final DO (mg/L) Experimental 8.1 4.6 Blank 7.8 6.7 hat...
4. A 20 mL wastewater sample was mixed with 280 mL deionized (DI) water in a...
4. A 20 mL wastewater sample was mixed with 280 mL deionized (DI) water in a standard BOD bottle. The 5- day BOD test provided an initial dissolved oxygen concentration (DO intial) = 9 mg/L and a final dissolved oxygen concentration (DO final) = 3.5 mg/L. The BOD experiment was carried out at 20°C and the BOD rate constant at 20°C is 0.22/day. a. Find the dilution factor (P) and BODs value at 20°CY b. Find the ultimate BOD of...
Question 13 Calculate the amount of dissolved oxygen (ppm 02) using the following data: • Sample...
Question 13 Calculate the amount of dissolved oxygen (ppm 02) using the following data: • Sample volume (mL) = 224.893 • Buret Reading, initial (mL) = 3.581 • Buret Reading, final (mL) = 18.25 • Volume of Na2S203 dispensed (mL) = • Average molar concentration of Na2S203 (mol/L) = 0.0219 • Moles of Na2S203 dispensed (mol) = • Moles of 13° reduced by S2032-(mol) = • Moles of O2 (mol) = • Mass of O2 (mg) • Dissolved oxygen, ppm...
De constant kiday"). 8. A standard BOD test is run using seeded dilution water. In one...
De constant kiday"). 8. A standard BOD test is run using seeded dilution water. In one bottle, the water sample is mixed with seeded dilution water giving a dilution of 1:30. Another bottle, the blank, contains just seeded dilution water. Both bottles begin the test with DO at the saturation value of 9.2 mg/L. After five days, the bottle containing waste has DO equal to 2.5 mg/L while that containing just seeded dilution water has DO equal to 8 mg/L....
The actual concentration was found to be 0.002364 M. 4. Knowing that 1 mole of dissolved...
The actual concentration was found to be 0.002364 M.
4. Knowing that 1 mole of dissolved oxygen ends up reacting with 4 moles of the titrant S2032, how do you calculate the amount of dissolved oxygen (in units of mg/L) from the volume of titrant added? Show your work. (2.5 pts) • Use X (mL) as your total volume of titrant added. • Use the actual concentration of the -0.0025 M NazS2O3 solution used during the titration (calculated in question...
Calculate the concentration in units of mg/L for oxygen dissolved in 1050.0 mL of water at...
Calculate the concentration in units of mg/L for oxygen dissolved in 1050.0 mL of water at 25 °C, if the partial pressure of O2 in gas phase in equilibrium with the water is 238.1 mm Hg. Assume the density of water =1 g/mL
5. The following data is for a BOD determination run on a raw sewage sample incubated...
5. The following data is for a BOD determination run on a raw sewage sample incubated for 5 days at 20°C in the dark; calculate the BOD5,20. Raw sewage contains a sufficient population of bacteria to conduct a BOD test. Therefore, this is an unseeded sample. Data: BOD bottle volume mL raw sewage added to BOD bottles DO, in bottle containing waste and dilution water DO in bottle containing waste and dilution water 300 mL 10.00 mL 7.70 mg/L 2.35...
ENV3006C HW#4 The following data provided to you. Calculate the BODS of the WW sample (mg/L)...
ENV3006C HW#4 The following data provided to you. Calculate the BODS of the WW sample (mg/L) are Seeded wastewater (WW): Initial DO 8.2 mg/L; Volume of wastewater Final DO (after 5 days) 2.5 mg/L Total volume in BOD bottle = 300 mL 5 mL Seeded dilution water: Final DO (after 5 days) 7.3 mg/L Initial DO 8.3 mg/L; Volume of seeded dilution water 300 mL
QUESTION1 A 15-mL sample is added to à 300-ml 8OD bottle. Dilution water is added to the sample bottle and the initial dissoved axygen concentration is measured as 8.5 mg/L. After the sample is sealed, the laboratory incorrectly takes a measurement of dissolved oxygen reading on day 6 of 3 mg/L. If the BOD reaction rate coefticient for the sample is 0.30/day, estimate what the 5-day BOD should have been QUESTION 2 A 15-mL sample is added to a 300-mL...
Q 1(a) [13.33 Marks] Levels of dissolved oxygen in water can be determined by the Winkler method in the Leaving Certificate Chemistry course as follows: Under alkaline conditions manganese(II) sulphate produces a white precipitate of manganese(II) hydroxide: Mn2+(aq) + 2OH-(aq) → Mn(OH)2(s) This reacts with the dissolved oxygen to produce a brown precipitate. 2Mn(OH)2(s) + O2(aq) → 2MnO(OH)2(s) Addition of concentrated H2SO4 releases free iodine from KI. MnO(OH)2(s) + 4H+(aq) + 21-(aq) + Mn2+(aq) + 12(aq) + 3H2O(1) The free...
While interning at a wastewater treatment plant, you conduct a seeded BOD5 test of the lant's effluent. For the experimental bottle, you insert 30 mL of test sample into the 300-mL ottle and fill the rest with seeded dilution water. You fill a blank bottle with seeded dilution ater. You measure the following dissolved oxygen (DO) concentrations at the start and end f the 5-day test: Bottle Initial DO (mg/L) Final DO (mg/L) Experimental 8.1 4.6 Blank 7.8 6.7 hat...
4. A 20 mL wastewater sample was mixed with 280 mL deionized (DI) water in a standard BOD bottle. The 5- day BOD test provided an initial dissolved oxygen concentration (DO intial) = 9 mg/L and a final dissolved oxygen concentration (DO final) = 3.5 mg/L. The BOD experiment was carried out at 20°C and the BOD rate constant at 20°C is 0.22/day. a. Find the dilution factor (P) and BODs value at 20°CY b. Find the ultimate BOD of...
Question 13 Calculate the amount of dissolved oxygen (ppm 02) using the following data: • Sample volume (mL) = 224.893 • Buret Reading, initial (mL) = 3.581 • Buret Reading, final (mL) = 18.25 • Volume of Na2S203 dispensed (mL) = • Average molar concentration of Na2S203 (mol/L) = 0.0219 • Moles of Na2S203 dispensed (mol) = • Moles of 13° reduced by S2032-(mol) = • Moles of O2 (mol) = • Mass of O2 (mg) • Dissolved oxygen, ppm...
De constant kiday"). 8. A standard BOD test is run using seeded dilution water. In one bottle, the water sample is mixed with seeded dilution water giving a dilution of 1:30. Another bottle, the blank, contains just seeded dilution water. Both bottles begin the test with DO at the saturation value of 9.2 mg/L. After five days, the bottle containing waste has DO equal to 2.5 mg/L while that containing just seeded dilution water has DO equal to 8 mg/L....
The actual concentration was found to be 0.002364 M.
4. Knowing that 1 mole of dissolved oxygen ends up reacting with 4 moles of the titrant S2032, how do you calculate the amount of dissolved oxygen (in units of mg/L) from the volume of titrant added? Show your work. (2.5 pts) • Use X (mL) as your total volume of titrant added. • Use the actual concentration of the -0.0025 M NazS2O3 solution used during the titration (calculated in question...
5. The following data is for a BOD determination run on a raw sewage sample incubated for 5 days at 20°C in the dark; calculate the BOD5,20. Raw sewage contains a sufficient population of bacteria to conduct a BOD test. Therefore, this is an unseeded sample. Data: BOD bottle volume mL raw sewage added to BOD bottles DO, in bottle containing waste and dilution water DO in bottle containing waste and dilution water 300 mL 10.00 mL 7.70 mg/L 2.35...
ENV3006C HW#4 The following data provided to you. Calculate the BODS of the WW sample (mg/L) are Seeded wastewater (WW): Initial DO 8.2 mg/L; Volume of wastewater Final DO (after 5 days) 2.5 mg/L Total volume in BOD bottle = 300 mL 5 mL Seeded dilution water: Final DO (after 5 days) 7.3 mg/L Initial DO 8.3 mg/L; Volume of seeded dilution water 300 mL
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