• Join over 1.2 million students every month
  • Accelerate your learning by 29%
  • Unlimited access from just £6.99 per month

Determination of DO & BOD in a water sample

Extracts from this document...

Introduction

Determination of DO & BOD in a water sample Objective Using volumetric analysis, the amount of dissolved oxygen (DO) in a water sample can be found. On standing another sample for 5 days, the difference between these two values is defined as biochemical oxygen demand in 5 days (BOD5). The aquatic life needs certain amount of DO for survival, whereas the value of BOD reflects the water quality. Principle In alkaline solution, dissolved oxygen will oxidize manganese(II) into manganese(III): 4Mn2+(aq) + 8OH-(aq) + O2(aq) + 2H2O(l) � 4Mn(OH)3(s) The amount of DO can be found by titrating the iodine produced from potassium iodide by manganese(III) with sodium thiosulphate: 2Mn(OH)3(s) + 2I-(aq) + 6H+(aq) � 2Mn2+(aq) + I2(aq) + 6H2O(l) I2(aq) + 2S2O32-(aq) � 2I-(aq) + S4O62-(aq) Chemicals water sample, MnSO4, conc.H2SO4, alkaline KI, standard Na2S2O3, starch solution Apparatus titration apparatus, pipette, 250cm3 volumetric flask x2, beaker, droppers, magnetic stirrer, white tile Procedure 1.> Collect a water sample with two 250cm3 volumetric flasks. ...read more.

Middle

= 0.04935 mmol = 0.04935 x 32 = 1.5792 mg Dissolved oxygen (DO0) in aquarium water in the first day at room temperature = 1.5792 � (200x10-3) = 7.896 mgdm-3 Concentration of standard Na2S2O3 after 5 days: 0.0127 M Volume of 0.0127M Na2S2O3 used = 34.40 - 22.95 = 11.45 cm3 Amount of 0.0127M Na2S2O3 used = 11.45 x 0.0127 = 0.1454 mmol Amount of I2 generated = 1/2 (0.1454) = 0.0727 mmol Amount of Mn(OH)3 generated = 0.0727 x 2 = 0.1454 mmol Amount of O2 present in 200 cm3 of aquarium water (stood for 5 days) at room temperature = 1/4 (0.1454) = 0.03635 mmol = 0.03635 x 32 = 1.1632 mg Dissolved oxygen (DO5) in aquarium water stood for 5 days at room temperature = 1.1632 � (200x10-3) = 5.816 mgdm-3 Biochemical oxygen demand for 5 days (BOD5) = DO5 - DO0 = 2.08mgdm-3 (c) KCl http://hk.geocities.com/fatherofchemistry Precaution 1.> At all stages, every method should be made to assure that oxygen is neither introduced to nor lost from the water sample. ...read more.

Conclusion

3.> Biochemical oxygen demand (BOD) is the amount of oxygen required to break down the organic matter present in a water sample. It can be used as an indicator of water pollution. If the water sample has a high BOD, it indicates that a lot of organic waste is present and a lot of oxygen is needed to break down the waste. Natural clean water has a BOD of about 1-4 mgdm-3 only. A low BOD indicates only small amount of organic matter is present but there is still a little organic pollution. If BOD is high and if the oxygen depleted is greater than the oxygen replenished, then some fishes may die, organic debris accumulates and anaerobic microorganisms begin to multiply, producing unpleasant taste and smell. 4.> Since air pump is applied to the water sample continuously, the DO is slightly higher than other water samples. Conclusion The value of DO and BOD can be used as an indicator of water pollution and the quality of the living environment of aquatic life. ...read more.

The above preview is unformatted text

This student written piece of work is one of many that can be found in our AS and A Level Inorganic Chemistry section.

Found what you're looking for?

  • Start learning 29% faster today
  • 150,000+ documents available
  • Just £6.99 a month

Not the one? Search for your essay title...
  • Join over 1.2 million students every month
  • Accelerate your learning by 29%
  • Unlimited access from just £6.99 per month

See related essaysSee related essays

Related AS and A Level Inorganic Chemistry essays

  1. Peer reviewed

    Determining the concentration of acid in a given solution

    5 star(s)

    run out of solution and have to take 4 readings which would double the percentage error. When I read the burette I also put a piece of white paper behind it so that the meniscus was more clearly visible. I made sure that the meniscus was just resting on the calibration line when I took the measurement.

  2. Peer reviewed

    Deducing the quantity of acid in a solution

    5 star(s)

    This reading needs to be more accurate than the reading of the conical flask. That's because the burette is the apparatus that we read to take into account the endpoint. As we need at least 3 concordant titres that are to within 0.1cm3, we need to be sure that we

  1. effects Concentration and Temperature on the Rate of Reaction

    32.1 = 32.1 O4 = 4 x 16.0 = 64.0 = 98.1 grams needed = (1 x 98.1 x 250) 1000 = 24.525g dissolved in enough distilled water to make 250ml of solution. 0.0001M phenol: C6H5OH: C = 6 x 12.0 = 72.0 H = 6 x 1.0 = 6.0

  2. Lab report Determination of Enthalpy Change of Neutralization

    experimental enthalpy change of neutralization is smaller or greater than the expected value. Questions Q.1: The temperature increased in neutralization between 25 cm3 of 2.0 M of HCl and 25 cm3 of 2.0 M of NaOH was 12.0 . The temperature increased in neutralization between 25 cm3 of 2.0 M

  1. Determination of Chemical Oxygen Demand (COD) of a Given Sample of Waste Water

    = 0.00 Final Volume of FAS (ml) = 17.20 Volume of FAS used (ml) = 17.20 Volume of FAS used to titrate with the blank solution = 17.20 ml * Data treatment and questions: 1. The reaction between K2Cr2O7 and FAS can be described by the following equation: 6Fe2+ + Cr2O72- + 14H+ --> 6Fe3+ + 2Cr3+

  2. Chemistry Iodine Clock

    like this: Rate of reaction against concentration Concentration against time Plan Making a standard solution A standard solution is a solution of known concentration which is mostly measured in mol dm-3 Standard solution of Sodium thiosulphate We need to calculate how many moles we need.

  1. Chemistry - Data Analysis

    Put the thermometer into the boiling water and wait until it is level at 90 degrees, and temperature does not continue to rise in the thermometer. iv) Write down the temperature every 2 minutes on the table of results. v)

  2. Aim To study the effect of concentration of iodide ion ...

    As the number of effective collisions decrease, the rate of reaction would decrease, signaled by the increase in time taken for formation of the blue-black solution. The presence of delayer - thiosulphate - would not have a net effect on the reaction because the concentration of thiosulphate is fixed throughout the experiment.

  • Over 160,000 pieces
    of student written work
  • Annotated by
    experienced teachers
  • Ideas and feedback to
    improve your own work