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

The aim of this experiment is to measure the BOD and DO of water.

Extracts from this document...


Experiment #14

December 17/2002

DO and BOD Winkler titration


The aim of this experiment is to measure the BOD and DO of water.


The quality of water depends on several factors including oxygen-demanding wastes, disease-causing pathogens, microorganisms affecting health, plant nutrients, suspended solids and dissolved minerals. Other pollutants may be excess acidity due to acid rain, thermal pollution and substances such as benzene, chromium and mercury that are all toxic to aquatic life.

As humans and land animals obtain oxygen from respiration from the air, for plants and animals to survive in aquatic systems, water must contain a minimum amount of dissolved oxygen. The dissolved oxygen (DO) content of a body of water is an important indicator of its quality. At 200C, DO content of 8 to 9 ppm O2 at sea level is considered to be water of good quality.

The Biochemical Oxygen Demand (BOD) is a measure of the amount of oxygen consumed by the biodegradable organic wastes and ammonia in a given amount of water over a time period, normally 5 days at 200C. The greater the oxygen demanding wastes, the higher the BOD.  

The BOD of a sample of water can be determined by the Winkler method. The sample of the water is saturated with oxygen so the initial concentration of dissolved oxygen is known.

...read more.



The chemical reactions involved in the Winkler Titration method are:

According to this method, an excess of manganese salt is added to the sample water. Since the manganese (II) ions from the salt – in this case MnSO4- are oxidised manganese (IV) oxide according to the first reaction, we expect the color of the solution to change. That is because manganese is a transition metal that its color changes according to the oxidation state that it is found. Thus, the initial color of the solution is pale yellow and transparent, unlike the final color that is brown. Indeed, the color of the solution becomes much darker and we may observe a precipitate forming, which is the solid MnO2. In the solution NaOH is added because under alkaline conditions the manganese (II) ions will oxidise to manganese (IV) oxide. Potassium iodide is then added which is oxidised by the manganese (IV) oxide in acidic solution to form iodine, making the solution dark green. So we add H2SO4 to create the acidic condition needed. The iodine released is then titrated with standard sodium thiosulfate solution according to the third reaction. The solution becomes blue when titrated and at the presence of starch – that acts as an indicator- the solution decolorizes. Sodium and potassium in this case are just spectator ions not taking part in the chemical reaction.

...read more.


-4 g of O2

In 1000ml of the sample there are     x              

x =  0.4g of O2


In 3.5ml of the sample there are 8 x 10-4g of O2

In 1000ml of the sample there are   x        

x = 0.23g of O2


DOinitial – DOfinal= 0.4 – 0.23 = 0.17g = 170mg/ppm of O2


The BOD value for the water sample under investigation was 170ppm, indicating that is comes from untreated sewage (which can range from 100 to 400ppm). Because of the high value, it can be deduced that the sample contains very high oxygen demanding wastes. The initial DO value is higher than that of the final. This decrease is because the biodegradable organic wastes consume the existing oxygen in the 5-days incubating period.

Experimental errors may have arisen during the procedure, involving the titration. The extreme value of 11 was excluded, however, had it been a correct measurement, the resulting BOD would have differed from the one used in the calculations.

If there is an increase in the temperature, the DO level will increase as well. That is in accordance to the collision theory. The rate of the chemical reactions will increase because of the higher kinetic energy that they will have, because of the higher temperature. The temperature was more or less stable (room temperature). However, it might have decreased a few oC during the 5-day period at nights. That would not affect though the DO levels.

Thus, since the BOD value was found 170mg/l, the sample is of unacceptable purity, most probably coming from untreated sewage.

...read more.

This student written piece of work is one of many that can be found in our AS and A Level Probability & Statistics 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 Probability & Statistics essays

  1. Differences in wealth and life expectancy of the countries of the world

    Also you can visibly see the correlation which is not possible with other methods. Reading a scatter graph: * A scatter graph describes a positive trend if, as one set of values increases, the other set tends to increase. * A scatter graph describes a negative trend if, as one set of values increases, the other set tends to decrease.

  2. Reaction Times

    My groups were decided as: - Reaction (cm's) 0-5 extremely fast 5-10 fairly fast 10-20 average 20-25 slow 25-30 Very slow The results for boys were as follows- Reaction Tally Frequency Cumulative Frequency 0 < r < 5 0 0 5 < r < 10 12 12 10 < r

  1. Identifying Relationships -Introduction to Statistical Inference.

    Features of this output: Analysis Conclusion: When the Comparison of Means procedure leads to uncertainty, we must complete further analysis and carry out the appropriate statistical test to enable us to make a decision, with a specified level of confidence in our results.

  2. Standard addition was used to accurately quantify for quinine in an unknown urine sample ...

    Accuracy of the measurement was ensured by the use of a Pasteur pipette to make the solution up to 100 cm3. A range of standard addition solutions were then prepared containing the same volume of diluted urine and varying concentrations of added standard quinine sulphate.

  1. I am going to design and then carry out an experiment to test people's ...

    Year group 1 (years7-8): 80/325 x 60= 14.769... Year group 2 (years8-9): 79/325 x 60= 14.584... Year group 3 (years9-10): 83/325 x 60= 15.323... Year group 4 (years10-11): 83/325 x 60= 15.323... As I cannot take part of a set of data, I will have to round these numbers to the nearest whole numbers.

  2. Data Analysis of American House Price

    is higher than the 3rd quartile of houses without a pool ($192,05). This implies that 75% of houses without a pool haves prices similar to the lowest 25% of houses with a pool. However, the standard deviation measures how spread the data set is.

  1. This is an investigation to identify some compounds containing oxygen AimThis experiment is to ...

    Mg + 2C2H5COOH (C2H5COO)2Mg + H2 Esters are used as artificial flavourings so they smell sweet and fruity; the ester in this case is no different so a simple test would be to have a smell at it.

  2. Investigating the Relationship Between the Amount of Money a Football Club Receives and its ...

    Division 3 - Barnet, Brentford, Cambridge Utd, Chester City, Darlington, Exeter City, Halifax Town, Hartlepool Utd, Hull City, Plymouth A, Rochdale, Southend Utd & Torquay Utd. I am happy that this sample is a fair sample as it is stratified and I expect it to give a good representation of the situation as a whole.

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