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

standard sodium thiosulfate solution. By knowing the number of moles of iodine produced, the amount of oxygen present in the sample of water can be calculated and hence its concentration.

Material and equipment: BOD bottle250ml conical flask50ml buretteH2SO4NaOHMnSO40.005M KI0.05M Na2S2O3Sample water

Method: We fill a BOD bottle to overflowing with the sample and replace the bottle stopper so that no air is trapped in the sample. It is important to minimize sample mixing and exposure to air. Then we add 0.7 ml concentrated H2SO4 to each sample bottle. This will arrest biological activity in the sample. We close the bottle again assuring that no air is trapped in the sample.

We water seal the sample bottles by covering the bottle cap with distilled water. We have to start the titration as soon as possible after sample collection. For sample analysis, we bring the sample to 200C. Then we add 1ml of a 30% MnSO4 solution followed by 1ml of 32% NaOH solution to the sample. We have to hold the pipette just above the sample surface when adding the reagents.

Then we close the sample bottle carefully to exclude air bubbles and mix by inverting the bottle several times. When the manganese hydroxide floc has settled to about half of the bottle volume add 1ml of concentrated H2SO4. We carefully reclose the bottle and mix by inverting it several times until the precipitate has completely dissolved. We measure a volume corresponding to 200ml of the original sample after correction for sample loss by displacement with reagents. We add an appropriate volume of 0.005M KI and titrate the 200ml sample with 0.05M Na2S2O3 solution to a pale color. Then we

add a few drops of starch solution and the sample should turn dark blue. We continue titrating until the first disappearance of the blue color. Biochemical Oxygen Demand (BOD)We fill a BOD bottle to overflowing with the sample and replace the bottle stopper so that no air is trapped in the sample.

It is important to minimize sample mixing and exposure to air. Water seal the sample bottles by covering the bottle cap with distilled water. The samples have to be stored in the dark. For the sample analysis, we have to measure the concentration of two duplicate samples. We incubate the samples for 5 days in a dark place and then we have to measure again the DO concentration so that we can calculate the total BOD.

Data and observations: 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 ions from the salt - in this case, MnSO4- are oxidized manganese oxide according to the first reaction, we expect the color of the solution to change. That is because manganese is a transition metal whose 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 ions will oxidize to manganese oxide. Potassium iodide is then added which is oxidized by the manganese oxide in an 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 - which acts as an indicator- the solution decolorizes.

Sodium and potassium in this case are just spectator ions not taking part in the chemical reaction. By knowing the number of moles of iodine produced, the amount of oxygen present in the sample of water can be calculated and hence its concentration.In the titration, the following volumes of Na2S2O3 were used: Titration Initial volumeFinal volumeNa2S2O3 used initial color final color1022BlueWhite2 (5 days after the 1st titration)01.51.50111105.55.5

Calculations and Results: MM (KI) = 166n = c x v = 0.005 x 0.5 = 2.5 x 10-3 molesm = n x MM = 2.5 x 10-3 x 166 = 0.415gMM (Na2S2O3) = 248.18n = C x V = 0.05 x 0.5 = 0.025 molesm = n x MM = 0.025 x 248.18 = 6.2gThe average amount of Na2S2O3 used for the 2nd titration is 3.5ml. The value of 11 was excluded, due to an experimental error during the procedure.

We expected the solution to decolorize totally but we had previously added an excess amount of starch. Thus, our solution would never turn totally transparent but instead it would remain turbid.

Do Initialamount of Na2S2O3n = c x v ? n = 0.05 x 2 x 10-3 = 1 x 10-4 molesI2 (aq) + 2 S2O32- (aq) --> S4O62- (aq) + 2 I- (aq)1mole 2 molesx 1 x 10-4 molesx = 5 x 10-5MnO2 (s)

+ 2 I- (aq) + 4 H+ (aq) --> Mn2+ (aq) + I2 (aq) + 2 H2O (l)1 mole 1molex 5 x 10-5 molesx = 5 x 10-5 moles2 Mn2+ (aq) + 4 OH- (aq) + O2 (aq) --> 2 MnO2 (s) + 2 H2O (l)2 moles 1 mole5 x 10-5 moles xx = 2.5 x 10-5In 2ml of the sample there are 8 x 10-4 g of O2In 1000ml of the sample there are xx = 0.4g of O2DO Finalin 3.5ml of the sample there are 8 x 10-4g of O2In 1000ml of the sample there are xx = 0.23g of O2BOD: DOinitial - DOfinal = 0.4 - 0.23 = 0.17g = 170mg/ppm of O2

Data analysis: 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 with 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 during the 5-day period at night. 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.