Experiment 6 Dissolved Oxygen PDF

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EXPERIMENT 6: DISSOLVED OXYGEN LEVELS IN NATURAL WATERS

De Vera, Ar Jay B. Gonzales, Joshua C. Lucas, Ana Lowela L. Sison, Herschel Annicole S. Velasco, Erikka Louisse C. CpE B12

General Chemistry Instructor (CM011L): Sir Rommel Galvan Mapua University

Abstract Just like most of the organisms, the animals in the water require oxygen in order to survive. They acquire oxygen from the water because water contains dissolved oxygen molecules. Dissolved oxygen molecules are simply oxygen that are mixed into a solution or dissolved in the water. There are many ways the oxygen can enter the water, one is through the atmosphere since these molecules make up 1/5 of the atmosphere and constantly enters the water. The amount of dissolved oxygen molecules depends on the water conditions like the saltiness and temperature of water. There would be no space for the oxygen molecules in the water if there is a high amount of salt, and if the temperature is too high, the oxygen molecules would vibrate which makes it easier for them to get out of water. Waters that contains a high level of dissolved oxygen are considered healthy or stable for marine life, while waters with low dissolved oxygen concentrations are considered unhealthy. In this experiment, the goal of the group is to determine the concentration of dissolved oxygen in a given water sample, while developing the right ways to obtain a natural water sample. The group also aims to identify the chemical reactions involved in fixing and analyzing a water sample for dissolved oxygen.

Introduction

Different types of bodies of water plays an essential part in everyone’s lives. They not only a source of food supplies but it also open doors for some water recreational activities. Furthermore, the larger bodies of water such as lakes and oceans have a big impact in earth’s seasonal weather changes and it is part of the conditions for hurricanes and typhoons. The quality of the water can be seen by just looking at the appearances of smaller bodies of water (e.g. streams and lakes). Color, surface development, and odor are the early markers of the nature of water and the idea of its marine life. In general, water supplies of bigger urban areas depend on the nearness of the surface water and water chemist must be accurately aware of the properties of water to provide safe and clean water to the buyers and consumers. Various water-quality parameter are essential enthusiasm for breaking down “natural” water sample: pH, dissolved oxygen, alkalinity etc. A quick test, pH, is generally determined with a formerly adjusted pH meter; dissolved oxygen concentrations can be completed with a dissolved oxygen meter although its availability is less likely than that of a pH meter. Alkalinity and hardness levels are determined using the titrimetric technique. The convergence of dissolved oxygen in a water sample is a vital marker of water quality. Waters with high oxygen concentrations demonstrate vigorous conditions: spotless, clear and unpolluted. Low oxygen fixations demonstrate anaerobic conditions: high turbidity, foul odors, and extensive plant growth on the surface. Dissolved oxygen levels that drop to under than 5 ppm can pressure the current oceanic life.

Methods Materials and Procedures Materials used in the experiment: 1. Burette 2. Burette Clamp 3. Beaker 4. Erlenmeyer Flask 5. Funnel 6. Bottle A Standard 0.025 M

Na2 S2 O 3 Solution

1. Preparation and standardization of 0.1 M

Na2 S2 O 3

solution. Prepare only 100 mL of

the Na2S2O3 of the solution and standardize the solution using standard solution Calculate the average concentration of the

KIO 3 Na2 S2 O 3

as the primary solution for

three trials. 2. Preparation of a standard 0.025 M volumetric flask, prepare a 0.025 M

Na2 S2 O 3

solution. Using a pipet and 100-mL

Na2 S2 O 3 solution from the standardized 0.1 M

Na2 S2 O 3 Collection of Water Sample 1. Prepare the flask for sampling. Thoroughly clean and rinse at least three 250-mL Erlenmeyer flasks and rubber stoppers to fit. Allow to air dry.

2. Collect the water sample. Gently lay the flask along the horizontal surface of the water. Slowly and gradually turn the flask upright as the flask fills being careful not to allow any air bubbles to form in the flask. Fill the flask to overflowing. 3. “Fix” the dissolved oxygen. Below the surface of the water sample, pipet ~1 mL of the basic 2.1 M

Mn SO 4 solution into the sample, some overflowing will occur. Similarly

pipet ~1 mL of the basic

KI−Na N 3

solution. A precipitate should form.

4. Secure the sample. a. Carefully stopper the sample to ensure that no air bubbles become entrapped beneath the stopper in the water sample. Again, some overflowing will occur. b. Invert and roll the flask to thoroughly mix the reagents. Once the precipitate settles, repeat the mixing process. c. Label the sample number for each of the flasks. Store the sample in the dark and, preferably, in a cool or cold location or on ice. 5. Read and record the temperature of the water at the sample site. Also, write a brief description of the sample site. 6. Analysis should begin within 6 hours of sampling. Sample Analysis

1. Prepare the titrant. Prepare a clean buret. Add 3 to 5 mL of the standard

Na2 S2 O 3

solution to the buret, roll the solution to wet the wall of the buret, and dispense through the buret tip and discard. Use a clean funnel to fill the buret—dispense a small portion through the buret tip. Read and record the volume of

Na2 S2 O 3

solution in the buret,

using all certain digits plus one uncertain digit. Place a white sheet of paper beneath the receiving flask.

2. Prepare sample a. Remove the stopper from the 250-mL Erlenmeyer flask. To the collected water sample, add ~1 mL of conc

H 2 SO 4 (Caution!) and stir or swirl to dissolve any

precipitate. The sample can now be handled in open vessels. b. Transfer a known, measured but exact volume (~200 mL, ±0.1 mL) to a receiving flask (either a beaker or Erlenmeyer flask) for the titrimetric analysis. 3. Titrate the sample. Slowly dispense the

Na2 S2 O 3

titrant into the water sample. Swirl

the flask as titrant is added. When the color of the analyte fades to a light yellow-brown, add ~1 mL of the starch solution. Continue slowly adding titrant—when one drop (ideally, half-drop) results in the disappearance of the deep-blue color of the I3–•starch complex, stop the titration and again (after ~15 seconds) read and record the volume of titrant in the buret. 4. Calculate the dissolved oxygen concentration for each sample expressed in ppm

O2

(mg O2/L sample)

Prelaboratory 1. For a natural water sample, what range of dissolved oxygen concentrations may you expect? Explain your reasoning. Answer: In common water there can be green growth and vegetation that reduce the dissolved oxygen. The dissolved oxygen levels of lower than 3 mg/L is stressful to fish and dissolved oxygen levels lower between 2 and 1 mg/L will not support survival for sea creatures. Therefore, reduced oxygen can extend from basically zero up to saturation for the temperature of the water.

At zero-degree Celsius, disintegrated oxygen has its most prominent solvency of 5.4 to 14.8 mg/L. 2. How does the dissolved oxygen concentration in a water sample change (if at all) with a. ambient temperature changes? Answer: the dissolved oxygen concentration in a water sample change in an inverse relation when ambient temperature change. b. atmospheric pressure changes? Answer: the dissolved oxygen concentration in a water sample change in a direct relation when atmospheric pressure change. c. the volume of the flask collecting the water sample? Answer: the dissolved oxygen concentration in a water sample change in a straight manner. d. the amount of organic matter in the water sample? Answer: the dissolved oxygen concentration in a water sample change in a reverse manner. e. the depth of the body of water (e.g., lake, river, or ocean)? Answer: the dissolved oxygen concentration in a water sample also changes in an inverse relation. 3. Experimental Procedure, Part A.1. A 100-mL volume of a primary standard 0.0110 M KI O 3 solution is prepared. A 25.0-mL aliquot of this solution is used to standardize a

prepared the

Na2 S2 O 3 solution. A 15.6-mL volume of the

KI O 3

Na2 S2 O 3 solution titrated

solution to the starch endpoint. What is the molar concentration of the

Na2 S2 O 3 solution?

−¿ ( aq )+ 3 H 2 O(l) +¿ ( aq) →3 I ¿3 ¿ −¿ ( aq )+6 H ¿ −¿ ( aq ) +8 I ¿ IO 3 2−¿(aq) −¿ ( aq) +S4 O¿6 ¿ +¿( aq ) → 3 I 2−¿( aq ) +6 H ¿ ¿ −¿ ( aq )+ 2 S2 O 3 I 3¿ Answer: 1 mole of KIO3 gives 3 moles of

−¿ ¿ I 3 and 1 mole

−¿ 2¿ I 3 requires 2 moles of S2O3

Formula: M1V1 = M2V2 M2 = 25 (0.011/15.6) = 0.01763M 4. Experimental Procedure, Part A.2. What is the procedure for preparing 250 mL of 0.0210 M

Na2 S2 O 3 for this experiment from a 100-mL volume of standard 0.106 M

Na2 S2 O 3 ? Answer: Formula: M1V1 = M2V2 V2 = 250 (0.0210/0.106) = 49.52mL By taking 49.52 mL of 0.106M solution and dilute it up to 250 mL by adding water.

5. Standard 0.025 M

Na2 S2 O 3

Solution

1. Sample volume (mL)

200.0

2. Buret reading, initial (mL)

3.85

3. Buret reading, final (mL)

18.25

4. Volume of

Na2 S2 O 3 dispensed (mL)

5. Average molar concentration of

14.4

Na2 S2 O 3

(mol/L)

0.0213

6. Moles of

Na2 S2 O 3 dispensed (mol)

7. Moles of

−¿ I 3¿ reduced by

8. Moles of

O 2 (mol)

0.00007668

9. Mass of

O 2 (mg)

2.454

10. Dissolved oxygen, ppm

0.0003067

2−¿ S 2 O¿3 (mol)

O 2 (mg/L)

0.001534

12.27

Calculation Zone Na2 S2 O 3 dispensed (mL) = 18.25mL – 3.85mL = 14.4 mL

Volume of

Na2 S2 O 3 dispensed (mol) = 14.4 mL (0.0213mol / 1000mL ) = 0.0003067mol

Moles of Na2 S2 O 3 Moles of

−¿ I 3¿ reduced by

2−¿ S 2 O¿3 (mol)

 2Na2S2O3 ≡ I32 mol

≡ 1mol

0.0003067 mol = ½ (0.0003067 mol) =0.0001534 mol Moles of

O 2 (mol)

 4 mol

S 2 O3

--

≡

1 mol O2

0.0003067 mol = ¼ (0.0003067 mol) = 0.00007668 mol Mass of

O 2 (mg)

Mass of O2 = (moles of O2 ) (molar mass of O2) 0.00007668 mol = 32g/mol 0.002454g x 1000mg = 2.454 mg Dissolved oxygen, ppm

O 2 (mg/L): Sample Volume = 200 mL  0.200L

0.200 L water contains 2.454 mg O2 1L of water contains (2.454 mg / 0.200 L) x 1L = 12.27mg Dissolved O2 = 12.27 mg/L

5.b

For Trials 2 and 3, the dissolved oxygen levels were 10.9 ppm and 11.1ppm respectively.

a. What is the average dissolved oxygen level in the water sample? Average ´x =

0.426+0.417 + 0.431+ 0.428+ 0.429 + 0.433 6

Average ´x =0.427 g/ g PPT = parts per thousand = 1g = 1000 parts g g Zn ppt=0.427 × g 1000 g −3

Zn ppt=0.427 ×10 g/kg

b. What are the standard deviation and relative standard deviation (%RSD) of the dissolved oxygen level in the water sample. ´x x−¿ ¿ ¿2 ¿ ∑¿ ¿ Standard Deviation=√ ¿

0.427 0.426−¿ ¿ 0.427 0.417−¿ ¿ 0.427 0.431−¿ ¿ 0.427 0.428−¿ ¿ 0.427 0.429−¿ ¿ 0.427 0.433−¿ ¿ ¿2 ¿ ¿ √¿ = 0.00561 Standard Deviation × 100 ´x 0.00561 × 100 Relative Standard Deviation= 0.429 Relative Standard Deviation = 1.3 Relative Standard Deviation=

Dissolved Oxygen Levels in Natural Waters Report Sheet

A. Sample Analysis

1. 2. 3. 4. 5.

Sample volume (mL) Buret reading, initial (mL) Buret reading, final (mL) Volume Na2 S2 O 3 dispensed (mL) Molar concentration of Na2 S2 O 3

500 mL 25 mL 18.5 mL 6.5 mL 0.02448M

(mol/L) 6. Moles of 7. Moles of

Na2 S2 O 3 dispensed −¿¿ 2−¿ I 3 reduced by S 2 O¿3 –

(mol) 8. Moles of O 2 (mol) 9. Mass of O 2 (mg) 10.Dissolved oxygen, ppm

O 2 (mg/L)

0.0001617 mol 0.00008085 mol

0.000040425 mol 1.2936 mg 6.468 mg/L

Calculation Zone: Volume Na2 S2 O 3 dispensed (mL): 25 mL – 18.5mL = 6.5 mL Moles of

Na2 S2 O 3 dispensed:

Moles of

−¿ ¿ I 3 reduced by

Moles of

O 2 (mol):

Mass of

6.5 ml ×

2−¿ S 2 O¿3 – (mol):

dissolved O2=

32 g 1000 mg =¿ 1.2936 mg × 1g 1 mol

O 2 (mg/L):

mgO 2 = L sample

Laboratory Questions

1 0.0001617 mol × =¿ 0.00008085 mol 2

2 1 0.00008085 mol × × =¿ 0.000040425 mol 1 4

O 2 (mg): 0.000040425 mol ×

Dissolved oxygen, ppm

0.02488 M =¿ 0.0001617 mol 1000 mL

1.2936 mg =6.468 mg/ L 1L 200 mL× 1000 mL

1. Part B. The water chemist waits until returning to the laboratory to fix the water sample for the dissolved oxygen analysis. Will the reported dissolved oxygen concentration be reported as too high, too low, or remain unchanged? Explain. Answer: The reported dissolved oxygen concentration will be reported as too low, as some of the dissolved oxygen molecules will escape from the sample. This is caused by various factors, such as organic materials, for instance, that are present in the water sample consuming the oxygen. If the oxygen is not fixed in the precipitate, its amount will continually deplete, hence the lower amount recorded. 2. Part B.3. A solution of MnSO4 is added to fix the dissolved oxygen in the collected sample. a. What is the meaning of the expression, “fix the dissolved oxygen,” and why is it so important for the analysis of dissolved oxygen in a water sample? Answer: Fixing the oxygen means trapping it in the precipitate formed by adding the Manganese Sulfate in the solution. This method prevents the dissolved oxygen from escaping or, in other words, trapping them in the solution. This is important because by fixing the oxygen, the accuracy of its dissolved amount is ensured, thus increasing the probability of accurate laboratory results. b. Only an approximate volume (~1 mL) of MnSO4 is required for fixing the dissolved oxygen in the sample. Explain why an exact volume is not critical. Answer: An exact volume of Manganese Sulfate is not critical because in fixing the dissolved oxygen, the Manganese Sulfate is the excess reactant and the Oxygen is the limiting reactant.

3. Part B.4. No precipitate forms! Assuming the reagents were properly prepared and dispensed into the sample, what might be predicted about its dissolved oxygen concentration? Explain. Answer: The precipitate forms when the oxygen molecules react with Manganese Sulfate. If no precipitate forms, assuming the proper preparation of the reagents, it means that there is extremely little to no presence of dissolved oxygen in the water sample. 4. Part B.5. A water chemist measured and recorded the air temperature at 27°C when he should have measured the water temperature, which was only 21°C. As a result of this error, will the dissolved oxygen concentration be reported as being higher or lower than it should be? Explain. Answer: Temperature is inversely proportional with the concentration of dissolved oxygen. If the recorded temperature was too high as compared to its true value, the recorded concentration of dissolved oxygen would be too low. 5. Part C.3. The color of the analyte did not fade to form the light yellow-brown color but

remained intense even after the addition of a full buret of the

2−¿ S 2 O¿3 titrant, even though

a precipitate formed in Part B.4. What can be stated about the dissolved oxygen concentration of the sample? Explain. Answer: The reason why this happened is because they did not add the

2−¿ S 2 O¿3 titrant gradually,

instead they poured it all at once that's why it resulted to a different color 6. Assuming a dissolved oxygen concentration of 7.0 ppm (mg/L) in a 200-mL water sample,

a. how many moles of Mn(OH)3 will be produced with the addition of the MnSO4 solution? Answer: 1.4 mol b. how many moles of I3– will be produced when the KI-NaN3 solution is added to the above solution? Answer: 0.467 mol c. how many moles of S2O32– will be needed to react with the I3– that is generated? Answer: 0.021 mol d. and also, assuming the concentration of the S2O32– titrant to be 0.025 M, how many milliliters of titrant will be predictably used for the analysis. Answer: 0.046 M 7. A nonscientist brings a water sample to your laboratory and asks you to determine why there was a fish kill in the nearby lake. Having recently finished this experiment, what might you tell that person about the legitimacy of a test for dissolved oxygen? What reasoning would you use to maintain the integrity of your laboratory? Answer: Marine animals also need oxygen in order to survive, they acquire oxygen from the water since water contains dissolved oxygen. Dissolved oxygen is basically oxygen that are dissolved in water. The amount of oxygen in water that stays dissolved depends water conditions like the saltiness and the temperature of water. It is important to test for dissolved oxygen to determine if the water is healthy or has a high amount of dissolved oxygen. 8. A. Fish kills are often found near the discharge point of water from cooling waters at electrical generating power plants. Explain why this occurrence may occur. Answer: Electrical generating power plants found in bodies of water heats up the surroundings, thus increasing water temperature. If the water temperature increases, dissolved oxygen molecules from the water would rapidly move, making it easier for them to escape the water.

Since the marine life also depends on the oxygen in the water, they would die they are near these power plants. B. Fish kills are often found in streams following heavy rainfall in a watershed dominated by farmland or denuded forestland. Explain why this occurrence may occur. Answer: Fish kills are often found in streams following heavy rainfall in a watershed dominated by farmland or denuded forestland, because waters in nearby farmlands are usually polluted by chemicals or other particles that occupies space in the water, and if there is no space in the water, there would be no space for dissolved oxygen molecules. 9. Explain how the dissolved oxygen concentrations may change starting at the headwaters of a river and ending at the ocean. Account for the changes. Answer: The amount of dissolved oxygen would be high from the headwaters of a river and would decrease as it reaches the ending of an ocean, since the level of saltiness will increase as water flows from the rivers to the oceans, and the amount of dissolved oxygen relies on the saltiness of water. If there is too much salt in the water, there would be less space for the dissolved oxygen molecules. 10. Salt (ocean) water generally has a lower dissolved oxygen concentration than freshwater at a given temperature. Explain why this is generally observed. Answer: One of the factors that affects the amount of dissolved oxygen in the water is temperature. If the water temperature is too high, the dissolved oxygen molecules would rapidly vibrate making it easier for them to escape from the water. Since the ocean is warmer than the fresh water, it has a lower dissolved oxygen concentration. Re...