J.N. Macabata / Chemistry 26.1 (2016)
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Quantitative Determination of Dissolved Oxygen Content by Winkler Redox Titration Jelica N. Macabata
National Institute of Geological Sciences. University of the Philippines- Diliman, 1101, Quezon City Performed 30 June 2016; Submitted 4 July 2016 a
ABSTRACT Dissolved oxygen
J.N. Macabata / Chemistry 26.1 (2016)
Introduction In studying environmental and marine science, it is important to have knowledge on dissolved oxygen (O2) content of a freshwater sample. Dissolved oxygen concentration is expressed as the number of millilitres of oxygen gas (O2) per litre of freshwater sample (mL or L-1). Water pollution control and waste water treatment process control are determined by the dissolve oxygen test2. It may be used by marine biologist to measure primary production or by physical oceanographers to study water masses in the ocean. In terms of chemistry, it provides measure of the redox potential of a water column. Winkler method, developed in 1888, is used to accurately measure the dissolved oxygen concentration of a particular sample. The method was then modified by Strickland and Parsons in 1968. This experiment provides the readers the knowledge to perform the water sampling and pre-treatment techniques for dissolved oxygen analysis. It will also determine the amount of dissolved oxygen in a water sample from a pond in the Institute of Mathematics, College of Science University of the Philippines-Diliman using Winkler redox titration and discuss the chemistry behind the Winkler method for dissolved oxygen determination. Materials and Methods In performing the experiment, iron stands, burette clamps and aluminium foil are needed. We also prepared clean glass bottle with cap, burettes (50 mL), beakers (50-, 100-mL), volumetric flasks (25-, 100-, 250-mL), volumetric pipettes (10-, 50-mL), Erlenmeyer flasks (250-ml), watch glass and syringe. We also used analytical balance, top-loading balance and hot plate for the experiment. The following solutions should also be prepared: 1. 25.0 mL 4.0 M MnSO 4 from MnSO4∙2H2O crystals – weigh 18.62 grams of the crystals and dissolve in about 10-ml distilled water then filter the solution into a 25-ml volumetric flask and dilute to mark. 2. 25.0 mL 18 M NaOH with 5.0 g KI and 0.15 g NaN3 – weigh the following solids and dissolve in a beaker with 10 mL distilled water while stirring in a hot plate then transfer to a 25-mL volumetric flask and dilute to mark. This must be done in a water bath, under the fume hood.
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3. 250.0 mL 0.125 M stock Na2S2O3 solution from Na2S2O3∙5H2O crystals – weigh g of crystals in a beaker and dissolve in about 100.0 mL boiled distilled water then transfer to a 250-mL volumetric flask. Wash the beaker with boiled distilled water and catch the washings into the flask. 4. 50.0 mL 0.5 M H2SO4 solution 5. Starch solution – weigh 1.0 g of starch in a beaker and dissolve until a smooth paste is obtained. Pour it in a 100.0 ml boiled distilled water and keep the solution in 90-100 degrees Celsius to avoid drying up. 6. 250.0 mL 0.0125 M standard Na 2S2O3 solution from the 0.125 M stock Na2S2O3 solution – pipette 0.125 M stock solution and transfer to a 250.0 ml volumetric flask and dilute to mark. Standardization of the titrant Na2S2O3 was done by weighing 0.15 grams of the primary standard KIO3 and dissolved in 50.0 mL of distilled water. Then the solution was transferred quantitatively into a 100-mL volumetric flask and diluted to mark. Three 10-mL aliquots was taken from the solution and transferred into three 250mL Erlenmeyer flasks. 20.0 ml of water, 1.0g of KI and 10.0 mL of 0.5 M H2SO4 was also added to each of the solution in the flask and immediately titrated with the standard Na2S2O3. When the pale yellow solution is obtained, we added 1 mL or 20 drops of starch solution that causes it to turn to blue. Immediately titrate it again until the blue colour disappears. Record the initial and final solution of the titrant in each sample solution. For the water analysis sample, we went to the Institute of Mathematics front yard and filled the empty glass bottle covered with aluminium foil to overflowing with the pond water. We covered the mouth of the glass bottle with its air-tight cap while it is still submerged in the water to avoid the inclusion of air bubbles. After getting the sample, we added the following reagents in succession: 0.5 mL of MnSO 4 and 0.5 mL of NaOH with KI and NaN3 solution. Then, close the bottle and shake it thoroughly. Remove the cover again and slowly add 2.0 mL of concentrated H 3PO4. All the addition of reagents performed was done below the surface of the water using measuring pipettes to avoid inclusion of air bubbles. Cover and shake the bottle and let it stand for 10 minutes as instructed. Then, take a 50.0 mL aliquot of the solution and transfer into a 250-mL Erlenmeyer flask. Titrate it with the standard Na2S2O3 solution until a pale yellow colour is obtained then add 1.0 ml or 20 drops of starch
J.N. Macabata / Chemistry 26.1 (2016)
solution. The pale yellow solution will turn into blue then continue titrating until the blue colour disappears. Record the initial and final volume of the titrant in each sample solution. Results and Discussion The standardization of Na 2S2O3 includes the following equations: I2 + I- → I3[1] 22S2O3 + I3- → S4O62- + 3I- [2] The stoichiometric ratio is 1mmol IO3-: 6 mmols S2O32-. H2SO4 is added to increase the acidity of the solution and will let the reaction with iodate [IO3-] to take place. Iodine is a common oxidizing titrant. Compared with other redox titrants like MnO4-, Ce 4+ and Cr2O7, iodine is a weak oxidizing agent and is only useful in analysis of analytes that are strong reducing agents1. Because of iodine’s poor solubility, solutions are prepared by adding an excess of I-. Acidic H2SO4 was added first to adjust the environment of the solution and increase the acidity then KI was added to provide the I- needed to increase the solubility of I2 and form a more soluble triiodide ion, I3-. Since direct titration couldn’t be performed to I- solutions because it is subjected to air oxidation when forming I3- from I-, excess KI is added to reduce the analyte and liberate a stoichiometric amount of I3-. In water sample analysis, the following equations are involved: MnSO4∙2H2O → Mn2+ + SO42- [3] Mn2+ + 2OH- → Mn(OH)2 [4] O2 + 4Mn(OH)2 + 2H2O ↔ 4Mn(OH)3(s) [5] OR O2 + 4Mn(OH)2 ↔ 4MnO(OH)(S) + 2H2O [6] The stoichiometric ratio is 1 mmol O 2: 4 mmols S2O32-. The sample solution to be analysed is treated with a solution of MnSO4 and then with a solution of NaOH and KI as shown in reactions 3 and 4. Under these alkaline conditions, dissolved oxygen (O2) oxidizes the Mn2+ to MnO2. Then, concentrated phosphoric acid (H3PO4) was added to the sample. Under the new conditions, I- is oxidized to I3- by MnO2. MnSO4 was added first to the solution in order to have a steady source of Mn 2+ ions needed for the reaction to occur. Then, 0.5 ml of NaOH with KI and NaN3 solution was added and together they form ammonium hydrogen carbonate (NH4HCO3) and sodium azide (NaN3). NH4HCO3 is added to prevent the interference of organic compounds in the solution while NaN3 was introduced to eliminate the interference of nitrite (NO2-) in the solution. Other than thiosulfate, nitrite is also oxidized by iodine that results to the reduction of the volume of the titrant used. At this point, precipitates were formed as stated in reactions 5 and 6. Phosphoric acid was also added to the solution to have an acidic environment for the reaction
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and dissolves the precipitates formed earlier. It also inactivates Fe3+ ions that may interfere and consume the oxygen in the sample. The solution now has a pH value ranging from 1 to 2.5 after the addition of the acid. The addition of every reagent is in a definite sequence because every added chemical reacts for the completion of the reaction or elimination of interfering substances. Random addition of the reagents may cause errors in results or the reaction may not proceed. The amount of I 3- formed is determined by titrating with S2O32- using starch as an indicator. If we can determine the amount of I3- by titrating it with S2O32-, we can also determine the amount of dissolved oxygen present in the sample. The starch is added towards the end of the titration to prevent the stabilization of the I2- starch complex. The starch has a helical structure and once the I 2 is added, it will be “encaged” within the helix and achieving the endpoint will be difficult. In this experiment, we used an iodometric titration. The difference between iodometric and iodimetric titration is the choice of analyte and titrant. In iodometric titration, sodium thiosulfate is used as the titrant and the analyte is iodine while in iodimetric titration, iodine is the titrant and the analyte is sodium thiosulfate. With all the data gathered, we are able to calculate the molarity of the standard Na 2S2O3 solution. Using the stoichiometric ratio, we converted the moles of the primary standard to the moles of the titrant divided by the total volume consumed for the titration and multiplied by the aliquot factor. Average molarity of Na 2S2O3 was
0.0111849157∨approximately 0.0112 . The
molarity of Na2S2O3 was then used to calculate the ppm of the water sample. We obtained a value of
1.550878041 ppmO2 and from the table below,
DO Content (ppm O2), 20∘ C 8-9 6.7 - 7.9 4.5 – 6.6
Water Quality Clean, good water Slightly polluted Moderately polluted, can sustain life of warm water Below 4.5 fishes 0-2 Highly polluted Cannot sustain life we can see that the water sample cannot sustain life and is highly polluted. Some factors may affect the results obtained from the experiment like instead of ten minutes, the sample solution is made to stand overnight before analysis. The effect of this on the dissolved oxygen content would be indeterminate because the sample might have heterotrophic and photosynthetic organisms that may perform cellular respiration and photosynthesis. In other cases wherein the MnSO4 is added and the solution made to stand for an hour before the
J.N. Macabata / Chemistry 26.1 (2016)
alkaline KI solution is added may result to a decrease to the dissolved oxygen content of the water sample. The manganese will undergo reduction since it is light sensitive and will cause a decrease to both the volume of the titrant and the concentration of dissolved oxygen. Errors in every experiment is inevitable and predictions of their possible effects on the results may help determine what to do to correct them and if possible, to avoid them. Possible sources of errors in the conducted experiment may be the incorrect order of adding reagents in the water sample. The addition of phosphoric acid before potassium iodide will lead to the formation of iodic acid. IO3- + H+ → HIO3 [7]
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Conclusion and Recommendations
References Analytical Chemistry Academic Group, Analytical Chemistry Laboratory Manual (Elementary Quantitative Inorganic Analysis Chem 26.1), 2013, 30 Winkler, L.W. Die Bestimmung des in Wasser gelösten Sauerstoffen. Berichte der Deutschen Chemischen Gesellschaft, 1888, 21: 2843–2855.
J.N. Macabata / Chemistry 26.1 (2016)
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Appendix (single column format A. Answers to Questions (if necessary) B. Figures (if necessary) C. Raw Data Table (if necessary) D. Working Equations E. Sample Calculations Standardization of Na2S2O3 Solution Primary standard used: KIO3 Formula Mass of 1∘ standard: 214.0 % Purity of 1∘ standard: 99.4 % Weigh of standard, g: 0.1584 Volume of standard solution: 100 mL
Trial
1
2
3
Volume of KIO3 standard, mL
10
10
10
Final volume of Na2S2O3, mL
38.8
42
40.7
Initial volume of Na2S2O3, mL
0
5
0
Net volume of Na2S2O3, mL
38.8
37.0
40.7
M Na2S2O3
Trial 1:
0.1584 g x
= Trial 2:
0.1584 g x
1mol 6 mol Na2 S 2 O3 x x 0.994 214.0 g 1 mol KI O3 10 x 0.0388 100 0.01137751228
≈ 0.0114
1mol 6 mol Na2 S 2 O3 x x 0.994 214.0 g 1 mol KI O3 10 x 0.0370 100
= 0.01133086133 ≈ 0.0113 Trial 3:
0.1584 g x
1mol 6 mol Na2 S 2 O3 x x 0.994 214.0 g 1 mol KI O3 10 x 0.0407 100
= 0.0108463735 ≈ 0.0108 Average M Na2S2O3
0.01137751228+0.01133086133+ 0.0108463735 =0.01118 3
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Sample Analysis Location of Sampling: Institute of Mathematics frontyard Total Volume of Sample: 500 mL
Trial
1
2
3
Final volume of Na2S2O3, mL
1.50
6.20
3
Initial volume of Na2S2O3, mL
0
5
1.50
Net volume of Na2S2O3, mL
1.50
1.20
1.50
DO content, ppm O2
Trial 1:
0.0015 L x
0.0111849157 mol 4 mol O2 31.998 g x x L 1 mol S2 O3 1mol O2 0.00150 ¿ 1.43157973≈ 1. 43
Trial 2:
0.0015 L x
Trial 3:
0.0015 L x
0.0111849157 mol 4 mol O2 31.998 g x x L 1 mol S2 O3 1mol O2 =1.7894 0.00120
0.0111849157 mol 4 mol O2 31.998 g x x L 1 mol S2 O3 1mol O2 0.00150 ¿ 1.43157973≈ 1. 43
Average DO content, ppm O2 RSD Confidence Interval
Other remarks:
1.43157973+1.789474663+1.4317973 =1.550878041 ppm O 2 3 RSD=
0.2065679575 x 1000 ppt=133.1879702 ppt 1.550950564 1.550878041 ± 0.2
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