EXPERIMENT - 2 DETERMINE THE PRODUCT OF A REDOX REACTION REACTION OF BROMATE AND HYDROXYLAMMONIUM IONS CHM110H5F

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1 EXPERIMENT - 2 DETERMINE THE PRODUCT OF A REDOX REACTION REACTION OF BROMATE AND HYDROXYLAMMONIUM IONS CHM110H5F EXPERIMENT PERFORMED ON: 03 OCTOBER, 2012 REPORT SUBMITTED ON: 10 OCTOBER, 2012 SUBMITTED TO: MATTHEW DACOSTA

2 P a g e 1 I. Purpose: The purpose of the experiment was to find the equation of the redox reaction with the help of quantitative and qualitative analysis of the reaction where Bromate ions and Hydroxylammonium ions are reacted to find the identity of the nitrogen containing product. II. Experimental method 2 1. Using a burette, 20mL solution of KBrO 3 was measured and transferred into a 250mL Erlenmeyer flask. 2. A 10mL of NH 3 OHCL solution was measured using a volumetric pipette and delivered it into the Erlenmeyer flask. 3. Then, with a graduated cylinder, 15mL of Hydrochloric acid was introduced into the mixture and the flask was kept under a fume hood to ensure any gas that evolved during the reaction escaped through the fume hood. 4. The mixture was mixed thoroughly by swirling and the flask and was allowed to rest under the fume hood with a watch glass covering the opening of the flask. 5. During the resting period, close observations were made with regards to color and temperature changes, any evidence of gas formation etc. 6. While the mixture was resting under the fume hood, a burette was taken and rinsed with sodium thiosulfate solution. Then, it was filled with the stock solution and the initial reading of the burette was noted. 7. After 10 minutes of reaction period, the flask was removed from the fume hood and 5mL of potassium iodate solution was added to the mixture and mixed well. 8. The mixture was immediately titrated with the sodium thiosulfate solution from the burette until the solution in the Erlenmeyer flask turned pale yellow. 9. A 2mL of starch solution was added to the flask. Since the titration was done accurately, the solution changed into a deep blue color

3 P a g e Following the color change, drop wise titration was performed to ensure that the equivalence point was reached as accurately as possible. 11. The final reading of the burette was noted down and the volume of Na 2 S 2 O 3 used to titrate was calculated by calculating the difference between the initial and the final burette readings. 12. The procedure was repeated 3 times and the results were averaged to arrive at the best possible answer. III. Collection of Data Refer to Data Sheet in the Appendix IV. Results and Calculations From conducting the experiment, the following data was obtained. Trial # Volume of KBrO 3 (0.0200M) Volume of Na 2 S 2 O 3 (0.1003M) Trial mL 12.00mL Trial mL 10.60mL Trial mL 10.75mL Table 1.1 Volume of KBrO 3 titrated with Na 2 S 2 O 3 (for more details, refer to appendix) Number of moles of Na 2 S 2 O 3 and KBrO 3 was calculated using where n is the number of moles and M is the Molarity of the given solution and v is the volume of the given solution 1. BrO I - + 6H 3 O + Br I H 2 O Equation1 [I S 2 O 3-3I - + S 4 O 6 - ] x Equation2 BrO S 2 O H 3 O + Br - + 9H 2 O + 3S 4 O Equation3 (Equation1 + 2) By adding Equation 1 and Equation 2, the following observations were made.

4 P a g e 3 1 mole of BrO reacts to give 3 moles of I 3 1 mole of I reacts with 2 moles of S 2 O 3 Therefore, 1 mole of BrO - 3 reacts with 6 moles of S 2 O - 3. By factoring the above observations into the reaction, the excess moles of BrO - 3 was calculated using the following method. Using the same steps as listed above, the excess KBrO 3 was calculated for Trial #2 and Trial #3 and the results are summarized as follows: Trial # Volume of KBrO 3 (0.0200M) Volume of Na 2 S 2 O 3 (0.1003M) n KBro3 (excess) Trial mL 12.00mL x 10-4 mol Trial mL 10.60mL x 10-4 mol Trial mL 10.75mL x 10-4 mol By taking the average of the excess KBrO3 from the three trials, a reasonably accurate measurement was calculated. = x 10-4 Therefore, # moles of KBrO 3 reacted = Then, by setting up the following equation, the # of moles of electrons transferred was found.

5 P a g e 4 6H 3 O + + 6e - + BrO 3 - Br - + 9H 2 O From the above oxidation half reaction, it was found that, 1 mole of BrO 3 - gains 6 electrons. Therefore the number of electrons transferred was calculated as, To find the # of moles of electrons released by NH 3 OHCl, = 6.35 can be rounded to 6 and using this, the oxidation number of the nitrogen containing product was found through the following calculation: This means the nitrogen containing compound must have an oxidation state of +5 - Therefore, the unknown product that contains nitrogen must be NO 3 V. Summary Reduction: 6H 3 O + + 6e - + BrO - 3 Br - + 9H 2 O Oxidation: 10H 2 O + NH 3 OH + NO e - + 8H 3 O + Balanced Equation: H 2 O + NH 3 OH + + BrO - 3 NO Br - + 2H 3 O + Therefore, for every 1 mole of electrons gained by BrO - 3 to become Br -, NH 3 OH + loses 1 mole of electrons VI. Experimental Technique While performing the experiment, some techniques were adopted which saved time and ensured accurate results. Some of the techniques were,

6 P a g e 5 1) Before beginning the experiment, all glassware were rinsed with distilled water to make no residue was left behind. Failure to do this would result in inaccurate results. 2) In order to save time, at the end of 10 minutes of resting period, the solutions for the next trial were measured and mixed together. Meanwhile, while the reaction in the second flask is in progress, the first one was titrated. This process ensured efficiency and accuracy of results since the solutions are being titrated immediately following the 10 minute resting period.

7 P a g e 6 VII. References 1. M. S. Silberberg. Chemistry: The Molecular Nature of Matter and Change, 6 th edition, McGraw-Hill, New York, N.Y. (2012) 2. Poe, J. Course Manual for CHM110H; CHM110H Blackboard site; 2012; p

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