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CHEM2090-LAB-EXP7-2010 - EXPERIMENT 7 Sodium Hypochlorite...

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7 – 1 EXPERIMENT 7 Sodium Hypochlorite in Bleach Objective: Commercially available bleaching solutions (such as Clorox™) are produced by dissolving chlorine gas in aqueous sodium hydroxide: Cl 2 ( g ) + 2NaOH( aq ) → NaOCl( aq ) + NaCl( aq ) + H 2 O( l ) (1) The active ingredient in such products is sodium hypochlorite, NaOCl, which is dissociated completely into Na + ( aq ) and OCl ( aq ) ions in aqueous solution. The hypochlorite ion, OCl , is an oxidizing agent. In bleaching laundry, clothing stains are oxidized by the hypochlorite, and hypochlorite ion is, in turn, reduced to chloride ion, Cl . In this experiment you will be using the ability of hypochlorite ion to serve as an oxidizing agent to determine the molarity of hypochlorite ion in commercially available bleach, described by the following two-step procedure. Introduction: In the first step, you will mix a diluted sample of bleach with a solution containing an excess of iodide ion, I , a reducing agent. Hypochlorite ion will become reduced to chloride ion, and iodide ion will become oxidized to iodine, I 2 , which has a brown color in aqueous solution. The net ionic equation is: OCl ( aq ) + 2I ( aq ) + 2H 3 O + ( aq ) → I 2 ( aq ) + Cl ( aq ) + 3H 2 O( l ) (2) As can be seen from equation (2), the number of moles of iodine produced in the reaction will be equal to the number of moles of hypochlorite ion in the initial sample (aliquot) of bleaching solution. Thus, if you can determine how many moles of iodine are produced you will have determined the number of moles of hypochlorite ion initially present in the diluted sample of bleach. The amount of iodide ion present is not important, so long as it is in excess. In the second step of the procedure you will measure the amount of iodine produced by reaction (2) by titrating the iodine with a standardized solution of thiosulfate ion, S 2 O 3 2– . This is an example of an oxidation-reduction (“redox”) titration: I 2 ( aq ) + 2S 2 O 3 2– ( aq ) → 2I ( aq ) + S 4 O 6 2– ( aq ) (3) brown colorless colorless colorless To carrying out the titration you will place the standardized thiosulfate solution in a buret and add it progressively to a flask containing aqueous iodine produced by reaction (2). As the thiosulfate combines with the iodine, the iodine solution will gradually become less and less brown because the reaction products, I ( aq ) and tetrathionate ion, S 4 O 6 2– ( aq ), are both colorless. When you get close to the equivalence point, the solution will turn from brown to a pale yellow color. In principle, you could continue the titration until the solution became completely colorless, but in practice this is very difficult to accomplish because it involves trying to identify the exact point when the solution turns from very slightly pale yellow to colorless. To circumvent this difficulty you will use a method that takes advantage of the fact that iodine combines with starch to form an intensely colored, dark blue complex. If a starch solution contains the slightest amount of iodine, it is easy to see the blue color. Iodide ion, on the other
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