Lab 3 Commercial bleach3
4 Pages

Lab 3 Commercial bleach3

Course Number: CHEM 102, Spring 2010

College/University: Canisius College

Word Count: 1373

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I Purpose: To find the amount of hypochlorite ion present in a solution of bleach using oxidation-reduction titration. II Procedure: 1) Dilute the concentrated bleach. Use a pipet bulb and a 5-mL transfer pipet to measure 5.00mL of a commercial bleach solution into a 100mL volumetric flask. Dilute to the mark with distilled water, stopper and mix well. 2) Measure the potassium iodide. Weigh out approximately 2 g...

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Purpose: I To find the amount of hypochlorite ion present in a solution of bleach using oxidation-reduction titration. II Procedure: 1) Dilute the concentrated bleach. Use a pipet bulb and a 5-mL transfer pipet to measure 5.00mL of a commercial bleach solution into a 100mL volumetric flask. Dilute to the mark with distilled water, stopper and mix well. 2) Measure the potassium iodide. Weigh out approximately 2 g solid KI. This is a large excess over that which is needed. 3) Oxidize the Iodide ion with Hypochlorite ion. Pipet 25mL of the dilute bleach into an Erlenmeyer flask. Add the solid KI and about 25mL of distilled water. Swirl to dissolve the KI. Work in a fume hood and slowly, with swirling, add approximately 2mL of 3 M HCl. The solution should be dark yellow to redbrown from the presence of the I3- complex ion. 4) Titrate the iodine. Record the initial buret reading. Titrate with a standard .10 M sodium thiosulfate solution until the iodine color becomes light yellow. Add one dropper of starch solution. The blue color of the starch-iodine complex should appear. Continue the titration until one drop of Na2S2O2 solution causes the blue color to disappear. Record the final buret reading. 5) Repeat Repeat the titration beginning with step 2 two more times. III Data Collection: Initial buret reading: 50 ml Final Buret reading 1 (Turned yellow): 26.0 ml Final Buret reading 2 (Turned yellow): 27.5 ml Final Buret reading 3 (Turned yellow): 25.1 ml Final Buret reading 4 (Turned yellow): 24.5 ml Final Buret reading 1 (turned blue): 0.5ml later Final Buret reading 2 (turned blue): 0.7ml later Final Buret reading 3 (turned blue): 0.5ml later Final Buret reading 4 (turned blue): 0.3ml later IV Data analysis: Data and Calculations 1) Use the equations given to determine the number of moles of sodium thiosulfate that are equivalent to one mole of sodium hypochlorite. 1 mol ClO(-) 1 mole I2 1 mol ClO (-) 1 mol I3 1 mol I2 2 mol S2O3 1 mol I3 = 2 mol S2O3 2) Calculate the average volume and molarity of Na2S2O2 needed for the titration of 25.00 mL of diluted bleach. - 27.5 48.1/ 4 = 25.8 25.1 24.5 +26.0 48.1 3) Use the average volume and molarity of Na2S2O2 to determine the molarity of the diluted bleach. - 25.8 mL Na2S2O2 1L .1 mol 1mol bleach 1mol 1000ml 1000mL 1L 2 mol Na2S2O2 25 ml 1L = .0516 M 4) Calculate the molarity of the commercial bleach. - 5 ml = X X= 1.25 ml 100ml 25mL 25.8 mL 1L .1 mol Na2S2O2 1 mol ClO (-) 1mol = 1.032 1.03 M 1000mL 1L Na2S2O2 2 mol Na2S2O2 .000125 ml 5) Assume that the density of the commercial bleach is 1.08g/ml. Calculate the percent by mass of NaClO in the commercial bleach. - 1.03 mol NaClO 1 L 1 ml 74.49g NaClO 100 = 7.10 % 1000ml 1.08g 1mol NaClO 6) Read the label of the commercial bleach to find the percent by mass NaClO that is reported. Calculate the percent error of your value, assuming that the label is correct. - abs( Actual Theoretical) 100 Theoretical Abs(7.1 -6) 100 = 18.3% 18% 6 7) Calculate the average deviation of the three values you obtained for the molarity of the dilute bleach. - abs ( 25.8 26.1) = .2 3.9/4 = .975 abs ( 25.8 27.5) = 1.7 abs( 25.8 25.1) = .7 abs( 25.8 24.5) =1.3 + 3.9 Discussion 1) Define oxidation and reduction. - Oxidation-A reaction in which the atoms in an element lose electrons and the valence of the element is correspondingly increased. - Reduction- A decrease in positive valence or an increase in negative valence by the gaining of electrons 2) Write balanced oxidation and reduction half reactions for the equations (1) and (3). For each Half-reaction, identify which substance is oxidized or reduced. - (1) 2H(+) + ClO(-) + 2I(-) Cl(-) + I2 + H2O ClO(-) Cl(-) 2I(-) I2 Oxidizing ) Iodine Reduced) Chlorine 2H(+) + ClO(-) Cl(-) + H2O 2e(-) +2H(+) + ClO(-) Cl(-) + H2O 2I(-) I2 + 2e(-) 2e(-) +2H(+) + ClO(-) Cl(-) + H2O 2H(+) + ClO(-) + 2I(-) Cl(-) + I2 + H2O -(3) I3(-) + 2S2O5 (2-) 3I (-) + S4O6 I3(-) 3I 2S2O5 (2-) S4O6 (2-) Oxidizing) Iodine Reduction) In Sulfur 3) this analysis, an aliquot or diluted fraction of the initial solution is used for the titration. What advantage is there in diluting the original solution for the analysis? - The original solution NaClO are made by bubbling chlorine gas into a sodium hydroxide solution. The advantage of diluting the solution is helping to lower the concentration of the solution to obtain the final product. If the solution is too concentrated, the polymer molecules might get close enough together to interact with each other. The final results will be ruined and there will be a huge error in the lab. 4) How many 25-mL aliquots can be measured from a 100-ml volumetric flask? Explain. - In this lab the average amount of solution needed was 25.8 for this lab. If we were to obtain 25-ml of the solution then 4, 25-ml aliquots, can be measured from a 100-mL volumetric flask. If we were to keep the result we obtained in this lab then 3.88, 25.8 aliquots, can be measured. 5) The reaction with thiosulfate ions produces the dithionate ion, S4O6 (2-). Calculate the oxidation number of sulfur in this ion. Do you think that sulfur atoms in the ion will all have the same oxidation number? What might the oxidation numbers be? - (+10) (-12) (+10) S4, S4 O6 (2-) S oxidation # (+2.5) - No, the sulfur ions would not have the same charge because 2.5 contain a decimal and oxidation numbers must be in whole numbers. - The oxidation numbers of sulfur could be 2,2,3,3 or 4,4,1,1 As long as the final total of the four numbers equals ten. 6) How would each of the following laboratory mistakes affect the calculate value of the present NaClO in the commercial bleach (too high, too low, no change)? Explain a) In step 1, the pipet was rinsed with distilled water immediately before being used to measure the commercial bleach solution. - No change, the solution was already going to be diluted after being measure with the pipet. b) In step 2, 3g of KI was used instead of 2g - Too High, the amount of KI is a large excess over that which is needed making the amount of KI crucial to the experiment. c) In step 3, some of the iodine that formed vaporized from the solution. - Too High. In order for the solution to turn from dark yellow to red-brown the presence of iodine (I3(-)) was needed. 7) What is the major source of error in this experiment? Explain. The major errors in this experiment were: 1) Thiosulfate Very Old 2) Buret was not coated or water was present in the buret. V Conclusion The purpose of this lab was to find the amount of hypochlorite ions present in the solution of bleach using redox titration. Our final results in this lab were in a range from 20-25 ml of the sodium thiosulfate solution used in the titration. The final product of this lab was the solution changing from a red-brown solution to a clear colored solution. The patterns found in out lab was only the range in how much sodium thiosulfate was used in titration (20-25ml) to acquire the clear colored solution we needed. The relevant chemical concepts of this lab were on Redox reactions and how common they are in real life. In conclusion the lab provided us with much knowledge of the behavior of everyday items that go under redox reactions to make life easier for us. VI Error Analysis 1) Two specific examples of sources of error in this lab are: a) The thiosulfate was very old b) The buret was not coated properly with water or water was present when titrating. 2) The error found in this lab would not change much of the results found in this lab too much. For example if the buret was not coated well and water was present, the solutions final product would not change too much because water will only dilute the thiosulfate used a little. If it was the old thiosulfate then the solutions final product would change because the solution would be contaminated, therefore causing the final product to change dramatically. 3) abs( Actual Theoretical) 100 Theoretical Abs(7.1 -6) 100 = 18.3% 18% 6