Analysis of two commercial brands of bleaching solution Essay

Published: 2019-10-10 12:24:29
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Objective To determine the concentrations of the active ingredients in 2 commercial bleaches. Introduction Sodium hypochlorite is usually found in bleaching solutions. It is the active ingredient of bleaching solutions. It bleaches by oxidation. When it is added to dye, the following reaction occurs: ClO- + dye ”””””-> Cl- + (dye + O) If the oxidized form of the dye is colorless, then the color of the dye would fade away. In the presence of acid, the hypochlorite ions from the bleaching solution reacts with the iodine ions from potassium iodide in the following way: ClO- + 2I- + 2H+ ”””>I2 + H2O + Cl-.

When sodium thiosulphate solution is added into this reacted solution, a further reaction occurs: I2 + 2S2O32- ””””>2I- + S4O62- This reaction could be used in titration to find out the number of moles of thiosulphate ions, thus the concentration of hypochlorite ions in the bleaching solution. Procedure 1. 10 cm3 of Kao Bleach was pipette into a volumetric flask. Distilled water was added until the meniscus reaches the graduation point. 2. 25 cm3 of the titrated bleach was pipette into a conical flask. About 10 cm3 of potassium iodide and dilute sulphuric acid was added into the conical flask. 3.

The solution was titrated with sodium thiosulphate solution until the brown colour of the iodine fades. 4. Starch solution was added into the conical flask, and the solution was further titrated until the dark-brown colour of the starch-iodine complex turns to colourless. The volume of sodium thiosulphate solution required to reach the end point was recorded. 5. Steps 1 to 4 were repeated 3 more times. 6. Steps 1 to 5 were repeated using Clorox Bleach. Data and Calculation Molarity of standard Na2S2O3 solution = 0. 05182M Brand A: Kao Price: $11. 9/ 1500ml Trial 1 2 3 Final reading/cm3 26. 8 23. 1 25. 7 26. 0 Initial reading/ cm3.

4. 1 0. 4 2. 9 3. 2 Volume of Na2S2O3 22. 7 22. 7 22. 8 22. 8 Average no. of moles of Na2S2O3 used in the titration: 0. 05182 X (22. 7 + 22. 8 X 2)/3 X 0. 001 = 1. 180 X 10-3 moles So, there are (1. 180 X 10-3 /2) =5. 90 X 10-4 moles of I2 in the reaction So, there are 5. 90 X 10-4 moles of ClO- ions in the diluted solution. Concentration of ClO- in Kao bleach = 5. 90 X 10-4 X 10 /10 X 1000 =0. 5899M Brand B: Clorox Price: $21. 9/ 2840ml Trial 1 2 3 Final reading/cm3 33. 5 32. 4 32. 0 33. 2 Initial reading/ cm3 2. 7 1. 3 0. 9 2. 3 Volume of Na2S2O3 30. 8 31. 1 31. 1 30. 9 Average no. of moles of Na2S2O3 used in the titration:

0. 05182 X (30. 9+ 31. 1 X 2)/3 X 0. 001 = 1. 608 X 10-3 mole So, there are (1. 608 X 10-3 /2) =8. 041 X 10-4 moles of I2 in the reaction So, there are 8. 041 X 10-4 moles of ClO- ions in the diluted solution. Concentration of ClO- in Kao bleach = 8. 041 X 10-4 X 10 /10 X 1000 =0. 8041M Conclusion The concentration of ClO- in Kao is 0. 5899M while that of Clorox is 0. 8041M. Discussion 1. When we add starch solution into the conical flask, the solution turns dark blue. After that, when we add a few drops of sodium thiosulphate, the colour of the solution would turn colourless. We must be careful when we are doing this step.

This is because the starch-iodine complex does not show graduation of color change. We may get pass the end point easily. The readings would be inaccurate. 2. Dilute sulphuric acid is irritating. So we must be extra careful in using it. How did I use sulphuric acid more safely? I used a larger measuring cylinder to measure out the amount of sulphuric acid. The likeliness of spilling the acid would be lower. 3. After I had done all the experiments, I found out that the tip of the pipette was broken. When I asked Mr. Leung, he said that the pipette could not be used anymore. Why? I could think of 2 reasons.

First, the broken tip of the pipette could cause danger when we are using the pipette. We would have a higher chance of getting our finger cut. Second, the broken tip of the pipette may cause the solution to be carried to leak. So, it is unreliable. Answers to Study Question 1. (a) Amount of active ingredient in Kao: 0. 5899 X (35. 5 + 16) = 30. 38 g /dm3 Amount of active ingredient in Clorox: 0. 8041 X (51. 5) = 41. 41 g/dm3 (b)Cost per gram of Kao: (11. 9 X 1000/1500) /30. 38 = $0. 2611 per gram Cost per gram of Clorox: (21. 9 X 1000/2840) / 41. 41 = $ 0. 186 per gram 2. As Clorox is of a lower price, it is the better buy.

3. Adding potassium iodide in excess ensures that all chlorate ions have reacted. Only when all the chlorate ions have been reacted that the amount of iodine formed can fully reflect the amount of chlorate ions in the solution. This ensures that the volume of sodium thiosulphate used in the titration can be used to determine the number of moles of chlorate ions in the solution. 4. When an acid is added into a solution of chlorate and iodine ions, iodine would be liberated. The iodine can then be used in titration to determine the amount of the chlorate ions. 5. The second way it may deteriorate is by decomposition by sunlight:

2H+ + 2ClO- ”””-> 2HCl + O2 The chlorate ions, under sunlight, decompose back to chlorine ions and the bleaching ability of the bleaching solution is reduced. 6. This is because before reaching the end point, starch solution will not show any signs of being close to the end point. Other indicators will. For example, if methyl orange is close to its end point, it will first change the color of the solution to orange. Then, the solution would turn back to its original colour. In this way, we will know that we are close to the end point and we would add the solution more slowly.

However, similar characteristics could not be found when we use starch as an indicator. So, we may get pass the end point easily. This problem is solved by titrating the iodine solution without any indicator first. After the brown color of iodine vanishes, we know that we are very close to the end point. At this stage, adding starch solution can tell us whether there is still iodine in the solution. As we know that we are already very close to the end point, we would add the solution more slowly. It would be lees likely to shoot pass the end point.

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