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# analysis of two commercial brands of bleaching solution

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Introduction

F.6A Lam Pik Sum (10) Title: analysis of two commercial brands of bleaching solution Date: 6/10/2008 Objective: to find out which of the two brands of bleach is cheaper base on their actual bleaching strength. Introduction Redox titration can be divided into two parts: in one half the reducing agent loses electrons and in other half the oxidizing agent gains electrons. The stoichiometric equation for a redox reaction may then derive using the fact that all the electrons lost by the reducing agent must be gained by the oxidizing agent. In this experiment, we need to find out the cost of active ingredient per gram in two different brands of household bleaches in other to find out with bleach is cheaper. The two bleaches we use are KAO (\$13.9 for 1.5L) and Best Buy (\$11.9 for 2L). Sodium hypochlorite forms the basis of most of commercial bleaches. In this analysis, the sodium hypochlorite is allowed to react with an excess of potassium iodide solution in the presence of acid, liberating iodine, which is then titrated against standard sodium thiosulphate solution. Reactions involved: ClO- + 2 I- + 2 H+ --> I2 + H2O + Cl- I2 + 2 S2O32- --> 2 I- + ...read more.

Middle

= 0.569860483 g Strength of the original bleach = 0.569860483 � 0.01 = 56.9860483 g dm-3 = 57.0 g dm-3 (corr. to 3 sig. fig.) Available bleach present in one bottle = 0.569860483 x 150 = 85.47907245 g = 85.5 g (corr. to 3 sig. fig.) Cost of bleach per gram = \$13.9 � 85.47907245 g = \$ 0.163 (corr. to 3 sig. fig.) Best Buy In the titration, iodine in the reaction mixture react with the sodium thiosulphate as the following equation: I2 + 2S2O32- � 2I- + S4O62- Mole ratio of I2 to S2O32- is 1: 2 As 14.167 cm3 of sodium thiosulphate is used, No. of mole of sodium thiosulphate = 0.07676M x 0.014167 dm3 = 0.00108745892 mol. No. of mole of iodine = 0.00108745892 � 2 = 0.00054372946 mol. Iodine present in the reaction mixture is come from the reaction between bleach and potassium iodide and dilute sulphuric acid as the following equation: OCl- + 2I- +2H+ � I2 + H2O + Cl- Mole ratio of OCl- to I2 is 1:1 No. of mole of diluted sodium hypochlorite (25 cm3) = 0.00054372946 mol. No. of mole of diluted sodium hypochlorite (250 cm3) = 0.00054372946 x 10 = 0.0054372946 mol. No. ...read more.

Conclusion

The addition of starch indicator at the end-point is to check that whether there is any iodine still present in the solution. When this titration is near to the end-point, the I2/I- solution will become very pale yellow that we may wrongly judge it as colourless. The addition of starch solution can clearly show the presence or absence of iodine in the solution as starch will form a dark blue aqueous complex with iodine. 7) Why do thiosulphate ions only reacts with I2 and not H+ in the experiment? Thiosulphate ions only react with I2 and not H+ in the experiment because I2 is a stronger oxidizing agent than. Therefore thopsulphate ions is more likely to react with I2 than H+. 8) Explain the action of bleaching. Colour in most dyes and pigments is produced by molecules, such as beta carotene, which contain chromophores. Household bleach solution is a kind of oxidizing bleach due to the hypochlorite ions. They react with many organic and inorganic compounds. An oxidizing bleach works by breaking the chemical bonds that make up the chromophore. This changes the molecule into a different substance that either does not contain a chromophore, or contains a chromophore that does not absorb visible light. End of Report ...read more.

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