Commercial Brands of Bleaching Solution & a Determination of the Best Buy Essay

Introduction Sodium chlorate(I) forms the basis of most commercial bleaching agentes. The amount present in bleaching settlement keister be determined by a volumetric technique. In this experiment, sodium chlorate(I) reacts with surfeit potassium iodide in the presence of acid to liberate. The iodine is titrated against standard sodium thiosulphate solution.Chemicals Bleach solution , 0.1 M Na2S2O3 ,1 M H2SO4 , KI , Starch indicatorProcedures 1. Determine the total volume of bleach in the commercial bottles provided.Note the tick off name, volume & set of each sample.2. Pipette 25 cm3 bleach solution into a 250 cm3 volumetric flask. Make up to the mark & mix well.3. Pipette 25 cm3 of this solution into a conical flask. Add 10 cm3 KI solution & 10 cm3 1 M sulphuric acid.4. Titrate this mixture against the standard sodium thiosulphate solution provided.Add starch indicator when the solution become pale yellow.5. Record your results and calculate the jettyarity of the orginal bleach solution.6. Determine which bleach has the lowest price per bulwarke of actual bleach.Results Brand A = ___________KAO_____________ Brand B = _________Best Buy____________Volume = ___________1.5L______________ Volume = ____________2L_______________Price = _______$12.9 / Bottle____________ Price = ________$10.9 / Bottle____________ gram moleculee = _________0.753 Mol_____________ mol = ___________1.152 Mol ___________$/mole = _______$17.13 / Mol ___________ $/mole = __________$9.46 / Mol__________Best buy is ________ Best Buy___________Titration of Brand A against the standard sodium thiosulphate solutionTitration1234Final buret rendering (ml)31.7035.7027.8032.20Initial Burette Reading (ml)2.909.804.408.30Volume of titrant (ml)28.8025.9025.4023.50Mean volume of titrant (ml) = __(25.90+25.40+23.50)3 = 25.1 ml__CalculationThe Result Of Brand ASodium chlorate(I) reacts with excess potassium iodide in the Bleaching solution, which is a acid medium. Iodine solution produced.Fo llowing equation 2H+-(aq) + OCl(aq) + 2I(aq) Cl(aq) + I2-(aq) + H2O-(l)Secondly, the iodine solution is titrated with sodium thiosulphate solutionFollowing equation I2(aq) + 2 S2O32- (aq) S4O62- (aq) + 2 I-(aq)The Molarity of Na2S2O3 = 0.1M bod of mole of S2O3- reacted with I2 = Molarity X Volume= 0.1 X (25.11000) = 0.00251 molThe Mole ratio of Na2S2O3 I2= 21 phone number of moles of I2 reacted with S2O3-= (12) X 0.00251 mol= 0.001255 molThe Mole ratio of NaOCl I2= 11Number of moles of NaOCl used in the titration= I2= 0.001255 molNumber of moles of NaOCl in the 250ml volumetic flask= 0.001255 mol X 10= 0.01255 molNumber of moles in 1.5L bleach solution= 0.01255 mol 25 X 1500= 0.753 molPrice of OCl- per mole in brand A= $12.9 0.753mol= $17.13 / molResults Titration of Brand B against the standard sodium thiosulphate solutionTitration1234Final Burette Reading (ml)30.9032.831.9035.20Initial Burette Reading (ml)2.103.903.306.90Volume of titrant (ml)28.828.9028.6028.90Mean volume of ti trant (ml) = __(28.90+28.60+28.90)3 = 28.8 ml__CalculationThe Result Of Brand BSodium chlorate(I) reacts with excess potassium iodide in the Bleaching solution, which is a acid medium. Iodine solution produced.Following ionic equation 2H+-(aq) + OCl(aq) + 2I(aq) Cl(aq) + I2-(aq) + H2O-(l)Secondly, the iodine solution is titrated with sodium thiosulphate solutionFollowing ionic equation I2(aq) + 2 S2O32- (aq) S4O62- (aq) + 2 I-(aq)The Molarity of Na2S2O3 = 0.1MNumber of mole of S2O3- reacted with I2 = Molarity X Volume= 0.1 X (28.81000)= 0.00288 molThe Mole ratio of Na2S2O3 I2= 21Number of moles of I2 reacted with S2O3-= (12) X 0.00288 mol= 0.00144 molThe Mole ratio of NaOCl I2= 11Number of moles of NaOCl used in the titration= I2= 0.00144 molNumber of moles of NaOCl in the 250ml volumetic flask= 0.00144 mol X 10= 0.0144 molNumber of moles in 2L bleach solution= 0.0144 mol 25 X 2000= 1.152 molPrice of OCl- per mole in brand B= $10.9 1.152 mol= $9.46 / molQuestions 1. why must the K I be present in excess ? If less than the specified quantity of KI is added, what effect will this have on the results ?The bleach solution contain sodium chlorate(I),which have ions OCl-.We kitty prepare the iodine solution by adding the sodium chlorate(I) to potassium iodide in a acidic medium. At first, the sodium chlorate(I) Is the limiting agent. When we add the excess potassium iodide into the bleach solution. Not only it will non effect the result, simply also the calculation can be more accurate. All the ions OCl- can be completely reacted.In addition, Iodine solution is only slightly soluble in water but it is very soluble in the solution, which contain I- ions.2. What is the function of the sulphuric acid ?The function of sulphuric acid is provide a acidic medium containing excess iodide to ionizes the iodine solution to triiodide ions. We must know that the Brown colour of iodine solution is cause for the triiodide ions(I3-).This colour in observation of titration is v ery important.3. Bleaching solutions may deteriorate for 2 reasons (a) react with CO2 in the air according to the equation 2 OCl- + CO2 CO32-+ H2 + Cl2(b) what is the other reason ?It must be effected by light. It is because the The hypochlorite ions OCl-( will be decompose quickly under light Following equation2NaOCl 2NaCl + O2, .While losing some OCl- ions, the result in calculation will not be accurate4. What should the starch indicator not be added too early ?The starch solution turns the iodine to blue black because of the formation of starch-iodine complex. Also, the complex is not reversible when the concentration of iodine is high. If we add the starch solution early, the attraction of starch molecules and iodine molecules will attract so strongly. Although we have add standard sodium thiosulphate solution, but we can not do completely finish the real result of titration and effecting the calculation. The above-mentioned tell us that the starch solution should be added wh en only a few of iodine solution left, near the end point of the titration.