Question

3. A student followed the procedure of this experiment to determine the percent NaOCI in a commercial bleaching solution that was found in the basement of an abandoned house. The student diluted 50.00 mL of the commercial bleaching solution to 250 ml in a volumetric flask, and titrated a 20-mi aliquot of the diluted bleaching solution. The titration required 35.46 mL of 0.1052M Na, S.O, solution. A faded price label on the gallon bottle read $0.79. The density of the bleaching solution was 1.10 g/mL. Refer to the post laboratory assignment calculations forh.j. a) Calculate the number of moles of 5,0 required for the titration. b) Calculate the number of moles of I, produced in the titration mixture c) Calculate the number of moles of Ocr ion present in the diluted bleaching solution titrated. d) Calculate the mass of Naoci present in the diluted bleaching solution titrated. e) Determine the volume of commercial bleaching solution present in the diluted bleaching solution titrated. f) Calculate the mass of commercial bleaching solution titrated, 8) Determine the percent Naocl in the commercial bleaching solution. h) Calculate the mass of one gallon of the commercial bleaching solution. i) Calculate the cost of 100 g of the commercial bleaching solution. D) Determine the cost of the amount of commercial bleaching solution required to supply 100 g of Naoci.

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Answer 1

Volume of commorceal = 50mL bleach solution Dilutions made upto =250mL The reaction taking place during tetration; Dclo + 2H+ + 21- ci+ H₂O + Iz 2) Iz + 2403 402 - +21

(a) The titration required 35.46mL of 0.1052m lags0₃ 1.1.8652x03 mol of I will be produed from 1.8652x10 and Я оu- 35. 46 mx x 0.1052 mol 3.7304 mmol of Naz Szoz Molarity - moles Volume Na₂S₂O3 Nat + 5₂03² : 1 mol of Na₂S2O3 produces 1 mol of = 1.8652 xoß mol 20 mL xlo7 S₂9²- 3. 7304 mmol of Na₂S2} produers 3.7304 mmol of S₂03² Moles of 50 required = 3. 7304 No3 mol Molarity = 0.09326M Thus mdarity of 250ml flask containing diluted Bleaching solution was found to be 0.09326m Calculating concentration before dilution Mir =M_V2 Stock diluted solution Solution b) I +2407 Sub² + 21- clear from the equation that 2 moles of S₂9² - require 1 mol of Iz 1: 3.7304 X163 mol 3.7304 103 MX 50 = 0.09326 X 250 1.8652 xlo mol of Iz M 40.4663M dilection of 50m stock/ commerical solution was done & moles= molarity X volume 1 Moles of Iz required = 1.865286 mol = 0.4663 mol x 50my analysis was Thus the titration carried out. = 23.315 mmol of Naocl 18 clo + 2H+ + 2I Cl + tho + Iz clearly from the equation that; Imd of. I are produced from 1 mol of " mass = moles x molar mass = 23.315 xlo ² mol X 74.44 g mor . - 1.735579 clo- .

I Mass of Naoci present in 50 ml of Commercial Bleaching solution 1.735579 Thus, the cost of the amount I of commercial bleaching solution required to suppy loog of Naoc is 180.6013] : 1.73557g of Naoci present in 50mL of Bleach Solution loog Naoci " "50mL xloog Volume of 2880.9 mL! Bleach Solution Density = mass Volume Mass of solution = llog x 2880.9mL mK = 3168.99 g I gallon = 3.785 L = 3785mL i loog of Naoci present in 2880.9 mL of Bleach solution 1 x 2980.9 gallon 3785 = 0.76114 gallon " I gallon bottle frice = $0.79 ~ 0.76114 gallon bottle =$0.7980.76114 = $0.6013
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