Instructions: Show your work CLEARLY ON A SEPARATE SHEET. WRITE THE ANSWERS to 2
ID: 698795 • Letter: I
Question
Instructions: Show your work CLEARLY ON A SEPARATE SHEET. WRITE THE ANSWERS to 2, 3 and 4 ON THIS SHEET. Put your name on the separate sheet as well. 1. On a separate sheet, write the four relevant equations and show how to arrive at the overall balanced net ionic equation for the analysis of sodium hypochlorite in a bleach solution, using sodium thiosulfate as the limiting reactant. Remember: to get the overall equation you need to apply the rules for full ionic equations as well as the rules for strong, weak, and nonelectrolytes. 2. Suppose you want to find out how much sodium hypochlorite is in a commercial bleach solution. From the commercial quart bottle, you take 20.00 mL, deliver that to a 200.00 mL volumetric flask, dilute to the mark with deionized water, and mix well. You then titrate a 20.00 mL aliquot (of the dilute solution) with 16.25 mL of a 0.1500 M sodium thiosulfate solution to the endpoint. How many moles of Naoci are in the dilute sample in the 200.00 mL? 3. Knowing that you made a 1:10 dilution of your original bleach solution to perform the analysis in question #2, calculate the grams of Naocl in the original 20.00 mL of concentrated bleach. 4. What is the % (by mass) sodium hypochlorite in a quart of your (original) bleach solution? Assume Dundil. bleach 1.095 g/mL . 1 qt = 946.35 mL.Explanation / Answer
1) Sodium hypochlorite (NaOCl) ionizes in aqueous solution to produce hypochlorite ion which oxidized iodide to iodine. Iodine is reduced by thiosulfate to iodide. The ionic equations are given below.
NaOCl (aq) ---------> Na+ (aq) + OCl- (aq) ……(1)
OCl- (aq) + 2 H+ (aq) + 2 I- (aq) --------> I2 (aq) + Cl- (aq) + H2O (l) ……(2)
I2 (aq) + I- (aq) -------> I3- (aq) ……(3)
I3- + 2 S2O32- --------> S4O62- (aq) + 3 I- (aq) ……(4)
2) Note the stoichiometry of the reactions above.
2 mole S2O32- = 1 mole I3- = 1 mole I2 = 1 mole OCl-.
Millimoles of S2O32- required for the titration = (16.25 mL)*(0.1500 M) = 2.4375 mmole.
Millimoles of OCl- in the diluted sample = (2.4375 mmole)*(1 mole OCl-/2 mole S2O32-) = 1.21875 mmole = (1.28175 mmole)*(1 mole/1000 mmole) = 1.21875*10-3 mole.
Note that 20.00 mL aliquot was taken from 200.00 mL solution prepared; therefore, mole(s) of OCl- in 200.00 mL solution prepared = (1.21875*10-3 mole)*(200.00 mL)/(20.00 mL) = 0.0121875 mole (ans).
3) The test solution was prepared by diluting the concentrated solution to 200.00 mL; the dilution factor is (200.00 mL)/(20.00 mL) = 10.00.
Mole(s) of OCl- in 20.00 mL of the concentrated solution = (0.0121875 mole)*(10.00) = 0.121875 mole.
Molar mass of NaOCl = (1*22.9897 + 1*15.9994 + 1*35.453) g/mol = 74.4421 g/mol.
Mass of NaOCl in the 20.00 mL concentrated solution = (0.121875 mole)*(74.4421 g/mol) = 9.072630 g 9.0726 g (ans).
4) 20.00 mL of the bleach solution was taken; the density of bleach solution is 1.095 g/mL. The bottle contains 1 qt = 946.35 mL.
Mass of bleach solution = (946.35 mL)*(1.095 g/mL) = 1036.25325 g.
Mass of OCl- in the quart bottle = (9.0726 g)*(946.35 mL)/(20.00 mL) = 429.2927 g.
%OCl- in the quart bottle of bleach = (429.2927 g)/(1036.25325 g)*100 = 41.42739% 41.427% (ans).