A 10.0 mL sample of liquid bleach is diluted to 100. mL in a volumetric flask....

A 10.0 mL sample of liquid bleach is diluted to 100. mL in a volumetric flask. A 25.0 mL aliquot of this solution is analyzed using the procedure in this experiment. If 10.4 mL of 0.30 M Na2S2O3 is needed to reach the equivalence point, what is the percent by mass NaClO in the bleach? Hint: Percent by mass is grams of NaClO / mass of bleach. Presume a density for the bleach of 1.00 g/mL.

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Expert Solution

First, in this experiment, NaOCl is made to react with HI, which produces I2, which gets converted to I3-

Next, this I3- reacts with the Na2S2O3 to determine the end point.

Twice the amount of Na2S2O3 is needed to reduce I2 to I- , and also twice the amount of NaOCl is needed to oxidize I- to I2

So effectively,

molar ratio of Na2SO3:NaOCl is 2:1 in this titration experiment.

Moles of Na2S2O3 used = Molarity*Volume = 0.0104*0.3 = 0.00312

So, moles of NaOCl present = 0.00312/2 = 0.00156

Mass of NaOCl present = Moles*MW = 0.00156*74.4 = 0.116 g

This is the mass present in 25 mL aliquot.

So mass present in the original 100 mL dilution = 4*0.116 = 0.464 g

Mass of bleach taken = Volume*density = 10*1 = 10 g

So. %mass = (0.464/10)*100 = 4.64%

queen_honey_blossom answered 1 year ago

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