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A 0.0900 g sample of household bleach is completely reacted with KI(s). The resulting solution is then titrated with 0.150 M NaS2O3 solution. 0.750 mL of the solution is required to reach the colorless endpoint. What is the mass percent of NaOCl (MM = 74.44 g/mole) in the bleach?

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Oseni

Answer:

9.305%

Explanation:

First, let us consider the balanced equation between the two compounds involved in titration:

[tex]NaClO + NaS_2O_3 --> Na_2S_2O_3 + ClO[/tex]

1 mole of each of NaClO and [tex]NaS_2O_3[/tex] is required for complete neutralization.

mole of [tex]NaS_2O_3[/tex] = molarity x volume

                              = 0.150 x 0.00075

                                 = 0.0001125‬ mole

0.0001125 mole‬ of [tex]NaS_2O_3[/tex] will therefore require 0.0001125 mole of NaClO for complete neutralization.

Mass of NaClO = mole x molar mass

                             = 0.0001125 x 74.44

                               = 0.0083745‬ g

Percentage by mass of NaClO in the bleach = 0.0083745‬/0.0900 x 100

            = 9.305%

The percentage by mass of NaClO in the bleach is 9.305%.

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