Question

A solution of household bleach contains 5.25% sodium hypochlorite, NaOCl, by mass. Assuming that the density of bleach is the same as water, calculate the volume of household bleach that should be diluted with water to make 500.0 mL of a pH = 10.10 solution.

Answer 1

Solution :-

pH= 10.10

pOH= 14 – pH

pOH= 14 – 10.10

        = 3.90

Now lets find the OH- using the pOH

[OH-] = antilog [-pOH]

           = antilog [-3.90]

          = 1.26*10^-4 M

Now lets find the initial concentration of the OCl- needed to give this OH- using the Kb of the OCl-

Ka of HOCl = 2.9*10^-8

Kb = kw / ka

   = 1*10^-14 / 2.9*10^-8

   = 3.45*10^-7

Kb= [HOCl][OH-] /[OCl-]

3.45*10^-7 = [1.26*10^-4][1.26*10^-4]/[OCl-]

  [OCl-] = [1.26*10^-4][1.26*10^-4]/ 3.45*10^-7

             = 0.046 M

Now lets find the moles of OCl-

Moles = molarity * volume in liter

           = 0.046 mol per L * 0.500 L

          = 0.023 mol

Now lets find the mass of NaOCl

Mass of NaOCl= moles * molar mass

                           = 0.023 mol * 74.44 g per mol

                          = 1.71 g

Now lets find the mass of the bleach solution

1.71 g NaOCl * 100 % / 5.25 % = 32.6 g

Since the density is same as water so the volume of the bleach solution needed = 32.6 ml