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 (in mL) of household bleach that should be diluted with water to make 500.0 mL of a pH = 10.18 solution.

Answer 1

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 (in mL) of household bleach that should be diluted with water to make 500.0 mL of a pH = 10.18 solution.

Solution :-

pH=10.18

pOH= 14 – pH

pOH= 14 – 10.18

pOH= 3.82

[OH-] = antilog [-pOH]

          = antilog [-3.82]

           = 0.000151 M

OCl- + H2O ------- > HOCl + OH-

We know the OH- concentration

Using this lets calculate the initial concentration of the OCl-

Ka og HOCl = 3.5*10^-8

Kb of OCl- = Kw/ Ka

                   = 1*10^-14 / 3.5*10^-8

                   = 2.86*10^-7

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

2.86*10^-7 = [1.51*10^-4][1.51*10^-4]/[x]

2.86*10^-7 = [1.51*10^-4]^2 /x

X=[1.51*10^-4]^2/2.86*10^-7

X= 0.07972 M

Lets calculate the moles of the OCl-

Moles = molarity * volume

           = 0.07972 mol per L * 0.500 L

           = 0.03986 mol

Now lets find the mass of NaOCl

Mass of NaOCl= moles * molar mass

                          =0.03986 mol * 74.44 g per mol

                         = 2.97 g

Now we know bleach is 5.25 % NaOCl

Means 100 ml solution contains 5.25 g NaOCl

We need 2.97 g NaOCl

So

Volume of bleach needed is

2.97 g * 100 ml / 5.25 g = 56.6 ml

So the volume of bleach needed is 56.6 ml