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 = 10.00 solution.

Answer 1

Kw = Ka * Kb
1.0 x 10^-14 = 4.0 x 10^-8 * Kb
Kb = 2.5 x 10^-7

(1.00 g/mL) x (1000 mL x 0.0525) x (1/molarmass NaClO) = about 0.7 M for the 5.25% bleach.

If pH = 10.00, then pOH = 4.00 and [OH^-] = 1*10^-4 M
initial ClO^- + HOH ==> HClO   +    OH^-
Equil     x          x            1*10^-4     1*10^-4

Kb = (HClO)(OH^-)/(ClO^-)

2.5 x 10^-7 = (1*10^-4)^2/x

x = 0.2

M1V1 = M2V2

0.7*V1 = 0.2*500

V1 = 142.85 mL