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 = 10.18 solution.

Answer 1

pH required is 10.18

pOH will be 14 - 10.18 = 3.82

pOH = -log[OH^-] =

[OH^-] = 10^-3.82 = 1.51 x 10^-4

Since NaOCl is the salt of a strong base of NaOH and the weak acid of HOCl, when it dissolves it produces Na⁺ and OCl^- ions. The hypochlorite ion hydrolyses in water.

OCl^- + H₂O HOCl + OH^-

For the above equilibrium reaction, the equilbrium constant Kb is,

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

The conjugate acid of hypochlorite ion is hypochlorous acid (HOCl). Ka value of HOCl is found to be 3.5x10^-8

Dissociation of acid:
HOCl H⁺ + OCl^-

Ka = [H⁺][OCl^-] / [HOCl] = 3.5x10^-8

We know for water

H₂O H⁺ + OH^-
Kw=[H⁺][OH^-] = 1x10^-14

Kw / Ka = Kb
Kb = 1x10^-14/ 3.5x10^-8= 2.85 x 10^-7

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

OCl^- + H₂O HOCl + OH^-

[OH^-] required = 1.51 x 10^-4

therefore HOCl = 1.51 x 10^-4

2.85 x 10^{-7 =} (1.51 x 10^-4 x 1.51 x 10^-4)/[OCl-]

[OCl^-] =  (1.51 x 10^-4 x 1.51 x 10^-4)/( 2.85 x 10^-7) = 0.080 M

1000 mL should contain 0.080moles so 500 mL should contain 0.040 moles

Mol wt of NaOCl = 74.4 so 0.040 moles is 2.976 g

Household bleach contains 5.25% by wt of NaOCl, so 100 g of household bleach will have 5.25 g of NaOCl.

So 2.976 g of NaOCl will be present in 56.68 g of household bleach