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 = 10.26 solution.

Answer 1

Assume that total mass of solution is 100 g.

Given that; 5.25% sodium hypochlorite, NaOCl, by mass means 5.25 g NaOCl .

Now calculate the molarity of this solution :

5.25 g NaOCl / molar mass =

5.25 g/74.44 g/mol = 0.071 moles.

1.00 g/mL x 1000 mL x 0.0525 x (1/74.44 )

= 0.7 1M for the 5.25% bleach.

Here pH = 10.26, then pOH = 3.74 and OH^- = 1.82*10^-4 M

NaClO <= > Na+     ClO-

ClO^- + H2O ==> HClO +    OH^-

At equilibrium .x.......x.....             1.82*10^-4.. 1.82*10^-4

Kb = [HClO][ OH^-]/[ ClO-]

Kb from your work above = (HClO)(OH^-)/(ClO^-)

The Kb(base dissociation constant) for this reaction will be

Kw/Ka or (the self ionization constant for water)/(acid dissociation constant for HClO)

(1 x 10 ^-14)/(3.0 x 10^-8) = 3.33 x 10 ^ -7

Kb = [HClO][ OH^-]/[ ClO-]

3.33 x 10 ^ -7 = 1.82*10^-4 * 1.82*10^-4 / [ ClO-]

[ ClO-]= 0.0995 M

Now calculate the volume (in mL) of household bleach that should be diluted with water to make 500.0 mL of a pH = 10.26 solution use the following expression:

M1V1=M2V2

0.0995 M*500ml = 0.71 *V2

V2= 70.0 ml