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. Use the Ka of hypochlorous acid found in the chempendix. (4.0e-8)

Answer 1

Ka of hypochlorous acid = (4.0e-8)

Given that pH = 10.00 solution

pH required is 10.00, therefore pOH = 14-10.00 = 4.00

[OH-] = 10^-4.00 = 1.00*10^-4 M

Set up ICE table
                           [ClO-]                    [HClO]                  [OH-]
Initial                     x                             0                           0
Change            -1.00*10^-4     +1.00*10^-4      +1.00*10^-4 Equilibrium       x-1.00*10^-4    1.00*10^-4     1.00*10^-4

Per online table Ka of hypochlorous acid is 4.0*10^-8 therefore Kb of hypochlorite = Kw/Ka
Kb = Kw/Ka = 10^-14 / 4.0*10^-8 = 2.5*10^-7

therefore Kb = [HClO]*[OH-]/[ClO-]
2.5*10^-7 = 1.00*10^-4 * 1.00*10^-4 / (x-1.00*10^-4)

x-1.00*10^-4 = 1.00*10^-4 * 1.00*10^-4 /2.5*10^-7

x = 0.0401

Therefore initial [ClO-] is 0.0401 M, but only need 500 mL
mols ClO- needed = 0.0401 mol/L * .5 L
mols ClO- needed = 0.02005 mols
Since NaClO dissociation is 1:1 ratio mols ClO- needed = mols NaClO needed
mols NaClO needed = 0.02005 mols

At 5.25% NaOCl, 1L bleach = 0.0525L NaOCl = 52.5 mL NaOCl
Per density of 1 g/mL then 52.5 mL NaOCl = 52.5 g
NaOCl molar mass = 74.439 g/mol
So, mol NaOCl = 52.5 g / 74.439 g/mol = 0.7053 mol
Since this is the amount of NaOCl in 1L of bleach then [NaOCl] = 0.7053 mol/L

[NaOCl] * volume of bleach = mols NaOCl needed
volume bleach = mols NaOCl needed / [NaOCl]
volume bleach = 0.02005 mols / 0.7053 mol/L = 0.02843 L = 28.43 mL