Question

A solution of household bleach contains 5.25% sodium hypochlorite, NaOCI, 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 = 10.26 solution. Use the K_a of hypochlorous acid found in the chempendix. 

Answer 1

pH = 10.26

pOh = 14-10.26 =3.74

[OH-] = 10^-pOH = 10^-3.74 = 0.00018197

note that

NaOCl = Na+ + OCl-

OCl-+ H2O <-> HOCl + OH-

in equilbirium

[HOCl] = [OH-] = 0.00018197

[OCl-] = M- 0.00018197

Kb = (kw/Ka) = (10^-14)/(10^-7.52) = 3.311*10^-7

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

3.311*10^-7 =(0.00018197)(0.00018197)/(M-0.00018197)

M = (0.00018197^2)/(3.311*10^-7) + 0.00018197

M = 0.100191

NaOCl = 0.100191 M

V = 500 mL = 0.L

mol = MV = 0.100191*0.5=0.0500955

mass = mol*MW = 0.0500955*74.44 = 3.729 g of NaOCl required

%solutoin = 5.25% m/V

V = 3.729/0.0525

V = 71.028 mL required