Question

25 - 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 = 10.26 solution. Use the Ka of hypochlorous acid found in the chempendix.

Answer 1

Assume a basis of V = 1 L solution

so

mass = 5.25/100*1000 = 52.5 g of NaOCl

mol = mass/MW = 52.5/58.44 = 0.89835 mol of NaOCl

M= mol/V = 0.89835 / 1 L = 0.89835 M

Mstock = 0.89835M, Vstock = ?

now...

Mfinal = ?

NaOCl <-> Na+ + OCl-

OCl- + H2O <-> HOCl + OH- Kb

Kb = Kw/KA = (10^-14)/(3.5*10^-8) = 2.85*10^-7

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

2.85*10^-7 = x*x/(M-x)

find x via:

pOH = 14-pH = 14-10.26 = 3.74

[OH-] = 10^-3.74 =1.819700*10^-4

so [HOCl] = [OH-] = 1.819700*10^-4

[OCl-] = M - 1.819700*10^-4

substitute in Kb

2.85*10^-7 = (1.819700*10^-4)(1.819700*10^-4) / (M-1.819700*10^-4)

M = (1.819700*10^-4)^2 / (2.85*10^-7) + (1.819700*10^-4) = 0.116368 M

so

M2 = 0.116368, V2 = 500 mL

Mstock = 0.89835 M, Vstock = ?

Mstock * Vstock = M2*V2

Vstock = M2*V2/Mstock = (0.116368*500)/0.89835 = 64.767 mL

we need 64.767mL from solution