Asolution 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. ka=4.0×10–8 

Chemical reaction 1: NaClO(aq) → Na⁺(aq) + ClO⁻(aq).
Chemical reaction 2: ClO⁻(aq) + H₂O(l) ⇄ HClO(aq)+ OH⁻(aq).
ω(NaClO) = 5,25% ÷ 100% = 0,0525.
d(H₂O) = 1g/mL.
m(NaClO) = 0,0525 · 1 g/ml · 1000 ml = 52,5 g in 1 L.
n(NaClO) = 52,5 g ÷ 74,44 g/mol = 0,705 mol.
c₁(NaClO) = 0,705 M.
pH = 10,26.
pH + pOH = 14.
pOH = 14 - 10,26 = 3,74.
[OH⁻] = [HClO] = 10∧(-3,74) = 1,8·10⁻⁴ M.
Ka(HClO) = 4·10⁻⁸.
Ka · Kb = 10⁻¹⁴.
Kb(ClO⁻) = 2,5·10⁻⁷.
Kb(ClO⁻) = [OH⁻] · [HClO] / [ClO⁻].
2,5·10⁻⁷ = (1,8·10⁻⁴ M)² / [ClO⁻].
[ClO⁻] = [NaClO] = c₂(NaClO) = 0,1296 M.
V₂ = 500 mL.
c₁V₁ = c₂V₂.
V₁ = 0,1296 M · 500 mL ÷ 0,705 M.
V₁(NaClO) = 91,9 mL.