Chemical reaction 1: NaClO(aq) → Na⁺(aq) + ClO⁻(aq).
Chemical reaction 2: ClO⁻(aq)
+ H₂O(l) ⇄ HClO(aq)+ OH⁻(aq).
c(NaClO) = 0,030 M.
[ClO⁻] =
0,03 M - x.
Ka(HClO) = 4·10⁻⁸.
Ka · Kb = 10⁻¹⁴.
Kb(ClO⁻) = 2,5·10⁻⁷.
Kb(ClO⁻) = [OH⁻] · [HClO] / [ClO⁻].
[OH⁻]
= [HClO] = x.
2,5·10⁻⁷ = x² / (0,03 M
-x).
Solve quadratic equation: x = [OH⁻] = 0.0000893 M.
pOH = -log(0.0000893 M) = 4.05.
pH = 14 - 4.05.
pH = 9.95.
