HClO is a weak acid (Ka = 4.0 × 10–8) and the salt NaClO acts as a weak base. What is the pH of a solution with 0.024 M in NaClO at 25 °C?

Kb of ClO⁻ = Kw / (Ka of HClO) = (1.0 × 10⁻¹⁴) / (4.0 × 10⁻⁸) = 2.5 × 10⁻⁷

____________ ClO⁻(aq) __ + H₂O __ ⇌ __ HClO(aq) __ + __ OH⁻(aq) ___ Kb = 2.5 × 10⁻⁷
Initial: _____ 0.0750 M ________________ 0 M __________ 0 M
Change: ______ -y M _________________ +y M _________ +y M
At eqm: __ (0.024 - y) M _______________ y M __________ y M

As Kb is very small, the dissociation of ClO⁻ would be to a very small extent.
It is assumed that 0.024 ≫ y, i.e. [H₂A] at equilibrium = (0.024 - y) M ≈ 0.024 M

Kb = [HClO] [OH⁻] / [ClO⁻]
2.5 × 10⁻⁷ = y² / 0.024
y = √[0.024 × (2.5 × 10⁻⁷)]
y = 7.8 × 10⁻⁵

pOH = -log[OH⁻] = -log(7.75 × 10⁻⁵) = 4.1
pH = Kw - pH = 14.0 - 4.1 = 9.9