HClO is a weak acid (Ka = 4.0 × 10–8) and so the salt NaClO acts as a weak base.?

Given:
HClO(aq) + H₂O(l) ⇌ ClO⁻(aq) + H₃O⁺(aq) …. Ka
Ka = [ClO⁻][ H₃O⁺]/[HClO] = 4.0 × 10⁻⁸

Then, calculate Kb for ClO⁻ ion.
ClO⁻(aq) + H₂O(l) ⇌ HClO(aq) + OH⁻(aq) …. Kb
Kb = [HClO][OH⁻]/[ClO⁻] = ([H₃O⁺][OH⁻]) * {[HClO]/([ClO⁻][ H₃O⁺])} = Kw/Ka

______________ ClO⁻(aq) + H₂O(l) ⇌ HClO(aq) + OH⁻(aq) …. Kb
start _________ 0.048M _________________ 0 M ________ 0 M
change ________ -y M _________________ +y M ______ +y M
at eqm __ (0.048 - y) M _______________ y M ________ y M

For Kb is small, assume that 0.048 ≫ y, and thus (0.048 - y) M ≈ 0.048 M

Kb = [HClO][OH⁻]/[ClO⁻] = Kw/Ka
y² / 0.048 = (1.0 × 10⁻¹⁴) / (4.0 × 10⁻⁸)
y = √{0.048 * (1.0 × 10⁻¹⁴) / (4.0 × 10⁻⁸)}
y = 1.1 × 10⁻⁴

pOH = -log[OH⁻] = -log(1.1 × 10⁻⁴) = 4.0
pH = pKw - pOH = 14 - 4.0 = 10.0