What is the pH of a 0.417 M solution of hypobromous acid (HBrO)? Ka = 2.8 ✕ 10^-9?

Consider the dissociation of HBrO.
_______________ HBrO(aq) + H₂O(l) ⇌ BrO⁻(aq) + H₃O⁺(aq) …… Ka = 2.8 × 10⁻⁹
Initial: _________ 0.417 M ___________ 0 M _____ 0 M
Change: _________ -y M ____________ +y M ____ +y M
Equilibrium: __ (0.417 - y) M _________ y M _____ y M

As Ka is very small, the dissociation of HBrO is to a very small extent.
We can assume that 0.417 ≫ y
and thus [HBrO] at equilibrium = (0.417 - y) M = 0.417 M

Ka = [BrO⁻] [H₃O⁺] / [HBrO]
2.8 × 10⁻⁹ = y² / 0.417
y = √{0.417 × (2.8 × 10⁻⁹)} = 3.42 × 10⁻⁵

pH = -log[H₃O⁺] = -log(3.42 × 10⁻⁵) = 4.47