What is the [OH-] in a 0.25M hypobromous acid(HOBr) solution?Ka=2.5x10^-9  Here are the choices:?

_________ HOBr + H₂O(ℓ) ⇌ OBr⁻(q) + H₃O⁺(aq) __ Ka = 2.5 × 10⁻⁹
Initial:         0.25 M                   0 M          0 M
Change:      -y M                    +y M        +y M
Eqm:       (0.25 - y) M               y M          y M
_              ≈ 0.25 M

At equilibrium :
Ka = [OBr⁻] [H₃O⁺] / [HOBr]
2.5 × 10⁻⁹ = y² / 0.25
y = √(0.25 × 2.5 × 10⁻⁹) = 1.4 × 10⁻³
[H₃O⁺] = 2.5 × 10⁻⁵ M
[OH⁻] = Kw/[H₃O⁺] = (1.0 × 10⁻¹⁴)/(2.5 × 10⁻⁵) M = 4.0 × 10⁻¹⁰ M

The answer: a. 4.0 × 10⁻¹⁰ M

 HBrO  <-->  H+  + BrO-

    Ka = [H+][BrO-] / [HBrO]
   2.5x10^-9  =  x^2 / 0.25
     x  =  2.5x10^-5 M  =  [H+]

   Kw = [H+][OH-]  = 1x10^-14
    [OH-] = 1x10^-14 / 2.5x10^-5  =  4.0x10^-10 M  <<<  answer