Answer is: 153.52 grams of hypobromous acid must be added.
Chemical dissociation: HBrO ⇄ H⁺ + BrO⁻.
pH = 4.25.
pH = -log[H⁺].
[H⁺] = 10∧(-pH).
[H⁺] = 10∧(-4.25).
[H⁺] = [BrO⁻] = 5.62·10⁻⁵ M.
Ka = [H⁺] · [BrO⁻] / [HBrO].
2.00·10⁻⁹ = (5.62·10⁻⁵ M)² / [HBrO].
[HBrO] = 3.16·10⁻⁹ M² / 2.00·10⁻⁹.
[HBrO] = 1.58 M.
m(HBrO) = n(HBrO) · M(HBrO).
m(HBrO) = 1.58 mol · 96.91 g/mol.
m(HBrO) = 153.52 g.
