Answer:
ka = 2x10⁻⁹
Explanation:
Using the pH we can calculate the molar concentration of H⁺, [H⁺]
Then we use the expression of Ka for the equilibrium of a weak monoprotic acid:
HA ↔ H⁺ + A⁻
Where:
ka = [tex]\frac{[H^+][A^-]}{[HA]}[/tex]
ka = [tex]\frac{(2.24*10^{-5})(2.24*10^{-5})}{0.25-2.24*10^{-5}}[/tex] = 2x10⁻⁹