Answer:
0.42%
Explanation:
∵ pH = - log[H⁺].
2.72 = - log[H⁺]
∴ [H⁺] = 1.905 x 10⁻³.
∵ [H⁺] = √Ka.C
∴ [H⁺]² = Ka.C
∴ ka = [H⁺]²/C = (1.905 x 10⁻³)²/(0.45) = 8.068 x 10⁻⁶.
∵ Ka = α²C.
Where, α is the degree of dissociation.
∴ α = √(Ka/C) = √(8.065 x 10⁻⁶/0.45) = 4.234 x 10⁻³.
∴ percentage ionization of the acid = α x 100 = (4.233 x 10⁻³)(100) = 0.4233% ≅ 0.42%.
