Answer to Question #170663 in Chemistry for Hyeri

Question

Computation ionization constant of hydrofluoric acid, 0.1 F solution is 7.13 per cent ionized.

Answer 1

Hydrofluoric acid dissociates as following:

HF + H₂O = H₃O⁺ + F^-

Ionization constant equals:

K_a = [H₃O⁺] [F^-] / [HF]

As 7.13 % of solution is ionized:

[F-] = [H₃O⁺] = 7.13 % × [HF]_initial / 100 % = 0.00713 F

[HF] = 0.1 - 0.00713 = 0.09287 F

Finally:

K_a = 0.00713 F × 0.00713 F / 0.09287 F = 5.47 × 10^-4

Answer: 5.47 × 10^-4