The K_a for hydrofluoric acid is 6.8 × 10^-4.What is K_b for the fluoride ion? A)1.5 × 10^-11 B)6.8 × 10^-4 C)1.5 × 10³ D)6.8 × 10¹⁰ E)6.8 × 10^-18

Question

Quizplus · Accepted Answer

The K_a for hydrofluoric acid can be written as: HF ↔ H+ + F- K_a = [H+][F-]/[HF] = 6.8 × 10^-4 To find the K_b for fluoride ion, we can manipulate this equation using the equation for dissociation constant of a conjugate base: K_b = [H+][OH-]/[HF] Since the concentration of [OH-] can be calculated using Kw (ion product constant for water) and [H+], we can substitute: K_b = Kw/[H+][HF] K_b = (1.0 × 10^-14)/(6.8 × 10^-4) K_b = 1.5 × 10^-11 Therefore, the answer is A.

The Ka for Hydrofluoric Acid Is 6

The K_a for Hydrofluoric Acid Is 6