Hydrofluoric acid, hf, has a ka of 6.8 * 10-4. what are [h3o+], [f-], and [oh-] in 0.710 m hf?

[H₃O⁺] = [F⁻] = 2.2 x 10⁻² M. & [OH⁻] = 4.55 x 10⁻¹³.

Explanation:

For a weak acid like HF, the dissociation of HF will be:

HF + H₂O ⇄ H₃O⁺ + F⁻.

[H₃O⁺] = [F⁻].

∵ [H₃O⁺] = √Ka. C,

Ka = 6.8 x 10⁻⁴, C = 0.710 M.

∴ [H₃O⁺] = √Ka. C = √ (6.8 x 10⁻⁴) (0.710) = 2.197 x 10⁻² M ≅ 2.2 x 10⁻² M.

∴ [H₃O⁺] = [F⁻] = 2.2 x 10⁻² M.

∵ [H₃O⁺][OH⁻] = 10⁻¹⁴.

∴ [OH⁻] = 10⁻¹⁴/[H₃O⁺] = 10⁻¹⁴ / (2.2 x 10⁻²) = 4.55 x 10⁻¹³.