Ask Question
23 March, 09:40

Ka for hydrofluoric acid, HF, is 7.20E-4. Ka for phenol (a weak acid), C6H5OH, is 1.00E-10. Ka for acetylsalicylic acid (aspirin), HC9H7O4, is 3.00E-4. What is the formula for the strongest conjugate base?

+3
Answers (1)
  1. 23 March, 11:39
    0
    C₆H₅OH

    Explanation:

    In general for a given acid we have the dissociation

    HA ⇄ H⁺ + A⁻

    A⁻ is the conjugate base of the weak acid HA:

    A ⁻ + H₂O ⇄ HA + OH⁻

    Kb = [HA][OH⁻] / [A⁻]

    If we multiply and divide the right side by [H⁺], we have

    Kb = [HA][OH⁻][H⁺] / ([A⁻][H⁺]) = [OH⁻][H⁺] / Ka = Kw/Ka

    This is a very useful equation worth remembering, KaKb = Kw

    Now since Kw = 1.0 x 10⁻¹⁴ we are now in position to calculate the Kbs in this question, and the biggest will correspond to the strongest conjugate base.

    Kb F⁻ = 10⁻¹⁴ / 7.20 x 10⁻⁴ = 1.4 x 10⁻¹¹

    Kb C₆H₅OH = 10⁻¹⁴ / 1.00 x 10⁻¹⁰ = 1.0 x 10⁻⁴

    Kb HC₉H₇O₄ = 10⁻¹⁴ / 3.00 x 10⁻⁴ = 3.3 x 10⁻¹¹

    Phenol, C₆H₅OH has the largest Kb and is the strongest conjugate base of the three. This is to be expected since it has the smallest Ka, and this is another way to solve the question,.
Know the Answer?
Not Sure About the Answer?
Find an answer to your question ✅ “Ka for hydrofluoric acid, HF, is 7.20E-4. Ka for phenol (a weak acid), C6H5OH, is 1.00E-10. Ka for acetylsalicylic acid (aspirin), HC9H7O4, ...” in 📘 Chemistry if you're in doubt about the correctness of the answers or there's no answer, then try to use the smart search and find answers to the similar questions.
Search for Other Answers