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?

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,.