Question

The value of ka for benzoic acid is 6.30×10-5. what is the value of kb, for its conjugate base, c6h5coo-?

Answer 1

Benzoic acid release protons in water:
C₆H₅COOH(aq) ⇄ C₆H₅COO⁻(aq) + H⁺(aq).
Benzoic acid conjugate base gain protons in water:
C₆H₅COO⁻(aq) + H⁺(aq) ⇄ C₆H₅COOH(aq).
Ka(C₆H₅COOH) = 6.3·10⁻⁵.
Ka · Kb = 1·10⁻¹⁴.
Kb(C₆H₅COO⁻) = 1·10⁻¹⁴ ÷ 6.3·10⁻⁵.
Kb(C₆H₅COO⁻) = 1.587·10⁻¹⁰.