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14-04-2018
Chemistry

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

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Dejanras Dejanras

24-04-2018

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.

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