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Prove that Kw equals product of Ka and Kb for an acid-base pair in water?
Wiki User
∙ 14y ago
dissociation of acid in water: A + H2O <-> A- + H3O+ with dissociation constant Ka = [A-][H3O+]/[A][H2O] = [A-][H3O+]/[A].
dissociation of base in water: B + H2O <-> HB+ + OH- with dissociation constant Kb = [HB+][OH-]/[B][H2O] = [HB+][OH-]/[B]
dissociation of water in itself: 2H2O <-> H3O+ + OH- with dissociation constant Kw = [H3O+][OH-]/[H2O]^2 = [H3O+][OH-] where [H2O] has been ommitted because it is a pure liquid.
substituting relations for Ka and Kb into Kw gives: Kw = [H3O+][OH-] = (Ka[A]/[A-])(Kb[B]/[HB+]) = KaKb where [A] = [HB+] and [B] = [A-].
Wiki User
∙ 14y ago
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