Acids and bases can be thought of as chemical opposites (acids are proton donors, and bases are proton acceptors). Therefore, one might think that \(K_{\mathrm{a}}=1 / K_{\mathrm{b}}\) . Why isn't this the case? What is the relationship between \(K_{\mathrm{a}}\) and \(K_{\mathrm{b}}\) ? Prove it with a derivation.

First, let's denote an acid as HA and its conjugate base as A^-. An acid dissociates into its conjugate base and a proton (H^+) in the following reaction: HA + H2O ↔ A^- + H3O^+ The base dissociation reaction involves the conjugate base (A^-) and water: A^- + H2O ↔ HA + OH^-

The expressions can be written based on the concentrations of the products and reactants at equilibrium: Ka = \(\frac{[A^{-}][H_{3}O^{+}]}{[HA][H_{2}O]}\) Kb = \(\frac{[HA][OH^{-}]}{[A^{-}][H_{2}O]}\)

We know that the ion product constant of water, Kw, can be expressed as: Kw = [H3O^+][OH^-] We can solve the previous expressions of Ka and Kb for [H3O^+] and [OH^-] respectively: [H3O^+] = \(\frac{K_{a}[HA][H_{2}O]}{[A^{-}]}\) [OH^-] = \(\frac{K_{b}[A^{-}][H_{2}O]}{[HA]}\) Now, multiply the equations for [H3O^+] and [OH^-]: ([H3O^+])([OH^-]) = \(\frac{K_{a}K_{b}[HA]^{2}[H_{2}O]^{2}}{[A^{-}]^{2}}\) Since the ion product constant of water, Kw, equals [H3O^+][OH^-], we can rewrite the equation as: Kw = \(\frac{K_{a}K_{b}[HA]^{2}[H_{2}O]^{2}}{[A^{-}]^{2}}\) This equation shows the relationship between Ka and Kb, involving the ion product constant of water, Kw. However, it doesn't say that Ka equals the inverse of Kb.

To have a clearer understanding of their relationship, we will make the equation solve for the product of Ka and Kb: \(K_aK_b = \frac{K_w[A^{-}]^{2}}{[HA]^2[H_{2}O]^2}\) This equation shows that the product of \(K_a\) and \(K_b\) is not directly equal to 1, rather it depends on the squares of the concentrations of the acid, the conjugate base, and water, along with the ion product constant for water, \(K_w\). Thus, the relationship between \(K_a\) and \(K_b\) is more complex than \(K_{a} \neq 1 / K_{b}\).