Write the reaction and the corresponding \(K_{\mathrm{b}}\) equilibrium expression for each of the following substances acting as bases in water. a. \(\mathrm{NH}_{3}\) b. \(\mathrm{C}_{5} \mathrm{H}_{5} \mathrm{~N}\)

When \(\mathrm{NH}_3\) acts as a base in water, it accepts a hydrogen ion (\(\mathrm{H}^{+}\)) from a water molecule, forming the ammonium ion \(\mathrm{NH}_4^{+}\) and a hydroxide ion \(\mathrm{OH}^{-}\). The chemical reaction can be written as: \[\mathrm{NH}_3 (aq) + \mathrm{H}_2\mathrm{O} (l) \rightleftharpoons \mathrm{NH}_4^{+} (aq) + \mathrm{OH}^{-} (aq)\] The base dissociation constant, \(K_{\mathrm{b}}\), is the equilibrium constant for this reaction. It can be expressed as a ratio of the concentration of the products to the concentration of the reactants: \[K_{\mathrm{b}} = \frac{[\mathrm{NH}_4^{+}][\mathrm{OH}^{-}]}{[\mathrm{NH}_3]}\]

When \(\mathrm{C}_5\mathrm{H}_5\mathrm{N}\) (pyridine) acts as a base in water, it also accepts a hydrogen ion (\(\mathrm{H}^{+}\)) from a water molecule, forming the \(\mathrm{C}_5\mathrm{H}_5\mathrm{NH}^{+}\) ion and a hydroxide ion \(\mathrm{OH}^{-}\). The chemical reaction can be written as: \[\mathrm{C}_5\mathrm{H}_5\mathrm{N} (aq) + \mathrm{H}_2\mathrm{O} (l) \rightleftharpoons \mathrm{C}_5\mathrm{H}_5\mathrm{NH}^{+} (aq) + \mathrm{OH}^{-} (aq)\] The base dissociation constant, \(K_{b}\), for this reaction can be expressed as a ratio of the concentration of the products to the concentration of the reactants: \[K_{\mathrm{b}} = \frac{[\mathrm{C}_5\mathrm{H}_5\mathrm{NH}^{+}][\mathrm{OH}^{-}]}{[\mathrm{C}_5\mathrm{H}_5\mathrm{N}]}\] To summarize, we have found the chemical reactions and the base dissociation constant expressions for both \(\mathrm{NH}_3\) and \(\mathrm{C}_5\mathrm{H}_5\mathrm{N}\) acting as bases in water.