Question

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. \(\mathrm{C}_{5} \mathrm{H}_{5} \mathrm{N}\)

Short answer

a. The reaction for \(\mathrm{NH}_{3}\) acting as a base in water is: \[\mathrm{NH_3}(aq) + \mathrm{H_2O}(l) \rightleftharpoons \mathrm{NH_4^+}(aq) + \mathrm{OH^-}(aq)\] and the corresponding \(K_b\) expression is: \[K_b = \frac{[\mathrm{NH_4^+}][\mathrm{OH^-}]}{[\mathrm{NH_3}]}\] b. The reaction for \(\mathrm{C}_{5} \mathrm{H}_{5} \mathrm{N}\) acting as a base in water is: \[\mathrm{C}_5\mathrm{H}_5\mathrm{N}(aq) + \mathrm{H_2O}(l) \rightleftharpoons \mathrm{C}_5\mathrm{H}_5\mathrm{NH^+}(aq) + \mathrm{OH^-}(aq)\] and the corresponding \(K_b\) expression is: \[K_b = \frac{[\mathrm{C}_5\mathrm{H}_5\mathrm{NH^+}][\mathrm{OH^-}]}{[\mathrm{C}_5\mathrm{H}_5\mathrm{N}]}\]

Step-by-step solution

a. Reaction and \(K_b\) for \(\mathrm{NH}_{3}\)

Step 1: Write the chemical equation When \(\mathrm{NH}_3\) accepts a proton from water, it forms its conjugate acid, the ammonium ion (\(\mathrm{NH_4^+}\)) and hydroxide ions (\(\mathrm{OH^-}\)). \[\mathrm{NH_3}(aq) + \mathrm{H_2O}(l) \rightleftharpoons \mathrm{NH_4^+}(aq) + \mathrm{OH^-}(aq)\] Step 2: Write the \(K_b\) expression The base dissociation constant, \(K_b\), can be written as the ratio of the concentrations of the products to the reactants, excluding water since it is a pure liquid. \[K_b = \frac{[\mathrm{NH_4^+}][\mathrm{OH^-}]}{[\mathrm{NH_3}]}\]

b. Reaction and \(K_b\) for \(\mathrm{C}_5\mathrm{H}_5\mathrm{N}\)

Step 1: Write the chemical equation When \(\mathrm{C}_5\mathrm{H}_5\mathrm{N}\) accepts a proton from water, it forms its conjugate acid (\(\mathrm{C}_5\mathrm{H}_5\mathrm{NH^+}\)) and hydroxide ions (\(\mathrm{OH^-}\)). \[\mathrm{C}_5\mathrm{H}_5\mathrm{N}(aq) + \mathrm{H_2O}(l) \rightleftharpoons \mathrm{C}_5\mathrm{H}_5\mathrm{NH^+}(aq) + \mathrm{OH^-}(aq)\] Step 2: Write the \(K_b\) expression As with the previous example, the base dissociation constant, \(K_b\), can be written as the ratio of the concentrations of the products to the reactants, excluding water. \[K_b = \frac{[\mathrm{C}_5\mathrm{H}_5\mathrm{NH^+}][\mathrm{OH^-}]}{[\mathrm{C}_5\mathrm{H}_5\mathrm{N}]}\]