Write the reaction and the corresponding \(K_{\mathrm{b}}\) equilibrium expression for each of the following substances acting as bases in water. a. aniline, \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NH}_{2}\) b. dimethylamine, \(\left(\mathrm{CH}_{3}\right)_{2} \mathrm{NH}\)

In general, a weak base \(\mathrm{B}\) reacts with water to form a corresponding weak acid \(\mathrm{BH^+}\) and a hydroxide ion (\(\mathrm{OH^-}\)). The general reaction can be represented as follows: \(\mathrm{B} + \mathrm{H_2O} \rightleftharpoons \mathrm{BH^+} + \mathrm{OH^-}\)

Now let's write the reaction for each substance: a. Aniline, \(\mathrm{C}_6\mathrm{H}_5\mathrm{NH}_2\): \(\mathrm{C}_6\mathrm{H}_5\mathrm{NH}_2 + \mathrm{H_2O} \rightleftharpoons \mathrm{C}_6\mathrm{H}_5\mathrm{NH}_3^+ + \mathrm{OH}^-\) b. Dimethylamine, \((\mathrm{CH}_3)_2\mathrm{NH}\): \((\mathrm{CH}_3)_2\mathrm{NH} + \mathrm{H_2O} \rightleftharpoons (\mathrm{CH}_3)_2\mathrm{NH}_2^+ + \mathrm{OH}^-\)

The equilibrium constant expression for a base in water (\(K_\mathrm{b}\)) is the product of the concentrations of the products of the reaction, divided by the product of the concentrations of the reactants, excluding water because it's a liquid: \(K_\mathrm{b} = \frac{[\mathrm{BH^+}][\mathrm{OH^-}]}{[\mathrm{B}]}\) Now let's write the \(K_\mathrm{b}\) expressions for each reaction: a. Aniline, \(\mathrm{C}_6\mathrm{H}_5\mathrm{NH}_2\): \(K_{\mathrm{b},\mathrm{aniline}} = \frac{ [\mathrm{C}_6\mathrm{H}_5\mathrm{NH}_3^+][\mathrm{OH}^-]}{[\mathrm{C}_6\mathrm{H}_5\mathrm{NH}_2]}\) b. Dimethylamine, \((\mathrm{CH}_3)_2\mathrm{NH}\): \(K_{\mathrm{b},\mathrm{dimethylamine}} = \frac{ [(\mathrm{CH}_3)_2\mathrm{NH}_2^+][\mathrm{OH}^-]}{[(\mathrm{CH}_3)_2\mathrm{NH]}\) These expressions represent the \(K_\mathrm{b}\) equilibrium expressions for aniline and dimethylamine acting as bases in water.