Write the chemical equation and the \(K_{b}\) expression for the reaction of each of the following bases with water: (a) trimethylamine, \(\left(\mathrm{CH}_{3}\right)_{3} \mathrm{~N} ;\) (b) sulfite, \(\mathrm{SO}_{3}^{2-}\); (c) cyanide, \(\mathrm{CN}^{-}\).

Let's write the chemical equation for the reaction of trimethylamine, \(\left(\mathrm{CH}_{3}\right)_{3} \mathrm{~N}\), with water, \(\mathrm{H}_2\mathrm{O}\). 1. Reaction with water: \(\left(\mathrm{CH}_{3}\right)_{3} \mathrm{~N} + \mathrm{H}_{2}\mathrm{O}\) 2. Balanced chemical equation: The trimethylamine reacts with water to form its conjugate acid, \(\left(\mathrm{CH}_{3}\right)_{3} \mathrm{~NH}^+\), and hydroxide ions, \(\mathrm{OH}^-\). $$\left(\mathrm{CH}_{3}\right)_{3} \mathrm{~N} + \mathrm{H}_{2}\mathrm{O} \rightleftharpoons \left(\mathrm{CH}_{3}\right)_{3} \mathrm{~NH}^+ + \mathrm{OH}^-$$ 3. \(K_b\) expression: let the \(K_b\) for the reaction be represented by \(K_{b1}\). The \(K_b\) expression is given as the product of ion concentrations divided by the concentration of the base: $$K_{b1} = \frac{[\left(\mathrm{CH}_{3}\right)_{3} \mathrm{~NH}^+][\mathrm{OH}^-]}{[\left(\mathrm{CH}_{3}\right)_{3} \mathrm{~N}]}$$

Now, let's write the chemical equation for the reaction of sulfite, \(\mathrm{SO}_{3}^{2-}\), with water, \(\mathrm{H}_2\mathrm{O}\). 1. Reaction with water: \(\mathrm{SO}_{3}^{2-} + \mathrm{H}_{2}\mathrm{O}\) 2. Balanced chemical equation: The sulfite reacts with water to form bisulfite ions, \(\mathrm{HSO}_{3}^-\), and hydroxide ions, \(\mathrm{OH}^-\). $$\mathrm{SO}_{3}^{2-} + \mathrm{H}_{2}\mathrm{O} \rightleftharpoons \mathrm{HSO}_{3}^- + \mathrm{OH}^-$$ 3. \(K_b\) expression: let the \(K_b\) for the reaction be represented by \(K_{b2}\). The \(K_b\) expression is given as the product of ion concentrations divided by the concentration of the base: $$K_{b2} = \frac{[\mathrm{HSO}_{3}^-][\mathrm{OH}^-]}{[\mathrm{SO}_{3}^{2-}]}$$

Finally, let's write the chemical equation for the reaction of cyanide, \(\mathrm{CN}^{-}\), with water, \(\mathrm{H}_2\mathrm{O}\). 1. Reaction with water: \(\mathrm{CN}^{-} + \mathrm{H}_{2}\mathrm{O}\) 2. Balanced chemical equation: The cyanide reacts with water to form hydrogen cyanide, \(\mathrm{HCN}\), and hydroxide ions, \(\mathrm{OH}^-\). $$\mathrm{CN}^- + \mathrm{H}_{2}\mathrm{O} \rightleftharpoons \mathrm{HCN} + \mathrm{OH}^-$$ 3. \(K_b\) expression: let the \(K_b\) for the reaction be represented by \(K_{b3}\). The \(K_b\) expression is given as the product of ion concentrations divided by the concentration of the base: $$K_{b3} = \frac{[\mathrm{HCN}][\mathrm{OH}^-]}{[\mathrm{CN}^-]}$$