Problem 69 - Chemistry: The Central Science in SI Units, Global Edition

Question

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, \(\mathrm{CN}^{-}\).

Step-by-Step Solution

Answer

The chemical equations and \(K_b\) expressions for the given bases reacting with water are: (a) Trimethylamine (\(\left(\mathrm{CH}_{3}\right)_{3} \mathrm{~N}\)): $$\left(\mathrm{CH}_{3}\right)_{3} \mathrm{~N} + \mathrm{H}_{2}\mathrm{O} \rightleftharpoons \left(\mathrm{CH}_{3}\right)_{3} \mathrm{~NH}^+ + \mathrm{OH}^-$$ $$K_{b1} = \frac{[\left(\mathrm{CH}_{3}\right)_{3} \mathrm{~NH}^+][\mathrm{OH}^-]}{[\left(\mathrm{CH}_{3}\right)_{3} \mathrm{~N}]}$$ (b) Sulfite (\(\mathrm{SO}_{3}^{2-}\)): $$\mathrm{SO}_{3}^{2-} + \mathrm{H}_{2}\mathrm{O} \rightleftharpoons \mathrm{HSO}_{3}^- + \mathrm{OH}^-$$ $$K_{b2} = \frac{[\mathrm{HSO}_{3}^-][\mathrm{OH}^-]}{[\mathrm{SO}_{3}^{2-}]}$$ (c) Cyanide (\(\mathrm{CN}^{-}\)): $$\mathrm{CN}^- + \mathrm{H}_{2}\mathrm{O} \rightleftharpoons \mathrm{HCN} + \mathrm{OH}^-$$ $$K_{b3} = \frac{[\mathrm{HCN}][\mathrm{OH}^-]}{[\mathrm{CN}^-]}$$

Step 1: (a) Trimethylamine with water

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: $$K_{b1} = \frac{[\left(\mathrm{CH}_{3}\right)_{3} \mathrm{~NH}^+][\mathrm{OH}^-]}{[\left(\mathrm{CH}_{3}\right)_{3} \mathrm{~N}]}$$

Step 2: (b) Sulfite with water

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: $$K_{b2} = \frac{[\mathrm{HSO}_{3}^-][\mathrm{OH}^-]}{[\mathrm{SO}_{3}^{2-}]}$$

Step 3: (c) Cyanide with water

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: $$K_{b3} = \frac{[\mathrm{HCN}][\mathrm{OH}^-]}{[\mathrm{CN}^-]}$$

Key Concepts

Chemical Equilibrium

Chemical equilibrium is a fundamental concept in chemistry that describes the state in which the concentrations of reactants and products remain constant over time. This is because the forward and reverse reactions occur at the same rate. In our exercise, the reactions of bases like trimethylamine, sulfite, and cyanide with water reach a state of equilibrium. For example, the equilibrium between cyanide ions reacting with water to produce hydrogen cyanide and hydroxide ions can be represented as:\[\mathrm{CN}^- + \mathrm{H}_{2}\mathrm{O} \rightleftharpoons \mathrm{HCN} + \mathrm{OH}^-\]This shows that both the forward and reverse reactions are taking place, and at equilibrium, the amounts of each substance remain constant. At equilibrium, the ratio of the concentration of products to reactants raised to the powers of their coefficients is known as the equilibrium constant, denoted as \(K_b\) for bases.

Conjugate Acid

In the context of acid-base reactions, a conjugate acid is formed when a base gains a proton (\(H^+\)). The conjugate acid of a base participates in the reversible reaction that achieves equilibrium. Consider when trimethylamine reacts with water; it results in its conjugate acid, trimethylammonium, \((\mathrm{CH}_3)_3\mathrm{NH}^+\), as shown in the reaction:\[(\mathrm{CH}_3)_3 \mathrm{~N} + \mathrm{H}_2\mathrm{O} \rightleftharpoons (\mathrm{CH}_3)_3 \mathrm{~NH}^+ + \mathrm{OH}^-\]Similarly, when the sulfite ion reacts with water, it forms bisulfite, \(\mathrm{HSO}_{3}^-\), which is its conjugate acid. Conjugate acids play a critical role in these reactions by maintaining equilibrium between the base and its associated ions. This role highlights the balance between accepting and donating protons, which is central to understanding acid-base chemistry.

Hydroxide Ion

The hydroxide ion (\(\mathrm{OH}^-\)) is a crucial player in the chemistry of bases. When a base reacts with water, it typically results in the formation of hydroxide ions, which are characteristic of basic solutions. For example, as shown in the exercise, reacting bases such as sulfite, cyanide, and trimethylamine with water results in the formation of hydroxide ions in chemical equilibrium:- Trimethylamine: \((\mathrm{CH}_3)_3\mathrm{N} + \mathrm{H}_{2}\mathrm{O} \rightleftharpoons (\mathrm{CH}_3)_3\mathrm{NH}^+ + \mathrm{OH}^-\)- Sulfite: \(\mathrm{SO}_{3}^{2-} + \mathrm{H}_{2}\mathrm{O} \rightleftharpoons \mathrm{HSO}_{3}^- + \mathrm{OH}^-\)- Cyanide: \(\mathrm{CN}^- + \mathrm{H}_{2}\mathrm{O} \rightleftharpoons \mathrm{HCN} + \mathrm{OH}^-\)Hydroxide ions are responsible for increasing the pH of a solution, giving it basic properties. They are pivotal in measuring the strength of a base through the base ionization constant (\(K_b\)), as this constant reflects the base's ability to produce hydroxide ions when in solution.