Question

Write the chemical equation and the \(K_{b}\) expression for the reaction of each of the following bases with water. (a) propylamine, \(\mathrm{C}_{3} \mathrm{H}_{7} \mathrm{NH}_{2}\) i (b) monohydrogen phosphate ion, \(\mathrm{HPO}_{4}{ }^{2-} ;\) (c) benzoate ion, \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CO}_{2}^{-}\)

Short answer

(a) Chemical equation: \(C_3H_7NH_2 + H_2O \rightleftharpoons C_3H_7NH_3^+ + OH^-\); \(K_b = \frac{[C_3H_7NH_3^+][OH^-]}{[C_3H_7NH_2]}\) (b) Chemical equation: \(HPO_4^{2-} + H_2O \rightleftharpoons H_2PO_4^- + OH^-\); \(K_b = \frac{[H_2PO_4^-][OH^-]}{[HPO_4^{2-}]}\) (c) Chemical equation: \(C_6H_5CO_2^- + H_2O \rightleftharpoons C_6H_5COOH + OH^-\); \(K_b = \frac{[C_6H_5COOH][OH^-]}{[C_6H_5CO_2^-]}\)

Step-by-step solution

Write the general formula for the reaction of a base with water

The general formula for the reaction of a base (B) with water (H\(_2\)O) to form the conjugate acid (BH\(^+\)) and hydroxide ions (OH\(^-\)) is given by: \[ B + H_2O \rightleftharpoons BH^+ + OH^-\]

Step 2a: Write the chemical equation for propylamine reacting with water

Propylamine (C\(_3\)H\(_7\)NH\(_2\)) reacts with water (H\(_2\)O) to form the conjugate acid (C\(_3\)H\(_7\)NH\(_3^+\)) and hydroxide ions (OH\(^-\)). The chemical equation for this reaction is: \[ C_3H_7NH_2 + H_2O \rightleftharpoons C_3H_7NH_3^+ + OH^-\]

Step 3a: Write the \(K_{b}\) expression for propylamine and water reaction

The \(K_{b}\) expression is obtained from the chemical equation by the law of mass action. The expression for propylamine reaction with water is: \[ K_b = \frac{[C_3H_7NH_3^+][OH^-]}{[C_3H_7NH_2]}\]

Step 2b: Write the chemical equation for monohydrogen phosphate ion reacting with water

Monohydrogen phosphate ion (HPO\(_4^{2-}\)) reacts with water (H\(_2\)O) to form the conjugate acid (H\(_2\)PO\(_4^-\)) and hydroxide ions (OH\(^-\)). The chemical equation for this reaction is: \[ HPO_4^{2-} + H_2O \rightleftharpoons H_2PO_4^- + OH^-\]

Step 3b: Write the \(K_{b}\) expression for monohydrogen phosphate ion reacting with water

The \(K_{b}\) expression is obtained from the chemical equation by the law of mass action. The expression for monohydrogen phosphate ion reacting with water is: \[ K_b = \frac{[H_2PO_4^-][OH^-]}{[HPO_4^{2-}]}\]

Step 2c: Write the chemical equation for benzoate ion reacting with water

Benzoate ion (C\(_6\)H\(_5\)CO\(_2^-\)) reacts with water (H\(_2\)O) to form the conjugate acid (C\(_6\)H\(_5\)COOH) and hydroxide ions (OH\(^-\)). The chemical equation for this reaction is: \[ C_6H_5CO_2^- + H_2O \rightleftharpoons C_6H_5COOH + OH^-\]

Step 3c: Write the \(K_{b}\) expression for the benzoate ion and water reaction

The \(K_{b}\) expression is obtained from the chemical equation by the law of mass action. The expression for the benzoate ion reacting with water is: \[ K_b = \frac{[C_6H_5COOH][OH^-]}{[C_6H_5CO_2^-]}\]

Key concepts

Chemical Equations

A chemical equation is a symbolic representation of a chemical reaction where the reactants are written on the left and the products are on the right, separated by an arrow indicating the direction of the reaction. For bases reacting with water, this equation usually showcases the production of hydroxide ions, which is a hallmark of base hydrolysis.

To exemplify, propylamine when mixed with water forms propylammonium ions and hydroxide ions. This could be visually depicted like this:
\[ C_3H_7NH_2 + H_2O \rightleftharpoons C_3H_7NH_3^+ + OH^- \]
Similarly, for the reaction of monohydrogen phosphate ion and the benzoate ion with water, the respective chemical equations are written aligning with this convention, indicating the reactants, products, and the reversible nature of these hydrolysis reactions.

Base Hydrolysis

Base hydrolysis is the reaction where a base accepts a proton from water, resulting in the production of hydroxide ions (OH-), which increases the pH of the solution. This is a typical behavior of a base in aqueous solution. Bases such as propylamine, monohydrogen phosphate ion, and the benzoate ion undergo hydrolysis when dissolved in water, which is depicted in the chemical equations explained previously.

The concept can be further understood by noting that these substances accept protons from water, the water acting as an acid in this context, demonstrating the amphiprotic nature of water.

Equilibrium Constant Expression

The equilibrium constant expression (\( K_b \)) for the reaction of a base with water is derived from the law of mass action, which states that at equilibrium, a certain ratio of the concentrations of reactants and products raised to the power of their stoichiometric coefficients is constant. For bases in water, the \( K_b \) expression quantifies the degree of hydrolysis and is a measure of the base's strength.

For the given bases, the \( K_b \) expressions would look like:
\[ K_b = \frac{[BH^+][OH^-]}{[B]} \]
where [BH+] represents the concentration of the conjugate acid, [OH-] the concentration of hydroxide ions, and [B] the concentration of the base. Larger \( K_b \) values indicate stronger bases, reflecting a greater tendency to accept protons from water.

Law of Mass Action

The law of mass action underpins the concept of chemical equilibrium. It maintains that for a reversible chemical reaction at equilibrium and at a constant temperature, the ratio of the product of concentrations of products to the product of concentrations of reactants is constant, known as the equilibrium constant (\( K \)). In the context of base hydrolysis in water, the law of mass action is used to derive the \( K_b \) expression, as seen in the previous section.

For instance, the law of mass action can determine how a shift in the ratio of concentrations of reactants or products could drive the reaction forward or backward, in accordance with Le Chatelier's principle. This helps to explain why certain substances act as stronger bases or weaker bases when they hydrolyze in aqueous solutions.