Problem 70 - Chemistry The Central Science

Question

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

Step-by-Step Solution

Answer

(a) Propylamine: \(\mathrm{C}_{3} \mathrm{H}_{7} \mathrm{NH}_{2} + \mathrm{H}_{2}\mathrm{O} \rightleftharpoons \mathrm{C}_{3} \mathrm{H}_{7} \mathrm{NH}_{3}^+ + \mathrm{OH}^-\) \(K_b = [\mathrm{C}_{3} \mathrm{H}_{7} \mathrm{NH}_{3}^+][\mathrm{OH}^-] / [\mathrm{C}_{3} \mathrm{H}_{7}\mathrm{NH}_{2}]\) (b) Monohydrogen phosphate ion: \(\mathrm{HPO}_{4}^{2-} + \mathrm{H}_{2}\mathrm{O} \rightleftharpoons \mathrm{H}_{2}\mathrm{PO}_{4}^- + \mathrm{OH}^-\) \(K_b = [\mathrm{H}_{2}\mathrm{PO}_{4}^-][\mathrm{OH}^-] / [\mathrm{HPO}_{4}^{2-}]\) (c) Benzoate ion: \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CO}_{2}^- + \mathrm{H}_{2}\mathrm{O} \rightleftharpoons \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{COOH} + \mathrm{OH}^-\) \(K_b = [\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{COOH}][\mathrm{OH}^-] / [\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CO}_{2}^-]\)

Step 1: Write the reaction

Propylamine reacts with water as follows: \(\mathrm{C}_{3} \mathrm{H}_{7} \mathrm{NH}_{2} + \mathrm{H}_{2}\mathrm{O} \rightleftharpoons \mathrm{C}_{3} \mathrm{H}_{7} \mathrm{NH}_{3}^+ + \mathrm{OH}^-\)

Step 2: Write the equilibrium expression

\[ [\mathrm{C}_{3} \mathrm{H}_{7} \mathrm{NH}_{3}^+][\mathrm{OH}^-] / [\mathrm{C}_{3} \mathrm{H}_{7}\mathrm{NH}_{2}][\mathrm{H}_{2}\mathrm{O}] \]

Step 3: Write the \(K_b\) expression

Since the concentration of water is considered constant, we can write: \[ K_b = [\mathrm{C}_{3} \mathrm{H}_{7} \mathrm{NH}_{3}^+][\mathrm{OH}^-] / [\mathrm{C}_{3} \mathrm{H}_{7}\mathrm{NH}_{2}] \] (b) Monohydrogen phosphate ion, \(\mathrm{HPO}_{4}^{2-}\)

Step 1: Write the reaction

Monohydrogen phosphate ion reacts with water as follows: \(\mathrm{HPO}_{4}^{2-} + \mathrm{H}_{2}\mathrm{O} \rightleftharpoons \mathrm{H}_{2}\mathrm{PO}_{4}^- + \mathrm{OH}^-\)

Step 2: Write the equilibrium expression

\[ [\mathrm{H}_{2}\mathrm{PO}_{4}^-][\mathrm{OH}^-] / [\mathrm{HPO}_{4}^{2-}][\mathrm{H}_{2}\mathrm{O}] \]

Step 3: Write the \(K_b\) expression

Since the concentration of water is considered constant, we can write: \[ K_b = [\mathrm{H}_{2}\mathrm{PO}_{4}^-][\mathrm{OH}^-] / [\mathrm{HPO}_{4}^{2-}] \] (c) Benzoate ion, \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CO}_{2}^{-}\)

Step 1: Write the reaction

Benzoate ion reacts with water as follows: \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CO}_{2}^- + \mathrm{H}_{2}\mathrm{O} \rightleftharpoons \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{COOH} + \mathrm{OH}^-\)

Step 2: Write the equilibrium expression

\[ [\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{COOH}][\mathrm{OH}^-] / [\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CO}_{2}^-][\mathrm{H}_{2}\mathrm{O}] \]

Step 3: Write the \(K_b\) expression

Since the concentration of water is considered constant, we can write: \[ K_b = [\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{COOH}][\mathrm{OH}^-] / [\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CO}_{2}^-] \]

Key Concepts

Chemical Equation

Understanding a chemical equation is essential in the study of chemistry. It depicts a chemical reaction where reactants transform into products. A chemical equation is balanced when the number of atoms for each element is the same on both the reactant and product sides.

For a base reacting with water, the general formula is Base + H₂O ⇌ Conjugate Acid + OH⁻. In the case of propylamine (C₃H₇NH₂), the chemical equation with water would be:
C₃H₇NH₂ + H₂O ⇌ C₃H₇NH₃⁺ + OH⁻.

This shows how propylamine accepts a proton from water, becoming its conjugate acid, propylammonium ion (C₃H₇NH₃⁺), and producing hydroxide ions.

Kb Expression

The equilibrium constant for a base, designated as Kb, quantifies a base's strength in water. It is calculated from the concentrations of the products divided by the concentration of the reactants. The stronger the base, the higher the Kb value.

For a base like propylamine, the Kb expression is formulated without water since its concentration is constant and does not affect the equilibrium in a dilute solution. Thus, the Kb for propylamine becomes:
Kb = [C₃H₇NH₃⁺][OH⁻] / [C₃H₇NH₂]

Remember, a comprehensible Kb expression is key to understanding how the base interacts with water and forms its conjugate acid and hydroxide ions.

Base Reaction with Water

Bases react with water in a process known as hydrolysis. This involves the base accepting a hydrogen ion (H⁺) from water, forming the conjugate acid of the base and hydroxide ions (OH⁻).

An example is the benzoate ion (C₆H₅CO₂⁻), which reacts with water to form benzoic acid (C₆H₅COOH) and hydroxide ions:
C₆H₅CO₂⁻ + H₂O ⇌ C₆H₅COOH + OH⁻

This interaction shows the base's ability to raise the pH of a solution by producing hydroxide ions.

Equilibrium Expression

An equilibrium expression represents the ratio of product concentrations to reactant concentrations at equilibrium. It is related to the nature of reversible reactions, where the forward and reverse reactions occur at equal rates.

For a base like monohydrogen phosphate ion (HPO₄²⁻), the equilibrium expression is:
[H₂PO₄⁻][OH⁻] / [HPO₄²⁻]

Importance of Water Concentration

In dilute aqueous solutions, water's concentration remains effectively constant, thus it is typically omitted from the expression. Ultimately, understanding this concept is vital for predicting the position of equilibrium and calculating the pH of the solution.