Problem 70 - 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) 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) Chemical equation for propylamine with water: \[\mathrm{C}_{3}\mathrm{H}_{7}\mathrm{NH}_{2} + \mathrm{H}_{2}\mathrm{O} \rightleftharpoons \mathrm{C}_{3}\mathrm{H}_{7}\mathrm{NH}_{3}^{+} + \mathrm{OH}^{-}\] \(K_{b}\) expression for propylamine: \[K_{b} = \frac{[\mathrm{C}_{3}\mathrm{H}_{7}\mathrm{NH}_{3}^{+}][\mathrm{OH}^{-}]}{[\mathrm{C}_{3}\mathrm{H}_{7}\mathrm{NH}_{2}]} \] (b) Chemical equation for monohydrogen phosphate ion with water: \[\mathrm{HPO}_{4}^{2-} + \mathrm{H}_{2}\mathrm{O} \rightleftharpoons \mathrm{H}_{2}\mathrm{PO}_{4}^{-} + \mathrm{OH}^{-}\] \(K_{b}\) expression for monohydrogen phosphate ion: \[K_{b} = \frac{[\mathrm{H}_{2}\mathrm{PO}_{4}^{-}][\mathrm{OH}^{-}]}{[\mathrm{HPO}_{4}^{2-}]} \] (c) Chemical equation for benzoate ion with water: \[\mathrm{C}_{6}\mathrm{H}_{5}\mathrm{CO}_{2}^{-} + \mathrm{H}_{2}\mathrm{O} \rightleftharpoons \mathrm{C}_{6}\mathrm{H}_{5}\mathrm{CO}_{2}\mathrm{H} + \mathrm{OH}^{-}\] \(K_{b}\) expression for benzoate ion: \[K_{b} = \frac{[\mathrm{C}_{6}\mathrm{H}_{5}\mathrm{CO}_{2}\mathrm{H}][\mathrm{OH}^{-}]}{[\mathrm{C}_{6}\mathrm{H}_{5}\mathrm{CO}_{2}^{-}]} \]

Step 1: (a) Chemical equation for propylamine with water

Propylamine, \(\mathrm{C}_{3} \mathrm{H}_{7} \mathrm{NH}_{2}\), is a weak base, which means it will accept a proton from water. The chemical equation for this reaction is: \[\mathrm{C}_{3}\mathrm{H}_{7}\mathrm{NH}_{2} + \mathrm{H}_{2}\mathrm{O} \rightleftharpoons \mathrm{C}_{3}\mathrm{H}_{7}\mathrm{NH}_{3}^{+} + \mathrm{OH}^{-}\]

Step 2: (a) \(K_{b}\) expression for propylamine

Now, we can write the base ionization constant (\(K_{b}\)) expression for propylamine: \[K_{b} = \frac{[\mathrm{C}_{3}\mathrm{H}_{7}\mathrm{NH}_{3}^{+}][\mathrm{OH}^{-}]}{[\mathrm{C}_{3}\mathrm{H}_{7}\mathrm{NH}_{2}]} \]

Step 3: (b) Chemical equation for monohydrogen phosphate ion with water

Monohydrogen phosphate ion, \(\mathrm{HPO}_{4}^{2-}\), is an amphoteric ion (can act as both acid and base). In this case, we will consider it as a weak base that accepts a proton from water: \[\mathrm{HPO}_{4}^{2-} + \mathrm{H}_{2}\mathrm{O} \rightleftharpoons \mathrm{H}_{2}\mathrm{PO}_{4}^{-} + \mathrm{OH}^{-}\]

Step 4: (b) \(K_{b}\) expression for monohydrogen phosphate ion

The base ionization constant (\(K_{b}\)) expression for monohydrogen phosphate ion is: \[K_{b} = \frac{[\mathrm{H}_{2}\mathrm{PO}_{4}^{-}][\mathrm{OH}^{-}]}{[\mathrm{HPO}_{4}^{2-}]} \]

Step 5: (c) Chemical equation for benzoate ion with water

Benzoate ion, \(\mathrm{C}_{6}\mathrm{H}_{5}\mathrm{CO}_{2}^{-}\), acts as a weak base and will accept a proton from water: \[\mathrm{C}_{6}\mathrm{H}_{5}\mathrm{CO}_{2}^{-} + \mathrm{H}_{2}\mathrm{O} \rightleftharpoons \mathrm{C}_{6}\mathrm{H}_{5}\mathrm{CO}_{2}\mathrm{H} + \mathrm{OH}^{-}\]

Step 6: (c) \(K_{b}\) expression for benzoate ion

Lastly, we can write the base ionization constant (\(K_{b}\)) expression for the benzoate ion: \[K_{b} = \frac{[\mathrm{C}_{6}\mathrm{H}_{5}\mathrm{CO}_{2}\mathrm{H}][\mathrm{OH}^{-}]}{[\mathrm{C}_{6}\mathrm{H}_{5}\mathrm{CO}_{2}^{-}]} \]

Key Concepts

Base Ionization Constant

The Base Ionization Constant, often represented as \( K_b \), is a crucial concept in understanding how a weak base behaves in water. It quantifies the extent to which a weak base will ionize in water, forming its conjugate acid and hydroxide ions.When a weak base dissolves in water, it partially accepts protons from water molecules, resulting in a mixture of the base's ionized form and hydroxide ions (\( \text{OH}^- \)). To calculate \( K_b \):

• Write the balanced chemical equation for the reaction of the base with water.
• Identify the concentration of each product and reactant at equilibrium.
• Insert these concentrations into the \( K_b \) expression: \( K_b = \frac{[\text{Conjugate Acid}][\text{OH}^-]}{[\text{Base}]} \).

A larger \( K_b \) value indicates a stronger base, while a smaller \( K_b \) suggests a weaker base. This constant helps predict the pH of a solution and understand the relative strength of the base in solution.

Weak Base

A weak base is a base that does not completely ionize in a solution. Unlike strong bases, which fully dissociate in water, weak bases partially accept protons from water, forming an equilibrium between the base, its conjugate acid, and hydroxide ions.Key characteristics of weak bases include:

• Partial ionization in water.
• Presence of a dynamic equilibrium between the reactants and products.
• Lower \( K_b \) values compared to strong bases.

These features make weak bases less alkaline than strong bases. For an example, consider propylamine in water:\[ \text{C}_3\text{H}_7\text{NH}_2 + \text{H}_2\text{O} \rightleftharpoons \text{C}_3\text{H}_7\text{NH}_3^+ + \text{OH}^- \]The equilibrium established in the solution means not all propylamine will convert into its ionized form. This partial ionization is also responsible for a weak base's often subtle effect on the solution's pH. Understanding weak bases is fundamental in predicting the behavior of various chemical systems and designing chemical products.

Chemical Equations

Chemical equations are a way to represent the reactions between different substances symbolically. They illustrate how reactants transform into products. Both reactants and products are denoted by their chemical formulas separated by an arrow (\(\rightarrow\)) signifying the direction of the reaction.Several key elements should be considered when dealing with chemical equations:

• The equation should be balanced, meaning the number of atoms of each element should be equal on both sides of the equation.
• States of substances (solid, liquid, gas, or aqueous solution) are usually shown as \((s)\), \((l)\), \((g)\), and \((aq)\) respectively.
• In equilibrium reactions, the double arrow (\(\rightleftharpoons\)) represents that the reaction can proceed in both directions, establishing an equilibrium state.

For example, the reaction of monohydrogen phosphate ion in water can be written as:\[ \text{HPO}_4^{2-} + \text{H}_2\text{O} \rightleftharpoons \text{H}_2\text{PO}_4^- + \text{OH}^- \]This balanced equation depicts monohydrogen phosphate acting as a weak base, accepting a proton and generating hydroxide ions. Chemical equations provide vital information about the stoichiometry and dynamics of a chemical reaction, forming the basis for quantitative analysis in chemistry.