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, \(\mathrm{HPO}_{4}{\underline{\phantom{xx}}}^{2-}\); (c) benzoate ion, \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CO}_{2}^{-}\).

Step-by-Step Solution

Answer

The chemical reactions and \(K_b\) expressions for each base reacting with water are as follows: (a) Propylamine: \(C_{3}H_{7}NH_{2} + H_{2}O \rightleftharpoons C_{3}H_{7}NH_{3}^{+} + OH^{-}\) and \(K_{b} = \frac{[C_{3}H_{7}NH_{3}^{+}][OH^{-}]}{[C_{3}H_{7}NH_{2}]}\) (b) Monohydrogen phosphate ion: \(HPO_{4}^{2-} + H_{2}O \rightleftharpoons H_{2}PO_{4}^{-} + OH^{-}\) and \(K_{b} = \frac{[H_{2}PO_{4}^{-}][OH^{-}]}{[HPO_{4}^{2-}]}\) (c) Benzoate ion: \(C_{6}H_{5}CO_{2}^{-} + H_{2}O \rightleftharpoons C_{6}H_{5}CO_{2}H + OH^{-}\) and \(K_{b} = \frac{[C_{6}H_{5}CO_{2}H][OH^{-}]}{[C_{6}H_{5}CO_{2}^{-}]}\)

Step 1: (a) Chemical Equation for Propylamine

The reaction of propylamine with water is: \[ C_{3}H_{7}NH_{2} + H_{2}O \rightleftharpoons C_{3}H_{7}NH_{3}^{+} + OH^{-} \]

Step 2: (a) \(K_{b}\) Expression for Propylamine

The base dissociation constant expression for propylamine is: \[ K_{b} = \frac{[C_{3}H_{7}NH_{3}^{+}][OH^{-}]}{[C_{3}H_{7}NH_{2}]} \]

Step 3: (b) Chemical Equation for Monohydrogen Phosphate Ion

The reaction of monohydrogen phosphate ion with water is: \[ HPO_{4}^{2-} + H_{2}O \rightleftharpoons H_{2}PO_{4}^{-} + OH^{-} \]

Step 4: (b) \(K_{b}\) Expression for Monohydrogen Phosphate Ion

The base dissociation constant expression for the monohydrogen phosphate ion is: \[ K_{b} = \frac{[H_{2}PO_{4}^{-}][OH^{-}]}{[HPO_{4}^{2-}]} \]

Step 5: (c) Chemical Equation for Benzoate Ion

The reaction of benzoate ion with water is: \[ C_{6}H_{5}CO_{2}^{-} + H_{2}O \rightleftharpoons C_{6}H_{5}CO_{2}H + OH^{-} \]

Step 6: (c) \(K_{b}\) Expression for Benzoate Ion

The base dissociation constant expression for benzoate ion is: \[ K_{b} = \frac{[C_{6}H_{5}CO_{2}H][OH^{-}]}{[C_{6}H_{5}CO_{2}^{-}]} \]

Key Concepts

Chemical Equations

Chemical equations are symbolic representations of chemical reactions, where the reactants are on the left side and the products are on the right. The arrow (\( \rightleftharpoons \) for reversible reactions) shows that substances can be converted into each other. Understanding how to construct chemical equations is essential as they show how atoms and molecules are rearranged in a reaction.

For the chemical equation involving a base, we often see a base reacting with water. This reaction might result in the generation of the hydroxide ion (\(OH^-\)) and a corresponding conjugate acid. It's critical to ensure equations are balanced—they must have the same number of each type of atom on both sides.

In this exercise, propylamine, the monohydrogen phosphate ion, and the benzoate ion each have their unique chemical equations with water as a reactant.

Propylamine

Propylamine (\(C_{3}H_{7}NH_{2}\)) is an organic compound. It acts as a base because it accepts a proton from water, producing the propylammonium ion (\(C_{3}H_{7}NH_{3}^{+}\)) and hydroxide ion (\(OH^{-}\)).
Its reaction with water is:

• \(C_{3}H_{7}NH_{2} + H_{2}O \rightleftharpoons C_{3}H_{7}NH_{3}^{+} + OH^{-}\)

This equation clearly shows the base (propylamine) gaining a hydrogen ion, thus forming its conjugate acid. When writing the chemical equation, it's important to identify the portions of each compound that will participate in the exchange of the hydrogen ion or electron pairs.

For propylamine's \(K_{b}\) expression:

• \(K_{b} = \frac{[C_{3}H_{7}NH_{3}^{+}][OH^{-}]}{[C_{3}H_{7}NH_{2}]}\)

This expression represents the equilibrium concentrations of products and reactants, with concentration denoted in square brackets. The larger the \(K_{b}\), the stronger the base.

Monohydrogen Phosphate Ion

The monohydrogen phosphate ion (\(HPO_{4}^{2-}\)) is a polyatomic ion with significant importance in biological systems. In this reaction, it acts as a base by accepting a proton from water. The chemical equation is:

• \(HPO_{4}^{2-} + H_{2}O \rightleftharpoons H_{2}PO_{4}^{-} + OH^{-}\)

Here, water donates a proton to \(HPO_{4}^{2-}\), forming \(H_{2}PO_{4}^{-}\), meaning that \(HPO_{4}^{2-}\) acts as the base, and its conjugate acid is \(H_{2}PO_{4}^{-}\).

The \(K_{b}\) expression is:

• \(K_{b} = \frac{[H_{2}PO_{4}^{-}][OH^{-}]}{[HPO_{4}^{2-}]}\)

This equation provides insight into the strength of the base. Since the \(K_{b}\) is a measure of how completely the base dissociates into ions in a solution, its value can predict the behavior of this ion in a natural system like human physiology.

Benzoate Ion

The benzoate ion (\(C_{6}H_{5}CO_{2}^{-}\)) is derived from benzoic acid and acts as a base in aqueous solutions. When it reacts with water, the following chemical equation occurs:

• \(C_{6}H_{5}CO_{2}^{-} + H_{2}O \rightleftharpoons C_{6}H_{5}CO_{2}H + OH^{-}\)

In this reaction, the benzoate ion accepts a hydrogen ion from water, resulting in the formation of benzoic acid (\(C_{6}H_{5}CO_{2}H\)) and a hydroxide ion (\(OH^-\)).

The \(K_{b}\) expression for this equilibrium is:

• \(K_{b} = \frac{[C_{6}H_{5}CO_{2}H][OH^{-}]}{[C_{6}H_{5}CO_{2}^{-}]}\)

By measuring the equilibrium concentrations, \(K_{b}\) indicates the base strength. In applications, understanding the dissociation behavior of benzoate can help in predicting how benzoic acid and its salts might behave in food and pharmaceuticals.