Problem 83 - Chemistry An Atoms First Approach

Question

Write the reaction and the corresponding \(K_{\mathrm{b}}\) equilibrium expression for each of the following substances acting as bases in water. a. \(\mathrm{NH}_{3}\) b. \(\mathrm{C}_{5} \mathrm{H}_{5} \mathrm{N}\)

Step-by-Step Solution

Answer

a. The reaction of ammonia (\(\mathrm{NH}_{3}\)) with water as a base is: \(\mathrm{NH}_{3} + \mathrm{H}_{2} \mathrm{O} \rightleftharpoons \mathrm{NH}_{4}^{+} + \mathrm{OH}^{-}\). The corresponding \(K_{\mathrm{b}}\) expression is: \(K_{\mathrm{b}} = \frac{[\mathrm{NH}_{4}^{+}][\mathrm{OH}^{-}]}{[\mathrm{NH}_{3}]}\). b. The reaction of pyridine (\(\mathrm{C}_{5} \mathrm{H}_{5} \mathrm{N}\)) with water as a base is: \(\mathrm{C}_{5} \mathrm{H}_{5} \mathrm{N} + \mathrm{H}_{2} \mathrm{O} \rightleftharpoons \mathrm{C}_{5} \mathrm{H}_{5} \mathrm{NH}^{+} + \mathrm{OH}^{-}\). The corresponding \(K_{\mathrm{b}}\) expression is: \(K_{\mathrm{b}} = \frac{[\mathrm{C}_{5} \mathrm{H}_{5} \mathrm{NH}^{+}][\mathrm{OH}^{-}]}{[\mathrm{C}_{5} \mathrm{H}_{5} \mathrm{N}]}\).

Step 1: Write the reaction with water

Write the reaction of ammonia (\(\mathrm{NH}_{3}\)) with water, acting as a base: \[ \mathrm{NH}_{3} + \mathrm{H}_{2} \mathrm{O} \rightleftharpoons \mathrm{NH}_{4}^{+} + \mathrm{OH}^{-} \]

Step 2: Write the \(K_{\mathrm{b}}\) expression

Using the previously derived formula for \(K_{\mathrm{b}}\), write the equilibrium expression for ammonia (\(\mathrm{NH}_{3}\)): \[ K_{\mathrm{b}} = \frac{[\mathrm{NH}_{4}^{+}][\mathrm{OH}^{-}]}{[\mathrm{NH}_{3}]} \] b. \(\mathrm{C}_{5} \mathrm{H}_{5} \mathrm{N}\)

Step 1: Write the reaction with water

Write the reaction of pyridine (\(\mathrm{C}_{5} \mathrm{H}_{5} \mathrm{N}\)) with water, acting as a base: \[ \mathrm{C}_{5} \mathrm{H}_{5} \mathrm{N} + \mathrm{H}_{2} \mathrm{O} \rightleftharpoons \mathrm{C}_{5} \mathrm{H}_{5} \mathrm{NH}^{+} + \mathrm{OH}^{-} \]

Step 2: Write the \(K_{\mathrm{b}}\) expression

Using the formula for \(K_{\mathrm{b}}\), write the equilibrium expression for pyridine (\(\mathrm{C}_{5} \mathrm{H}_{5} \mathrm{N}\)): \[ K_{\mathrm{b}} = \frac{[\mathrm{C}_{5} \mathrm{H}_{5} \mathrm{NH}^{+}][\mathrm{OH}^{-}]}{[\mathrm{C}_{5} \mathrm{H}_{5} \mathrm{N}]} \]

Key Concepts

Ammonia Reaction with Water

When ammonia, denoted as \( \mathrm{NH}_{3} \), reacts with water, this involves a basic reaction. Ammonia is a well-known weak base that partially dissociates in an aqueous solution. The reaction can be depicted as follows:

• \( \mathrm{NH}_{3} + \mathrm{H}_{2} \mathrm{O} \rightleftharpoons \mathrm{NH}_{4}^{+} + \mathrm{OH}^{-} \)

Ammonia accepts a proton (\( \mathrm{H}^{+} \)) from water, resulting in the formation of ammonium \( \mathrm{NH}_{4}^{+} \) and hydroxide ions \( \mathrm{OH}^{-} \). In this equation, the double arrow signifies that this is a reversible process.
This means the forward reaction happens at the same rate as the reverse reaction at equilibrium.
Ammonia acting as a base is crucial in many chemical reactions because it slightly increases the pH of the solution by producing \( \mathrm{OH}^{-} \) ions.

Pyridine Reaction with Water

Pyridine, labeled as \( \mathrm{C}_{5} \mathrm{H}_{5} \mathrm{N} \), is another weak base that interacts with water similarly to ammonia. Just like ammonia, it tends to accept a proton from water"s" dom="urn:oasis:names:tc:xacml:3.0:core:schema:wd-17">pyridine reaction with water "sersions:

• \( \mathrm{C}_{5} \mathrm{H}_{5} \mathrm{N} + \mathrm{H}_{2} \mathrm{O} \rightleftharpoons \mathrm{C}_{5} \mathrm{H}_{5} \mathrm{NH}^{+} + \mathrm{OH}^{-} \)

In this scenario, pyridine acts as a Lewis base, accepting a proton to form the pyridinium ion \( \mathrm{C}_{5} \mathrm{H}_{5} \mathrm{NH}^{+} \) along with \( \mathrm{OH}^{-} \). This reversible process denotes the pyridine"s basic property.
Understanding pyridine"s reaction is significant in chemistry due to its applications in synthesizing other compounds and elucidating reaction mechanisms in various fields.

Equilibrium Constant Kb

The equilibrium constant, \( K_{\mathrm{b}} \), provides insight into the extent of base ionization in water. For any base, a \( K_{\mathrm{b}} \) value is derived from the concentration of the products and reactants at equilibrium.

• For ammonia: \[ K_{\mathrm{b}} = \frac{[\mathrm{NH}_{4}^{+}][\mathrm{OH}^{-}]}{[\mathrm{NH}_{3}]} \]
• For pyridine: \[ K_{\mathrm{b}} = \frac{[\mathrm{C}_{5} \mathrm{H}_{5} \mathrm{NH}^{+}][\mathrm{OH}^{-}]}{[\mathrm{C}_{5} \mathrm{H}_{5} \mathrm{N}]} \]

A higher \( K_{\mathrm{b}} \) value suggests a stronger base, indicating that more base has dissociated to form products. Conversely, a lower \( K_{\mathrm{b}} \) indicates a weaker base.
These expressions are fundamental to predicting the behavior and strength of bases in various solutions, contributing to our understanding of chemical equilibria in a broad spectrum of contexts.