Problem 78 - Chemistry The Central Science In Si Units

Question

(a) Given that \(K_{b}\) for ammonia is \(1.8 \times 10^{-5}\) and that for hydroxylamine is \(1.1 \times 10^{-8}\), which is the stronger base? (b) Which is the stronger acid, the ammonium ion or the hydroxylammonium ion? (c) Calculate \(K_{a}\) values for \(\mathrm{NH}_{4}^{+}\) and \(\mathrm{H}_{3} \mathrm{NOH}^{+}\).

Step-by-Step Solution

Answer

Ammonia is the stronger base; hydroxylammonium ion is the stronger acid. \( K_a(\mathrm{NH}_4^+) = 5.56 \times 10^{-10} \), \( K_a(\mathrm{H}_3\mathrm{NOH}^+) = 9.09 \times 10^{-7} \).

Step 1: Compare Base Ionization Constants (Kb)

Given the base ionization constants:- For ammonia, \( K_{b} = 1.8 \times 10^{-5} \).- For hydroxylamine, \( K_{b} = 1.1 \times 10^{-8} \).The stronger base is the one with the higher \( K_{b} \) value because it indicates a greater tendency to accept protons. Therefore, ammonia is the stronger base.

Step 2: Determine the Stronger Acid Using Conjugate Acid-Base Pairs

According to the relationship between \( K_a \) and \( K_b \):\[K_{a} \times K_{b} = K_{w} = 1.0 \times 10^{-14}\]The stronger acid is the conjugate acid of the weaker base. Since hydroxylamine is the weaker base (lower \( K_{b} \)), its conjugate acid, hydroxylammonium ion, is the stronger acid.

Step 3: Calculate \( K_{a} \) for \( \mathrm{NH}_4^+ \)

Using \( K_{a} \times K_{b} = 1.0 \times 10^{-14} \), for ammonia's conjugate acid \( \mathrm{NH}_4^+ \):\[K_{a}(\mathrm{NH}_4^+) = \frac{1.0 \times 10^{-14}}{1.8 \times 10^{-5}} = 5.56 \times 10^{-10}\]This calculation gives \( K_{a} \) for the ammonium ion.

Step 4: Calculate \( K_{a} \) for \( \mathrm{H}_3\mathrm{NOH}^+ \)

Similarly, for hydroxylamine's conjugate acid \( \mathrm{H}_3\mathrm{NOH}^+ \):\[K_{a}(\mathrm{H}_3\mathrm{NOH}^+) = \frac{1.0 \times 10^{-14}}{1.1 \times 10^{-8}} = 9.09 \times 10^{-7}\]This calculation gives \( K_{a} \) for the hydroxylammonium ion.

Key Concepts

Base Ionization Constant

In acid-base chemistry, bases have a special ability to accept protons, and the measure of this ability is known as the base ionization constant, or \( K_{b} \). The \( K_{b} \) value tells us how effectively a base can accept hydrogen ions from a solution.
For instance, when comparing ammonia and hydroxylamine, we consider their \( K_{b} \) values: - Ammonia has a \( K_{b} = 1.8 \times 10^{-5} \). - Hydroxylamine has a \( K_{b} = 1.1 \times 10^{-8} \).The base with the higher \( K_{b} \) value is stronger. This is because a higher \( K_{b} \) means the base more effectively captures protons, making ammonia the stronger base in this comparison.
When working with bases, it's crucial to understand these concepts and recognize how the scale of \( K_{b} \) impacts the strength of a base.

Conjugate Acids and Bases

A fundamental idea in acid-base chemistry is the concept of conjugate acids and bases. When a base gains a proton, it forms its conjugate acid.
Similarly, when an acid donates a proton, it forms its conjugate base.
This relationship helps us determine which acid is stronger.
To find the stronger acid, we first need to identify the conjugate acid of the weaker base. This is because the acid strength of a conjugate acid is inversely related to the base strength of its conjugate base.

• Hydroxylamine, being the weaker base (lower \( K_{b} \)), produces a stronger conjugate acid, the hydroxylammonium ion (\( H_3NOH^+ \)).
• This is opposed to the ammonium ion (\( NH_4^+ \)), which forms from the stronger base, ammonia.

Understanding conjugate acid-base pairs can easily tell us about the strength of an acid or base and provides a deeper insight into acid-base equilibrium.

Acid Dissociation Constant

The acid dissociation constant, \( K_{a} \), measures how well an acid donates its proton in a solution.
The \( K_{a} \) can be calculated using the relationship with its base counterpart \( K_{b} \): \[ K_{a} \times K_{b} = K_{w} = 1.0 \times 10^{-14} \] Using this equation, we can calculate \( K_{a} \) values for the conjugate acids of both ammonia and hydroxylamine:

• For ammonia's conjugate acid, ammonium ion (\( NH_4^+ \)): \[ K_{a} = \frac{1.0 \times 10^{-14}}{1.8 \times 10^{-5}} = 5.56 \times 10^{-10} \]
• For hydroxylamine's conjugate acid, hydroxylammonium ion (\( H_3NOH^+ \)): \[ K_{a} = \frac{1.0 \times 10^{-14}}{1.1 \times 10^{-8}} = 9.09 \times 10^{-7} \]

In this context, the hydroxylammonium ion, with its higher \( K_{a} \), stands out as the stronger acid. Knowing how to find these \( K_{a} \) values allows us to better understand acid strength and behavior in solutions.