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 \(\mathrm{H}_{3} \mathrm{NOH}^{+}\).

Short answer

(a) Ammonia is the stronger base since its \(K_{b}\) value is larger (\(1.8 \times 10^{-5} > 1.1 \times 10^{-8}\)). (b) The stronger acid is the hydroxylammonium ion (H3NOH+), as ammonia's conjugate acid, NH4+, is the weaker acid. (c) The Ka values for NH4+ and H3NOH+ are \(5.56 \times 10^{-10}\) and \(9.09 \times 10^{-7}\), respectively.

Step-by-step solution

(a) Determine the stronger base

First, we need to identify which compound is the stronger base by comparing their Kb values. Recall that the higher the value of Kb, the stronger the base. Given: \[K_{b}(\mathrm{NH}_{3}) = 1.8 \times 10^{-5}\] and \[K_{b}(\mathrm{H}_{2}\mathrm{NOH}) = 1.1 \times 10^{-8}\] \(1.8 \times 10^{-5} > 1.1 \times 10^{-8}\) Since the Kb value for ammonia (NH3) is larger than hydroxylamine (H2NOH), ammonia is the stronger base.

(b) Determine the stronger acid

To determine which is the stronger acid between the ammonium ion (NH4+) and the hydroxylammonium ion (H3NOH+), we need to understand the relationship between Ka, Kb, and pKa/pKb values. The relationship between Ka and Kb can be written as: \[K_{a}\times K_{b} = K_{w}\] where Kw is the ion product of water which is \(1\times10^{-14}\) at 25°C. A stronger base has a weaker conjugate acid, and a stronger acid has a weaker conjugate base. Since ammonia is the stronger base, its conjugate acid, the ammonium ion (NH4+), is the weaker acid. Therefore, the stronger acid is the hydroxylammonium ion (H3NOH+).

(c) Calculate Ka values for NH4+ and H3NOH+

Following the relationship between Ka, Kb, and Kw: \(K_{a}\times K_{b} = K_{w}\) We need to find the Ka values for both NH4+ and H3NOH+ using the provided Kb values. For NH4+: \[K_{a} = \frac{K_{w}}{K_{b}(\mathrm{NH}_{3})} = \frac{1 \times 10^{-14}}{1.8 \times 10^{-5}}\] \[K_{a}(\mathrm{NH}_{4}^{+}) = 5.56 \times 10^{-10}\] For H3NOH+: \[K_{a} = \frac{K_{w}}{K_{b}(\mathrm{H}_{2}\mathrm{NOH})} = \frac{1 \times 10^{-14}}{1.1 \times 10^{-8}}\] \[K_{a}(\mathrm{H}_{3}\mathrm{NOH}^{+}) = 9.09 \times 10^{-7}\] The Ka values for NH4+ and H3NOH+ are \(5.56 \times 10^{-10}\) and \(9.09 \times 10^{-7}\), respectively.

Key concepts

Understanding Ammonia

Ammonia, represented chemically as \( NH_3 \), plays a crucial role in acid-base chemistry. It is a common weak base in chemical reactions.

• The compound consists of one nitrogen atom bonded to three hydrogen atoms.
• In aqueous solutions, ammonia can accept a proton (\(H^+\)) from water, forming the ammonium ion \( NH_4^+ \).

Ammonia’s basicity is attributed to the lone pair of electrons on the nitrogen atom. This allows it to readily accept protons.
Its effectiveness as a base is often measured by its base dissociation constant (\( K_b \)). When \( K_b \) is higher, the substance is a stronger base. In the example, ammonia's \( K_b \) of \(1.8 \times 10^{-5}\) indicates it is a stronger base compared to hydroxylamine. This can be essential in reactions where a strong base is required.

Exploring Hydroxylamine

Hydroxylamine, denoted as \( H_2NOH \), is another intriguing compound in acid-base studies.

• It features a nitrogen atom connected to an OH group and a hydrogen atom.
• Like ammonia, it also has a lone pair on nitrogen, enabling it to act as a base.

Hydroxylamine's \( K_b \) value is \(1.1 \times 10^{-8}\), much lower than that of ammonia.
This lower value signifies that hydroxylamine is a weaker base.
The difference in \( K_b \) values highlights how structural differences can significantly impact basic strength. Furthermore, hydroxylamine forms its conjugate acid, the hydroxylammonium ion \( H_3NOH^+ \), by accepting protons in solution.

Equilibrium Constants: \( K_a \) and \( K_b \)

In acid-base chemistry, equilibrium constants such as \( K_a \) and \( K_b \) indicate the strengths of acids and bases. \( K_b \), the base dissociation constant, measures the strength of a base.

• The greater the \( K_b \), the stronger the base.
• Conversely, \( K_a \), the acid dissociation constant, shows how well an acid donates protons.

In the solution of this exercise, we use the relation \( K_a \times K_b = K_w \) to find the \( K_a \) of ammonium and hydroxylammonium ions.
Here, \( K_w \), the ion product of water, is \(1 \times 10^{-14}\) at room temperature.
The calculations reveal that ammonium \( NH_4^+ \) has a \( K_a \) of \(5.56 \times 10^{-10}\), indicating it is a weaker acid compared to hydroxylammonium \( H_3NOH^+ \), which has a \( K_a \) of \(9.09 \times 10^{-7}\). Using these constants, chemists can predict the behavior of acids and bases in solution.

Analyzing Conjugate Acid-Base Pairs

Conjugate acid-base pairs are fundamental concepts in understanding acid-base reactions. Each base has a conjugate acid, formed by the addition of a proton, and each acid has a conjugate base, formed by losing a proton.

• For ammonia \( NH_3 \), the conjugate acid is the ammonium ion \( NH_4^+ \).
• For hydroxylamine \( H_2NOH \), the conjugate acid is \( H_3NOH^+ \).

A key relationship in acid-base chemistry is that a strong base has a weak conjugate acid and vice versa.
This is exhibited by the ammonium ion being a weaker acid due to ammonia being a stronger base. Conversely, as hydroxylamine is a weaker base, its conjugate acid, the hydroxylammonium ion, is relatively stronger.
Understanding these pairs helps in predicting the outcome of acid-base reactions and in identifying the relative strengths and weaknesses of acids and bases.