Question

Aqueous ammonium hydroxide, \(\mathrm{NH}_{4} \mathrm{OH},\) is used as a household cleaning solution. If the hydroxide ion concentration of a \(0.245 \mathrm{M}\) solution is \(2.1 \times 10^{-3} \mathrm{M}\), what is the ionization constant for the base?

Short answer

The ionization constant \( K_b \) for ammonium hydroxide is approximately \( 1.80 \times 10^{-5} \).

Step-by-step solution

Write the Ionization Equation

The dissociation of ammonium hydroxide in water can be expressed as \( \text{NH}_4\text{OH} \rightleftharpoons \text{NH}_4^+ + \text{OH}^- \). This equilibrium reaction shows the dissociation of NH4OH into ammonium (\(\text{NH}_4^+\)) and hydroxide (\(\text{OH}^-\)) ions.

Understand the Ionization Constant

The base ionization constant, \(K_b\), is a measure of the strength of a base in solution. It is calculated using the concentration of the ions produced at equilibrium. \(K_b = \frac{[\text{NH}_4^+][\text{OH}^-]}{[\text{NH}_4\text{OH}]} \).

Substitute Known Values

Given that \([\text{OH}^-] = 2.1 \times 10^{-3} \text{ M}\), \([\text{NH}_4^+]\) would also be \(2.1 \times 10^{-3} \text{ M}\) due to the 1:1 stoichiometric ratio in the ionization equation. The initial concentration \([\text{NH}_4\text{OH}]\) is \(0.245 \text{ M}\), but at equilibrium, it is \(0.245 - 2.1 \times 10^{-3} \text{ M}\), slightly changing due to ionization.

Calculate Equilibrium Concentration

Determine \([\text{NH}_4\text{OH}]\) at equilibrium: \([\text{NH}_4\text{OH}]\) is approximately \(0.245 \text{ M}\), since \(2.1 \times 10^{-3} \text{ M}\) ionized is a relatively small amount compared with the initial concentration. So, use \(0.245\) in the equation.

Compute the Ionization Constant \(K_b\)

Substitute values into the equation for \(K_b\): \(K_b = \frac{(2.1 \times 10^{-3})(2.1 \times 10^{-3})}{0.245} \). Calculate this to find \(K_b \approx 1.80 \times 10^{-5}\).

Final Verification

Double-check the values and calculations to ensure accuracy: all values and units align according to the steps above, confirming the computed ionization constant is consistent with the context of the problem.

Key concepts

Base Ionization

When discussing base ionization, we refer to the process by which a base dissociates into ions when dissolved in water. Consider ammonium hydroxide, often represented by the chemical formula \( \text{NH}_4\text{OH} \). It is a weak base, and its ionization in water is a reversible process. This can be expressed through the equation \( \text{NH}_4\text{OH} \rightleftharpoons \text{NH}_4^+ + \text{OH}^- \).

Understanding the nature of this reaction is essential. Not all \( \text{NH}_4\text{OH} \) will dissociate into ions because it is not a strong base. This incomplete ionization leads to an equilibrium where the rate at which the base molecules dissociate into ions is equal to the rate at which the ions recombine to form the base. As such, the concentration of ions in the solution is relatively low compared to the initial concentration of the base.

The extent to which a base ionizes in solution is quantified by the base ionization constant, denoted as \( K_b \). This constant reflects the base's strength in solution, where a larger \( K_b \) value indicates a stronger base with a higher tendency to donate hydroxide ions. It's important to remember that for weak bases like ammonium hydroxide, the \( K_b \) value is quite small.

Equilibrium Concentration

Equilibrium concentration pertains to how the concentrations of reactants and products stabilize at equilibrium in a chemical reaction. In the case of ammonium hydroxide, at equilibrium, the concentration of \( \text{NH}_4\text{OH} \) that remains un-ionized, along with the concentrations of \( \text{NH}_4^+ \) and \( \text{OH}^- \), establishes the system's state.

For instance, if you begin with a solution of \( 0.245 \text{ M} \) ammonium hydroxide, and it ionizes to a certain extent, you'll have new concentrations for \( \text{NH}_4^+ \) and \( \text{OH}^- \). The equilibrium situation is described through the expression:
\[K_b = \frac{[\text{NH}_4^+][\text{OH}^-]}{[\text{NH}_4\text{OH}]}\]

Here, \([\text{OH}^-] = 2.1 \times 10^{-3} \text{ M}\), \([\text{NH}_4^+]\) mirrors this due to a 1:1 ratio, and the equilibrium concentration of \([\text{NH}_4\text{OH}]\) nearly equals the initial concentration due to its weak base nature. Since only a minute fraction of the base ionizes, we typically consider \([\text{NH}_4\text{OH}]\) unchanged at \(0.245 \text{ M}\) when calculating \( K_b \).

In practical use, knowing the equilibrium concentrations of these species allows chemists to predict reaction behaviors in different scenarios and adjust conditions for desired results.

Ammonium Hydroxide

Ammonium hydroxide, \( \text{NH}_4\text{OH} \), is a solution of ammonia in water and is used for various purposes, including household cleaning. This solution is categorized as a weak base, meaning it does not fully dissociate into its ions in water.

When dissolved, it partially separates into ammonium ions (\(\text{NH}_4^+\)) and hydroxide ions (\(\text{OH}^-\)). The dissociation is represented by \( \text{NH}_4\text{OH} \rightleftharpoons \text{NH}_4^+ + \text{OH}^- \), highlighting its reversible nature. The presence of \(\text{OH}^-\) ions in solution confirms its basic nature.

A solution of ammonium hydroxide is known for removing grease and dirt due to its ability to neutralize acids, which is why it is popular in cleaning products. Its faint pungent smell is due to ammonia, which readily mixes into the solution.

The understanding of ammonium hydroxide's behavior in water is important in both practical applications and academic contexts. It exemplifies the concepts of weak bases and equilibrium, allowing students to explore various chemical phenomena.