Question

What is the \(\mathrm{pH}\) of \(0.5 \mathrm{M}\) ammonia \(\left(\mathrm{NH}_{3}\right) ? K_{b}=1.8 \times 10^{-5}\)

Short answer

The pH of 0.5 M ammonia is approximately 11.48.

Step-by-step solution

Write the Ionization Equation

Ammonia (\(\text{NH}_3\)) reacts with water to form ammonium ions (\(\text{NH}_4^+\)) and hydroxide ions (\(\text{OH}^-\)). The ionization equation is:\[\text{NH}_3 + \text{H}_2\text{O} \rightleftharpoons \text{NH}_4^+ + \text{OH}^-\]

Write the Expression for Kb

The equilibrium expression for the base dissociation constant \( K_b \) of ammonia is given by:\[K_b = \frac{[\text{NH}_4^+][\text{OH}^-]}{[\text{NH}_3]}\]We know \( K_b = 1.8 \times 10^{-5} \).

Set Up the ICE Table

Initially, \([\text{NH}_3] = 0.5 \text{ M}\), and \([\text{NH}_4^+] = [\text{OH}^-] = 0 \text{ M}\). Let \(x\) be the change in concentration at equilibrium. The ICE table is:\[\begin{array}{c|c|c|c} & [\text{NH}_3] & [\text{NH}_4^+] & [\text{OH}^-] \\hline\text{Initial} & 0.5 & 0 & 0 \\text{Change} & -x & +x & +x \\text{Equilibrium} & 0.5 - x & x & x \\end{array}\]

Substitute Equilibrium Values into Kb Expression

Substitute the equilibrium concentrations into the \( K_b \) expression:\[K_b = \frac{x \cdot x}{0.5 - x} = 1.8 \times 10^{-5}\]which simplifies to \(x^2 = 1.8 \times 10^{-5} (0.5 - x)\).

Approximate and Solve for x

Assume \(x\) is small compared to 0.5, so \(0.5 - x \approx 0.5\). This gives:\[x^2 = 1.8 \times 10^{-5} \times 0.5\]Solving for \(x\) gives:\[x = \sqrt{9 \times 10^{-6}} = 3 \times 10^{-3}\]Thus, the concentration of \(\text{OH}^-\) is \(3 \times 10^{-3} \text{ M}\).

Calculate pOH

The \(\text{pOH}\) is calculated from the hydroxide ion concentration:\[\text{pOH} = -\log[\text{OH}^-] = -\log(3 \times 10^{-3}) \approx 2.52\]

Calculate pH

Using the relationship \(\text{pH} + \text{pOH} = 14\), we find:\[\text{pH} = 14 - \text{pOH} = 14 - 2.52 = 11.48\]

Key concepts

Ionization Equation

When dealing with ammonia (\(\text{NH}_3\)), we need to understand how it interacts with water. In a solution, ammonia acts as a base, meaning it can accept hydrogen ions from water. This reaction can be represented by the ionization equation:\[\text{NH}_3 + \text{H}_2\text{O} \rightleftharpoons \text{NH}_4^+ + \text{OH}^-\]These equations are crucial because they show how substances in a solution transform, giving insight into the concentrations of compounds at equilibrium.
To determine the pH of a solution, recognizing this transformation is essential, as the concentration of hydroxide ions (\(\text{OH}^-\)) directly impacts whether the solution is acidic or basic.

Base Dissociation Constant

The base dissociation constant (\(K_b\)) represents the strength of a base in solution. For ammonia, \(K_b\) indicates how effectively it converts into ammonium ions and hydroxide ions in water. The expression for the base dissociation constant of ammonia is:\[K_b = \frac{[\text{NH}_4^+][\text{OH}^-]}{[\text{NH}_3]}\]A higher \(K_b\) value means the base dissociates more in solution, producing more hydroxide ions.
This is critical because calculating \(K_b\) allows us to find the equilibrium concentrations of ions, central to understanding the resulting basicity of the solution.

ICE Table

The ICE table (Initial, Change, Equilibrium) helps in organizing and visualizing the changes in concentrations of substances in a chemical reaction. For ammonia in water, the table starts with the initial concentrations, shows the changes as the reaction proceeds, and ends with the equilibrium concentrations.

• **Initial**: Concentrations before the reaction (\([\text{NH}_3] = 0.5\text{ M}, [\text{NH}_4^+] = [\text{OH}^-] = 0\text{ M}\)).
• **Change**: Indicates the increase or decrease (\(-x, +x, +x\)).
• **Equilibrium**: Final concentrations (\(0.5 - x, x, x\)).

The ICE table allows us to see how much of each substance is present at each stage, essential for solving for unknown concentrations using the equilibrium expression of \(K_b\).

Equilibrium Concentrations

Equilibrium concentrations reflect the amounts of each ion or molecule present when the reaction has reached a state of balance. Knowing these concentrations helps us calculate the pH of the solution. After setting up the ICE table and applying the \(K_b\) expression:\[x^2 = 1.8 \times 10^{-5} (0.5 - x)\]this equation is solved to find \(x\), representing the hydroxide ion (\(\text{OH}^-\)) concentration at equilibrium.
Assumptions like \(x\) being small compared to the initial concentration simplify calculations, giving us a manageable approach to finding the solution's pH. By understanding equilibrium concentrations, we complete the picture of how basic the solution is, directly relating to ammonia's behavior in water.