Problem 52 - Chemistry and Chemical Reactivity

Question

A hypothetical weak base has \(K_{\mathrm{b}}=5.0 \times 10^{-4} .\) Calculate the equilibrium concentrations of the base, its conjugate acid, and OH \(^{-}\) in a 0.15 M solution of the base.

Step-by-Step Solution

Answer

[BH^+] = [OH^-] ≈ 8.66 × 10^{-3} M, [B] ≈ 0.141 M

Step 1: Write the Dissociation Equation

Consider the dissociation of the weak base \( B \) in water: \[ B + H_2O \rightleftharpoons BH^+ + OH^- \] Here, \( B \) is the weak base, \( BH^+ \) is its conjugate acid, and \( OH^- \) is the hydroxide ion.

Step 2: Set Up Initial Concentrations

Initially, we have a concentration of \( 0.15 \, M \) for the weak base \( B \), and both \( BH^+ \) and \( OH^- \) are \( 0 \, M \).

Step 3: Define Changes in Concentrations

As the reaction reaches equilibrium, let \( x \) be the change in concentration for \( BH^+ \) and \( OH^- \), which are produced. Thus, the equilibrium concentrations are:\[ [B] = 0.15 - x \]\[ [BH^+] = x \]\[ [OH^-] = x \]

Step 4: Apply the Equilibrium Constant Expression

The equilibrium constant expression for the base dissociation is given by:\[ K_b = \frac{[BH^+][OH^-]}{[B]} \]Substituting the known values, we have:\[ 5.0 \times 10^{-4} = \frac{x^2}{0.15 - x} \]

Step 5: Simplify and Solve the Equation

Assuming \( x \) is small compared to 0.15, the expression simplifies to:\[ 5.0 \times 10^{-4} = \frac{x^2}{0.15} \]Solving for \( x \), we get:\[ x^2 = 5.0 \times 10^{-4} \times 0.15 \]\[ x^2 = 7.5 \times 10^{-5} \]\[ x = \sqrt{7.5 \times 10^{-5}} \approx 8.66 \times 10^{-3} \]

Step 6: Calculate Equilibrium Concentrations

The equilibrium concentrations are:\[ [BH^+] = [OH^-] = x \approx 8.66 \times 10^{-3} \, M \]\[ [B] = 0.15 - x \approx 0.15 - 8.66 \times 10^{-3} \approx 0.141 \, M \]

Key Concepts

Dissociation Equation

When discussing weak bases, the dissociation equation is crucial for understanding how the base interacts with water. A weak base, like our example base \( B \), does not completely dissociate in water. Instead, it establishes an equilibrium with its products. For the reaction:
\[ B + H_2O \rightleftharpoons BH^+ + OH^- \]
\( B \) is the weak base, \( BH^+ \) is the conjugate acid, and \( OH^- \) is the hydroxide ion. This equation shows that as \( B \) reacts with water, "gaining" an \( H^+ \) ion, it forms \( BH^+ \). Simultaneously, an \( OH^- \) ion is produced, contributing to the basicity of the solution. Therefore, understanding this equation provides insight into the nature of weak bases, which only partially ionize in solution and form an equilibrium, rather than ionizing completely.

Equilibrium Concentration

Equilibrium concentrations reflect the amounts of each species present after a weak base's dissociation process has reached a balance. Starting with an initial concentration of \(0.15 \, M\) for the base \( B \), the reaction shifts towards equilibrium:
- Initial: \( [B] = 0.15 \, M \), \( [BH^+] = 0 \, M \), \( [OH^-] = 0 \, M \)- Change: \( [B] = 0.15 - x \), \( [BH^+] = x \), \( [OH^-] = x \)
During equilibrium, for each molecule of \( B \) that dissociates, one molecule of \( BH^+ \) and one of \( OH^- \) forms."\( x \)" represents the concentration change, assumed small in this situation. The calculated equilibrium concentrations are:

• \( [B] \approx 0.141 \, M \)
• \( [BH^+] \approx 8.66 \times 10^{-3} \, M \)
• \( [OH^-] \approx 8.66 \times 10^{-3} \, M \)

This small value of \( x \) demonstrates minimal dissociation, characteristic of a weak base.

Base Dissociation Constant (Kb)

The Base Dissociation Constant \( K_b \) quantifies a base's strength by measuring its ability to dissociate into ions in solution. For weak bases, like our example \( B \), this constant is significantly less than one (\( 5.0 \times 10^{-4} \) here), denoting limited dissociation. To calculate \( K_b \), you use the equilibrium expression:
\[ K_b = \frac{[BH^+][OH^-]}{[B]} \]
This formula considers the equilibrium concentrations of the conjugate acid \( BH^+ \) and the hydroxide ion \( OH^- \) relative to the undissociated base \( B \). For a weak base, since it doesn't dissociate much, \( [BH^+] \) and \( [OH^-] \) values remain small, affirming a small \( K_b \). Solving:

• Substitute values: \( 5.0 \times 10^{-4} = \frac{x^2}{0.15} \)
• Solve for \( x \) and assess concentrations and base strength efficiently

Thus, the \( K_b \) helps demonstrate the intrinsic properties of the weak base, confirming its behavior in aqueous solutions.