Problem 74 - Chemistry

Question

Determine the \(K_{\mathrm{b}}\) of a weak base if a \(0.19-M\) aqueous solution of the base at \(25^{\circ} \mathrm{C}\) has a \(\mathrm{pH}\) of 10.88 .

Step-by-Step Solution

Answer

The value of \( K_{\mathrm{b}} \) is approximately \( 3.03 \times 10^{-6} \).

Step 1: Understand the Relationship Between pH and pOH

First, recall that pH and pOH are related by the equation: \( \text{pH} + \text{pOH} = 14 \) at \(25^{\circ} \text{C}\). In this case, since the pH is given as 10.88, we can find the pOH as follows: \( \text{pOH} = 14 - 10.88 = 3.12 \).

Step 2: Convert pOH to OH- Concentration

The concentration of hydroxide ions, \([\text{OH}^-]\), can be determined using the expression \( [\text{OH}^-] = 10^{-\text{pOH}} \). Substitute the pOH value into the equation: \( [\text{OH}^-] = 10^{-3.12} \approx 7.59 \times 10^{-4} \text{ M} \).

Step 3: Write the Equilibrium Expression for the Weak Base

Consider the equilibrium equation for a weak base, \( \text{B} + \text{H}_2\text{O} \rightleftharpoons \text{BH}^+ + \text{OH}^- \). The equilibrium constant for this reaction, called \( K_{\text{b}} \), is given by the formula: \( K_{\text{b}} = \frac{[\text{BH}^+][\text{OH}^-]}{[\text{B}]} \).

Step 4: Determine Values to Substitute into the Equilibrium Expression

Initially, the base concentration, \([\text{B}]\), is provided as 0.19 M and \([\text{OH}^-]\) has been calculated as \( 7.59 \times 10^{-4} \text{ M} \). We assume \([\text{BH}^+]\) is also \( 7.59 \times 10^{-4} \text{ M} \) because each OH- comes from one dissociated BH+. Substitute these values into the expression \( K_{\text{b}} = \frac{(7.59 \times 10^{-4})(7.59 \times 10^{-4})}{0.19 - 7.59 \times 10^{-4}} \).

Step 5: Simplify the Expression and Calculate Kb

Simplify the expression: \( K_{\text{b}} = \frac{(7.59 \times 10^{-4})^2}{0.19} \approx \frac{5.76 \times 10^{-7}}{0.19} \approx 3.03 \times 10^{-6} \). Thus, \( K_{\text{b}} \approx 3.03 \times 10^{-6} \).

Key Concepts

Weak Base

A weak base is a substance that partially ionizes in water, releasing some, but not all, of its base particles as hydroxide ions \(\text{OH}^-\). This partial ionization is what differentiates weak bases from strong bases, which completely dissociate in solution.

Examples of common weak bases include ammonia (NH₃) and methylamine (CH₃NH₂). Weak bases play an important role in chemistry since they're often involved in buffer solutions and various biological processes.

• Weak bases do not release many hydroxide ions in solution, which is why they have higher pH values than neutral water but lower than strong bases.
• The degree of ionization of a weak base is quantified by its equilibrium constant \(K_{\text{b}}\).

Honing in on the behavior of a weak base can help us predict and calculate other important properties, such as its concentration and equilibrium conditions.

pH and pOH Relationship

Understanding the relationship between pH and pOH is essential in the study of acidity and basicity in solutions. These two parameters are inversely related through the formula:
\[ \text{pH} + \text{pOH} = 14 \]
This equation is valid at 25°C, which is standard room temperature. It serves as a bridge between the concentration of hydrogen ions \([\text{H}^+]\) and hydroxide ions \([\text{OH}^-]\).

Knowing the pH can help you find the pOH, and vice versa. For example, if the pH of a solution is given as 10.88, you can directly calculate its pOH by subtracting the pH from 14:

• This allows chemists to evaluate the balance of acid and base in a solution.
• The relationship helps in determining the concentrations of ions in the solution.

By knowing how to manipulate these numbers, you become adept at analyzing the nature of any aqueous solution.

Equilibrium Constant (Kb)

The equilibrium constant for a weak base, denoted \(K_{\text{b}}\), quantifies how far the equilibrium of a weak base reaction proceeds. The equilibrium expression for a generic weak base \(\text{B}\) reacting with water can be written as:
\[ \text{B} + \text{H}_2\text{O} \rightleftharpoons \text{BH}^+ + \text{OH}^- \]
The equilibrium constant \(K_{\text{b}}\) is given by:
\[ K_{\text{b}} = \frac{[\text{BH}^+][\text{OH}^-]}{[\text{B}]} \]
This expression provides a ratio that indicates the strength of a weak base: a higher \(K_{\text{b}}\) means a stronger base.

• \(K_{\text{b}}\) is determined experimentally and varies for different bases.
• It allows us to calculate ion concentrations given an initial concentration of the weak base.

Understanding \(K_{\text{b}}\) helps predict the behavior of weak bases in various chemical environments.

Hydroxide Ion Concentration

The concentration of hydroxide ions \([\text{OH}^-]\) is pivotal in determining the basicity of a solution. For weak bases, the ionization in water results in the release of \(\text{OH}^-\) ions, which in turn affects the pOH and pH of the solution.

Hydroxide ion concentration can be calculated when the pOH is known using the expression:
\[ [\text{OH}^-] = 10^{-\text{pOH}} \]
This formula allows us to transition from the logarithmic scale of pOH to a more tangible concentration value.

• It accounts for the ionization of weak bases and contributes to the equilibrium constant \(K_{\text{b}}\).
• A higher \([\text{OH}^-]\) indicates a more basic solution.

By calculating \([\text{OH}^-]\), one can gain critical insights into the equilibrium dynamics and the reactivity of the solution.