Problem 48 - Chemistry and Chemical Reactivity

Question

Methylamine, \(\mathrm{CH}_{3} \mathrm{NH}_{2},\) is a weak base. \(\mathrm{CH}_{3} \mathrm{NH}_{2}(\mathrm{aq})+\mathrm{H}_{2} \mathrm{O}(\ell) \rightleftharpoons \mathrm{CH}_{3} \mathrm{NH}_{3}^{+}(\mathrm{aq})+\mathrm{OH}^{-}(\mathrm{aq})\) If the \(p H\) of a \(0.065 \mathrm{M}\) solution of the amine is \(11.70,\) what is the value of \(K_{\mathrm{b}} ?\)

Step-by-Step Solution

Answer

The value of \(K_b\) is approximately \(4.18 \times 10^{-4}.\)

Step 1: Find the Concentration of OH^-

The given pH of the solution is 11.70. Use the relationship between pH and pOH to find pOH: \[pOH = 14 - pH = 14 - 11.70 = 2.30.\]

Step 2: Calculate [OH^-] from pOH

Use the formula to find [OH^-] from pOH: \[ [OH^-] = 10^{-pOH} = 10^{-2.30} \] Calculating gives \[ [OH^-] = 5.01 \times 10^{-3} \; \text{M}.\]

Step 3: Assume Initial Conditions for Ice Chart

For methylamine's dissociation, the initial concentration of \([\mathrm{CH}_3\mathrm{NH}_2]\) is 0.065 M, and the initial concentrations of \([\mathrm{CH}_3\mathrm{NH}_3^+ ]\) and \([\mathrm{OH}^- ]\) are both 0 M. At equilibrium, the concentrations change by \(-x\) for \([\mathrm{CH}_3\mathrm{NH}_2 ]\), and \(+x\) for both products.

Step 4: Write Equilibrium Concentrations

The equilibrium concentrations are as follows:- \([\mathrm{CH}_3\mathrm{NH}_2 ] = 0.065 - x\)- \([\mathrm{CH}_3\mathrm{NH}_3^+ ] = x\)- \([\mathrm{OH}^- ] = x\)Since \([\mathrm{OH}^-] = 5.01 \times 10^{-3} \; \text{M}\), assume \(x = [\mathrm{OH}^-]\).

Step 5: Write Kb Expression

The expression for the base dissociation constant \(K_b\) is:\[ K_b = \frac{{[\mathrm{CH}_3\mathrm{NH}_3^+][\mathrm{OH}^-]}}{[\mathrm{CH}_3\mathrm{NH}_2]} \] Using the equilibrium concentrations, this becomes:\[ K_b = \frac{(5.01 \times 10^{-3})(5.01 \times 10^{-3})}{0.065 - 5.01 \times 10^{-3}}.\]

Step 6: Solve for Kb

Calculate \(K_b\) using the expression from Step 5:\[ K_b = \frac{(5.01 \times 10^{-3})^2}{0.065 - 5.01 \times 10^{-3}} \]\[ K_b \approx \frac{25.10 \times 10^{-6}}{0.05999} \approx 4.18 \times 10^{-4}.\]

Key Concepts

Methylamine

Methylamine is an organic compound with the formula \( \mathrm{CH}_3\mathrm{NH}_2 \). It is classified as a weak base. Understanding its behavior in water is essential for acid-base equilibrium studies. Methylamine accepts protons, meaning it's a prototypical Bronsted-Lowry base. When it dissolves in water, methylamine partially dissociates to form its conjugate acid, \( \mathrm{CH}_3\mathrm{NH}_3^+ \), and hydroxide ions, \( \mathrm{OH}^- \). This partial dissociation is a key feature of weak bases. In solutions like the one described in the exercise, where the initial concentration is 0.065 M, the behavior of methylamine can be analyzed using equilibrium concepts to determine its dissociation constant, \( K_b \). Understanding methylamine's equilibrium helps in predicting the solution's pH and efficiency in accepting protons during reactions.

pH and pOH

The pH and pOH scales are critical for measuring the acidity and basicity of solutions. They are related by the equation: \[ pH + pOH = 14 \] at 25°C. In the original exercise, the pH is given as 11.70. This high pH value indicates that the solution is basic, as values above 7 are characteristic of bases. From the pH, we can find the pOH by subtracting the pH from 14, resulting in a pOH of 2.30. This calculation highlights the inverse relationship: as pH increases, indicating more basicity, pOH decreases.
Understanding pH and pOH helps predict the behavior of solutions like methylamine. In basic solutions, fewer hydrogen ions are present, meaning more hydroxide ions exist, impacting shifts in chemical equilibrium.

Base Dissociation Constant

The base dissociation constant, \( K_b \), is a measure of a base's strength in water. It provides insight into how well a base, such as methylamine, dissociates into its ions. The expression for \( K_b \) is formulated from the equilibrium concentrations of the ions, represented as:
\[ K_b = \frac{{[\mathrm{CH}_3\mathrm{NH}_3^+][\mathrm{OH}^-]}}{[\mathrm{CH}_3\mathrm{NH}_2]} \]
In this equation, \([\mathrm{CH}_3\mathrm{NH}_3^+]\) and \([\mathrm{OH}^-]\) denote the concentrations of the products, while \([\mathrm{CH}_3\mathrm{NH}_2]\) is the concentration of the undissociated base at equilibrium. By substituting these equilibrium values, derived from the ICE chart, into the expression, we can calculate \( K_b \). The resulting value provides a quantitative measure of methylamine's basicity, helping predict its behavior in various chemical scenarios.

ICE Chart Method

The ICE chart method is a systematic approach for solving chemical equilibrium problems. It stands for Initial, Change, and Equilibrium, accounting for concentration changes as a reaction reaches equilibrium.

• Initial: Start by identifying initial concentrations. In our methylamine example, 0.065 M for methylamine, and 0 M for products.
• Change: Define the changes in concentration as reaction progress. A typical notation is \( -x \) for reactants and \( +x \) for products.
• Equilibrium: Combine initial concentrations and changes to find final equilibrium concentrations.
  Using these values to express the \( K_b \) formula, we solve for the equilibrium constant value.
  ICE charts are a powerful tool enabling clear visualization of shifts in equilibrium, crucial for accurately determining unknown quantities like \( K_b \) from experimental data.