Problem 53 - Chemistry and Chemical Reactivity

Question

The weak base methylamine, \(\mathrm{CH}_{3} \mathrm{NH}_{2},\) has \(K_{\mathrm{b}}=\) \(4.2 \times 10^{-4} .\) It reacts with water according to the equation \(\mathrm{CH}_{3} \mathrm{NH}_{2}(\mathrm{aq})+\mathrm{H}_{2} \mathrm{O}(\ell) \rightleftarrows \mathrm{CH}_{3} \mathrm{NH}_{3}^{+}(\mathrm{aq})+\mathrm{OH}^{-}(\mathrm{aq})\) Calculate the equilibrium hydroxide ion concentration in a \(0.25 \mathrm{M}\) solution of the base. What are the \(\mathrm{pH}\) and pOH of the solution?

Step-by-Step Solution

Answer

[OH^-] = 0.01025 M, pH = 11.01, pOH = 2.99

Step 1: Write the Base Ionization Equation

The given base, \(\mathrm{CH}_{3} \mathrm{NH}_{2},\) ionizes in water as follows:\[\mathrm{CH}_{3} \mathrm{NH}_{2}(\mathrm{aq}) + \mathrm{H}_{2}\mathrm{O}(\ell) \rightleftarrows \mathrm{CH}_{3} \mathrm{NH}_{3}^{+}(\mathrm{aq}) + \mathrm{OH}^{-}(\mathrm{aq})\]

Step 2: Set up the Expression for Kb

The expression for the base dissociation constant, \(K_b\), is:\[K_b = \frac{[\mathrm{CH}_{3} \mathrm{NH}_{3}^{+}][\mathrm{OH}^{-}]}{[\mathrm{CH}_{3} \mathrm{NH}_{2}]}\]Given \(K_b = 4.2 \times 10^{-4}\) and initial concentration \([\mathrm{CH}_{3} \mathrm{NH}_{2}] = 0.25\,M\).

Step 3: Define Initial and Change Concentrations

Initially, \([\mathrm{OH}^-] = 0\) and \([\mathrm{CH}_{3} \mathrm{NH}_{3}^+] = 0\). Let \(x\) be the change in concentration at equilibrium. Thus, at equilibrium:\([\mathrm{OH}^{-}] = x\), \([\mathrm{CH}_{3} \mathrm{NH}_{3}^{+}] = x\), and \([\mathrm{CH}_{3} \mathrm{NH}_{2}] = 0.25 - x\).

Step 4: Insert Equilibrium Concentrations into Kb

Substitute the equilibrium concentrations into the \(K_b\) expression:\[4.2 \times 10^{-4} = \frac{x^2}{0.25 - x}\]Assume \(x\) is much smaller than 0.25, so \(0.25 - x \approx 0.25\).

Step 5: Solve for x

Solving \((4.2 \times 10^{-4} = \frac{x^2}{0.25})\), we find:\[x^2 = 4.2 \times 10^{-4} \times 0.25\]\[x^2 = 1.05 \times 10^{-4}\]\[x = \sqrt{1.05 \times 10^{-4}}\]\[x \approx 0.01025\,M\] (hydroxide ion concentration).

Step 6: Calculate pOH

The pOH is calculated as follows:\[pOH = -\log[\mathrm{OH}^-] = -\log(0.01025)\approx 2.99\]

Step 7: Calculate pH

Use the relation \(pH + pOH = 14\) to find pH:\[pH = 14 - pOH = 14 - 2.99 = 11.01\]

Key Concepts

Weak Bases

Weak bases are chemical compounds that do not completely ionize in solution. Unlike strong bases, which fully dissociate in water, a weak base's ionization is reversible and establishes an equilibrium between the ionized species and the un-ionized base. For example, methylamine (\(\text{CH}_3\text{NH}_2\)) is a weak base that does not completely dissociate in water. When it reacts with water, the reaction is reversible and can be represented as follows:\[\text{CH}_3\text{NH}_2(\text{aq}) + \text{H}_2\text{O}(\ell) \rightleftarrows \text{CH}_3\text{NH}_3^+(\text{aq}) + \text{OH}^-(\text{aq})\]

• The forward reaction produces the methylammonium ion (\(\text{CH}_3\text{NH}_3^+\)) and hydroxide ions (\(\text{OH}^-\)).
• The backward reaction allows these ions to recombine to form methylamine and water.

Since the ionization is not complete, weak bases have a lower concentration of \(\text{OH}^-\) ions compared to strong bases, leading to a less basic or higher pOH of the solution.

Equilibrium Calculations

To determine the concentration of ions at equilibrium in a solution of a weak base, we use equilibrium calculations. These calculations take into account the base dissociation constant (\(K_b\)), which is a measure of the strength of the base in an aqueous solution. For methylamine, \(K_b\) is given as \(4.2 \times 10^{-4}\).

• The initial concentration of methylamine is known, in this case, \(0.25 \, M\).
• This concentration decreases slightly as some of the base ionizes. Let \(x\) be the change in concentration at equilibrium.
• Thus, \([ ext{OH}^-] = x\) and \([ ext{CH}_3 ext{NH}_3^+] = x\).

Assumptions can simplify these calculations, such as considering x to be much smaller than the initial concentration, allowing us to approximate \(0.25 - x \approx 0.25\).Substituting the equilibrium concentrations into the \(K_b\) expression allows us to solve for x, which represents the hydroxide ion concentration at equilibrium.

pH and pOH Calculations

Understanding the pH and pOH of a solution helps in describing its acidity or basicity. The pH scale ranges from 0 to 14, where lower values indicate acidity, and higher values indicate basicity. For a weak base like methylamine, the focus is more on the pOH, which is directly calculated from the hydroxide ion concentration.To find the pOH, use the formula:\[pOH = -\log[ ext{OH}^-]\]For our methylamine solution, with \([ ext{OH}^-] \approx 0.01025 \, M\), we find:

• \(pOH \approx 2.99\)

To find the pH, use the relation between pH and pOH:

• \(pH + pOH = 14\)
• Thus, \(pH = 14 - 2.99 = 11.01\)

Identifying these values helps us understand that a solution of a weak base has a relatively high pH, indicating its basic nature. While the calculations may seem a bit complex, understanding the fundamentals of pH, pOH, and equilibrium principles can significantly enhance comprehension.