Problem 44 - Chemistry and Chemical Reactivity

Question

Methylamine, \(\mathrm{CH}_{3} \mathrm{NH}_{2},\) is a weak base. $$\mathrm{CH}_{3} \mathrm{NH}_{2}(\mathrm{aq})+\mathrm{H}_{2} \mathrm{O}(\ell) \rightleftarrows \mathrm{CH}_{3} \mathrm{NH}_{3}^{+}(\mathrm{aq})+\mathrm{OH}^{-}(\mathrm{aq})$$ If the pH of a 0.065 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: Calculate \\([\text{OH}^- ]\\) from pH

Since the given pH is 11.70, we can find pOH using the equation \( ext{pOH} = 14 - ext{pH} \). Thus, \( \text{pOH} = 14 - 11.70 = 2.30 \). The concentration of hydroxide ions, \( \text{[OH]}^- \), can then be determined using \( \text{[OH]}^- = 10^{-\text{pOH}} \), which gives \( \text{[OH]}^- = 10^{-2.30} \approx 5.01 \times 10^{-3} \; M \).

Step 2: Write the expression for \( K_b \)

For the given equilibrium reaction, \( K_b \) is given by the expression \[ K_b = \frac{[\text{CH}_3\text{NH}_3^+][\text{OH}^-]}{[\text{CH}_3\text{NH}_2]} \] where \([\text{CH}_3\text{NH}_3^+]\) and \([\text{OH}^-]\) are both equal to the concentration of \( \text{OH}^- \) calculated previously, and \([\text{CH}_3\text{NH}_2]\) is the initial concentration minus \([\text{OH}^-]\).

Step 3: Calculate \( \text{CH}_3\text{NH}_2 \) at equilibrium

Since \([\text{OH}^-] = [\text{CH}_3\text{NH}_3^+] = 5.01 \times 10^{-3} \; M\), the concentration of \([\text{CH}_3\text{NH}_2]\) at equilibrium is \(0.065 - 5.01 \times 10^{-3} = 0.05999 \; M\).

Step 4: Solve for \( K_b \)

Substitute the values into the \( K_b \) expression: \[ K_b = \frac{(5.01 \times 10^{-3})(5.01 \times 10^{-3})}{0.05999} \approx 4.18 \times 10^{-4} \].

Step 5: Conclusion

Thus, the equilibrium constant \( K_b \) for the given methylamine solution is \( 4.18 \times 10^{-4} \).

Key Concepts

Equilibrium Constant

The equilibrium constant, denoted as \(K_b\) for bases, is a crucial concept in understanding chemical reactions at equilibrium. It specifically applies to weak bases, like methylamine in the exercise, which do not completely dissociate in water.
The equilibrium constant \(K_b\) provides valuable insights into the position of equilibrium—or balance—between the reactants and products:

• A large \(K_b\) value indicates that the products are favored, meaning the weak base dissociates significantly to produce hydroxide ions.
• A small \(K_b\) suggests that the reactants are favored, meaning less dissociation occurs.

For methylamine, the exercise shows that \(K_b\) equals \(4.18 \times 10^{-4}\). This relatively small value indicates that there isn't complete dissociation, which is typical for a weak base.

Hydroxide Ion Concentration

The concentration of hydroxide ions \([ ext{OH}^-]\) is a key part of understanding the basicity of a solution. In a basic solution like the methylamine scenario, hydroxide ions are produced as the weak base reacts with water.
To find \([ ext{OH}^-]\), we first use the concept of pOH which relates to hydroxide ions in the same way that pH relates to hydrogen ions. We calculate pOH from the pH value using the relationship \( ext{pOH} = 14 - ext{pH} \).

Once we have the pOH, we can convert it to \([ ext{OH}^-]\) concentration using the formula \([ ext{OH}^-] = 10^{- ext{pOH}}\). For example, in the exercise, a pOH of 2.30 gives us \([ ext{OH}^-] \approx 5.01 \times 10^{-3} \, M\). This reveals the concentration of hydroxide ions in the solution.

pH and pOH Relationships

The pH and pOH are related concepts used to understand the acidity and basicity of a solution. Together, they help us quantify the concentration of hydrogen ions (for pH) and hydroxide ions (for pOH) in a solution.

The pH scale ranges from 0 to 14, with values below 7 indicating acidity and values above 7 indicating basicity. The pOH scale operates in the opposite fashion: a low pOH indicates high hydroxide ion concentration, thus a basic solution.

The relationship between pH and pOH is given by the equation:

• \( ext{pH} + ext{pOH} = 14 \)

This fundamental correlation helps us determine one if the other is known. In the exercise, knowing the pH (11.70) allows us to calculate the pOH (2.30), enabling further calculations regarding hydroxide and equilibrium constants.

Chemical Equilibrium Equations

Chemical equilibrium equations describe the state of a reaction at balance, where the rate of the forward reaction equals the rate of the reverse reaction. For weak bases like methylamine, these equations are essential in calculating equilibrium constants and assessing the degree of dissociation in solution.

Based on the chemical equation:

• \( \text{CH}_3\text{NH}_2(\text{aq}) + \text{H}_2\text{O}(\ell) \rightleftharpoons \text{CH}_3\text{NH}_3^+(\text{aq}) + \text{OH}^-(\text{aq}) \)

The equilibrium constant expression for a base, \(K_b\), is derived from the concentrations of products and reactants at equilibrium:

• \[ K_b = \frac{[\text{CH}_3\text{NH}_3^+][\text{OH}^-]}{[\text{CH}_3\text{NH}_2]} \]

This relation allows us to figure out how much of a weak base ionizes in water and the strength of its basic properties. The solution to the exercise determines this equilibrium constant as \(4.18 \times 10^{-4}\), highlighting the basicity level of methylamine in water.