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 \(\mathrm{CH}_{3} \mathrm{NH}_{2}(\mathrm{aq})+\mathrm{H}_{2} \mathrm{O}(\ell) \rightleftharpoons \mathrm{CH}_{3} \mathrm{NH}_{3}^{+}(\mathrm{aq})+\mathrm{OH}^{-}(\mathrm{aq})\) Calculate the equilibrium hydroxide ion concentration in a \(0.25 \mathrm{M}\) solution of the base. What are the \(\mathrm{pH}\) and \(\mathrm{pOH}\) of the solution?

Short answer

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Step-by-step solution

Write the Equilibrium Expression

For a weak base like methylamine dissociating in water, the equilibrium expression based on the reaction \( \mathrm{CH}_3\mathrm{NH}_2(\mathrm{aq}) + \mathrm{H}_2\mathrm{O}(\ell) \rightleftharpoons \mathrm{CH}_3\mathrm{NH}_3^+(\mathrm{aq}) + \mathrm{OH}^- (\mathrm{aq}) \) is given by:\[ K_b = \frac{[\mathrm{CH}_3\mathrm{NH}_3^+][\mathrm{OH}^-]}{[\mathrm{CH}_3\mathrm{NH}_2]} \]

Key concepts

Weak Base

Weak bases, unlike strong bases, do not completely dissociate in water. Instead, they establish an equilibrium between the base itself and the ions it forms. This is an important feature that distinguishes weak bases from strong bases. One classic example of a weak base is methylamine (\( \text{CH}_3\text{NH}_2 \)). When methylamine is added to water, it undergoes a reversible reaction, forming methylammonium ions (\( \text{CH}_3\text{NH}_3^+ \)) and hydroxide ions (\( \text{OH}^- \)). This reversible nature of the reaction sets up an equilibrium condition, typical for weak bases.

For weak bases, the extent of ionization is less than 100%, meaning that not all molecules of the base dissociate to form ions. This partial ionization is quantified using the equilibrium constant known as the base dissociation constant (\( K_b \)). The value of \( K_b \) provides insight into the strength of the weak base; a smaller \( K_b \) indicates a weaker base. Methylamine has a \( K_b \) of \( 4.2 \times 10^{-4} \), reflecting its partial ionization in water and confirming its status as a weak base.

When dealing with weak base equilibria, it is essential to write down the balanced chemical equation and recognize the equilibrium that exists between the reactants and the products.

Equilibrium Constant

The equilibrium constant (\( K_b \)) for weak bases provides an essential understanding of how the base behaves in an aqueous solution. For methylamine reacting with water, the equilibrium is described by the reaction:

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

The associated equilibrium expression using the base dissociation constant \( K_b \) is:

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

This expression tells us how the concentrations of the ions and molecules are related at equilibrium. The numerator consists of the concentrations of the products, while the denominator has the concentration of the un-ionized weak base.

In practical terms, knowing \( K_b \) allows us to calculate the concentration of hydroxide ions (\( [\text{OH}^-] \)) when the initial concentration of the base is known. This is vital for further calculations related to the solution's pH and pOH values.

It's important to remember that equilibrium constants are temperature-dependent, meaning the value of \( K_b \) can change with varying temperatures, but for most textbook problems, it is given as a constant value to be used at standard conditions.

pH and pOH Calculation

After determining the concentration of hydroxide ions formed at equilibrium, the next step is to calculate the pH and pOH of the solution. These calculations allow us to understand the acid-base nature of the solution better. For a solution where the \( [\text{OH}^-] = x \), the pOH can be calculated using the formula:

• \( \text{pOH} = -\log_{10}[\text{OH}^-] \)

Calculating the pOH provides insight into the basicity of the solution; a lower pOH indicates a higher concentration of hydroxide ions, hence a more basic solution.

Once the pOH is known, we can easily find the pH using the relation:

• \( \text{pH} + \text{pOH} = 14 \)

This equation derives from the water ion product constant at 25°C, which states that the sum of pH and pOH of a solution must equal 14. Therefore, solving for pH gives us an understanding of the overall hydrogen ion concentration, quantifying the acidity or basicity of the solution.

Understanding how to move from equilibrium concentrations to pH and pOH values is critical for mastering acid-base equilibria in chemistry, enabling one to assess a solution's acidic or basic character easily.