Problem 61 - Chemistry The Molecular Science

Question

Calculate the \(\left[\mathrm{OH}^{-}\right]\) and the \(\mathrm{pH}\) of a \(0.024-\mathrm{M}\) methylamine solution; \(K_{\mathrm{h}}=5.0 \times 10^{-4}\).

Step-by-Step Solution

Answer

[OH^-] is approximately 3.46 x 10^-3 M and the pH is about 11.54.

Step 1: Understanding the Problem

We need to find the hydroxide ion concentration, \( [\text{OH}^-] \), and the pH of a 0.024 M methylamine solution. Given \( K_b \) for methylamine is \( 5.0 \times 10^{-4} \).

Step 2: Writing the Equilibrium Reaction

Methylamine \((CH_3NH_2)\) in water accepts a proton to form \((CH_3NH_3^+)\) and hydroxide ions \((\text{OH}^-)\). The equilibrium equation is: \[ CH_3NH_2 + H_2O \rightleftharpoons CH_3NH_3^+ + OH^- \]

Step 3: Applying the Equilibrium Expression

For the reaction, the base dissociation constant \(K_b\) can be written as:\[ K_b = \frac{[CH_3NH_3^+][OH^-]}{[CH_3NH_2]} \]

Step 4: Setting up the ICE Table

Initial concentrations are: - \( [CH_3NH_2] = 0.024 \text{ M} \) - \( [OH^-] = 0 \) - \( [CH_3NH_3^+] = 0 \)Change concentrations are: - \( [CH_3NH_2] = -x \)- \( [OH^-] = +x \)- \( [CH_3NH_3^+] = +x \) At equilibrium: - \( [CH_3NH_2] = 0.024 - x \)- \( [OH^-] = x \)- \( [CH_3NH_3^+] = x \)

Step 5: Solving for x

Substitute into the \(K_b\) expression:\[ 5.0 \times 10^{-4} = \frac{x^2}{0.024 - x} \]Assume \(x\) is small enough that \(0.024 - x \approx 0.024\):\[ 5.0 \times 10^{-4} = \frac{x^2}{0.024} \]Solve for \(x\):\[ x^2 = 5.0 \times 10^{-4} \times 0.024 \]\[ x = \sqrt{1.2 \times 10^{-5}} \approx 3.46 \times 10^{-3} \text{ M} \]

Step 6: Calculating pH

Since \([OH^-] = x = 3.46 \times 10^{-3} \text{ M}\), use the relationship:\[ pOH = -\log([OH^-]) = -\log(3.46 \times 10^{-3}) \approx 2.46 \]Use \( pH + pOH = 14 \) to find pH:\[ pH = 14 - 2.46 = 11.54 \]

Key Concepts

Understanding Hydroxide Ion Concentration

The hydroxide ion concentration, denoted as \([\text{OH}^-]\), is a fundamental aspect of understanding the basic nature of a solution. In essence, it tells us how many hydroxide ions are present in a given volume of solution. The hydroxide ion is a significant player in determining the overall behavior of bases in solution since bases dissociate to produce this ion. By knowing the concentration of hydroxide ions, we can infer how strongly basic the solution is.

In a typical equilibrium reaction involving a weak base like methylamine (\(\text{CH}_3\text{NH}_2\)), the base accepts protons from water, producing \(\text{OH}^-\) ions. The chemical reaction can be represented as follows:

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

This equation indicates that as methylamine dissolves and reaches equilibrium, it transforms into hydroxide ions and \(\text{CH}_3\text{NH}_3^+\) ions.

To calculate \([\text{OH}^-]\), we need to consider the base dissociation constant (\(K_b\)) and initial concentration of the base. This setup is typically addressed using the ICE (Initial, Change, Equilibrium) table that helps us organize and solve for the varying concentrations throughout the reaction.

The Art of pH Calculation

Calculating the pH of a solution is an integral part of understanding its acidity or basicity. The pH scale runs from 0 to 14, with lower values signifying acidic solutions, higher values indicating basic solutions, and a pH of 7 represent a neutral solution. The key relationship that links the concentration of hydroxide ions to find the pH involves calculating the \(pOH\) first, and then using the simple mathematical relationship between \(pH\) and \(pOH\).

For basic solutions, we often start by calculating the \(pOH\) using the formula:

• \(pOH = -\log([\text{OH}^-])\)

If you've determined that \([\text{OH}^-]\) is approximately \(3.46 \times 10^{-3} \, \text{M}\), you can calculate \(pOH\) as follows:

• \(pOH = -\log(3.46 \times 10^{-3}) \approx 2.46\)

Once we know \(pOH\), we can easily find the \(pH\) because:

• \(pH + pOH = 14\)

This means the pH of the solution can be found by subtracting the \(pOH\) from 14, thus:

• \(pH = 14 - 2.46 = 11.54\)

The higher the pH value, the more basic the solution is. A pH of 11.54 indicates a strongly basic solution.

Decoding the Base Dissociation Constant

The base dissociation constant, represented as \(K_b\), is a key factor in understanding the strength of a base in solution. It gives us insight into how effectively a base can dissociate to accept protons and produce \(\text{OH}^-\) ions. A higher \(K_b\) value means the base will dissociate more completely, making it stronger.

In our scenario with methylamine, we know that the \(K_b\) is given as \(5.0 \times 10^{-4}\). This relatively small value suggests that methylamine is a weak base, as it does not fully dissociate in water. The reaction equilibrium can be expressed using \(K_b\) as follows:

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

This equation enables us to set up calculations related to initial and equilibrium concentrations, helping us find the concentrations of ions formed. Utilizing an ICE table and substituting known values into the \(K_b\) equation allows us to find \([\text{OH}^-]\) by solving for 'x'.

Solving for 'x' in our calculations, we assume simplifying approximations initially, such as considering \(0.024 - x \approx 0.024\) under the assumption that 'x' is much smaller than the initial concentration of methylamine. Through this process, we observe how with a weak base like methylamine, the \(K_b\) effectively governs the equilibrium position of the reaction.