Problem 97 - Chemistry and Chemical Reactivity

Question

The base ethylamine \(\left(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{NH}_{2}\right)\) has a \(K_{\mathrm{b}}\) of \(4.3 \times 10^{-4} .\) A closely related base, ethanolamine \(\left(\mathrm{HOCH}_{2} \mathrm{CH}_{2} \mathrm{NH}_{2}\right),\) has a \(K_{\mathrm{b}}\) of \(3.2 \times 10^{-5}\) (a) Which of the two bases is stronger? (b) Calculate the \(\mathrm{pH}\) of a \(0.10 \mathrm{M}\) solution of the stronger base.

Step-by-Step Solution

Answer

(a) Ethylamine is stronger. (b) pH = 11.82

Step 1: Compare Base Strength

To determine which base is stronger, compare their base dissociation constants, \( K_b \). A higher \( K_b \) value indicates a stronger base because it means a greater degree of ionization in solution.Given:\[ K_{b, \text{ethylamine}} = 4.3 \times 10^{-4} \]\[ K_{b, \text{ethanolamine}} = 3.2 \times 10^{-5} \]Since \( 4.3 \times 10^{-4} > 3.2 \times 10^{-5} \), ethylamine is the stronger base.

Step 2: Set Up for pH Calculation of Ethylamine

Since ethylamine is the stronger base, calculate the \( \text{pH} \) of a \( 0.10 \text{ M} \) solution.First, write the base dissociation equation for ethylamine:\[ \text{CH}_3\text{CH}_2\text{NH}_2 + \text{H}_2\text{O} \rightleftharpoons \text{CH}_3\text{CH}_2\text{NH}_3^+ + \text{OH}^- \] Use the expression for \( K_b \):\[ K_b = \frac{[\text{CH}_3\text{CH}_2\text{NH}_3^+][\text{OH}^-]}{[\text{CH}_3\text{CH}_2\text{NH}_2]} \]

Step 3: Calculate Hydroxide Ion Concentration

Assuming \( x \) is the concentration of \( \text{OH}^- \) ions formed:\[ K_b = \frac{x^2}{0.10 - x} \approx \frac{x^2}{0.10} \] Solving for \( x \) when \( K_b = 4.3 \times 10^{-4} \):\[ x^2 = 4.3 \times 10^{-4} \times 0.10 \]\[ x^2 = 4.3 \times 10^{-5} \] \[ x = \sqrt{4.3 \times 10^{-5}} \approx 6.56 \times 10^{-3} \] Thus, the concentration of \( \text{OH}^- \) is \( 6.56 \times 10^{-3} \text{ M} \).

Step 4: Determine pOH and pH

Calculate \( \text{pOH} \) from \( \text{OH}^- \) concentration:\[ \text{pOH} = -\log[\text{OH}^-] = -\log(6.56 \times 10^{-3}) \approx 2.18 \] Convert \( \text{pOH} \) to \( \text{pH} \):\[ \text{pH} = 14 - \text{pOH} = 14 - 2.18 = 11.82 \] Therefore, the \( \text{pH} \) of the solution is approximately \( 11.82 \).

Key Concepts

Base Dissociation Constant

The Base Dissociation Constant, denoted as \( K_b \), is a crucial concept in acid-base chemistry. It measures the extent to which a base can ionize in a solution. A higher \( K_b \) value indicates a stronger base because it suggests that the base is more likely to dissociate and produce hydroxide ions (\( OH^- \)). This is especially helpful when comparing the strength of bases.
For example, in the given exercise, ethylamine and ethanolamine have \( K_b \) values of \( 4.3 \times 10^{-4} \) and \( 3.2 \times 10^{-5} \), respectively. Since ethylamine has a larger \( K_b \), it's the stronger base, meaning it can produce more hydroxide ions in a solution compared to ethanolamine.

pH Calculation

pH is an important measure in chemistry that tells us how acidic or basic a solution is. It is calculated as the negative logarithm of the hydrogen ion concentration (\( H^+ \)) in a solution. However, for basic solutions, we often find the pOH first, using the hydroxide ion concentration, and then convert it to pH by using the formula:

• \( \text{pH} = 14 - \text{pOH} \)

In the exercise, for ethylamine, we ended up calculating the pOH as \( 2.18 \) (derived from the hydroxide concentration), and subsequently, the pH was found to be \( 11.82 \). This high pH value reflects that the ethylamine solution is basic.

Hydroxide Ion Concentration

The hydroxide ion concentration (\([OH^-]\)) is key in determining the basicity of a solution. It's a direct result of the dissociation of the base in water. For bases, the increase in \([OH^-]\) leads to an increase in pH, indicating a more basic solution.
In this exercise, the \( K_b \) expression for ethylamine is given as:

• \( K_b = \frac{[CH_3CH_2NH_3^+][OH^-]}{[CH_3CH_2NH_2]} \)

Using this, we solve for \([OH^-]\) when ethylamine dissociates in a 0.10 M solution. The concentration \([OH^-]\) was calculated to be approximately \( 6.56 \times 10^{-3} \) M. This informs us about the solution’s baseness and helps in further pH calculations.

Chemical Equilibrium

Chemical equilibrium is the state in which the concentrations of reactants and products remain constant over time, indicating a balance in a chemical reaction. For base dissociation in water, such as with ethylamine, this equilibrium balances between undissociated molecules and the ions produced.
The equation for the dissociation of ethylamine is:

• \( CH_3CH_2NH_2 + H_2O \rightleftharpoons CH_3CH_2NH_3^+ + OH^- \)

This equilibrium dynamics is significant because the stronger the base (ethylamine in this case), the further to the right the equilibrium position will be, meaning more products (ions) are formed, indicating a higher pH and \([OH^-]\). Understanding this concept helps predict how changes in conditions (like concentration changes) can affect ion concentrations and pH.