Question

Calculate the molar concentration of \(\mathrm{OH}^{-}\) in a \(0.050 \mathrm{M}\) solution of ethylamine \(\left(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{NH}_{2} ; K_{b}=6.4 \times 10^{-4}\right) .\) Calculate the \(\mathrm{pH}\) of this solution.

Short answer

The concentration of \(\mathrm{OH}^{-}\) ions at equilibrium is \(1.8\times 10^{-2}\ \mathrm{M}\), and the \(\mathrm{pH}\) of the solution is \(12.26\).

Step-by-step solution

Write the reaction equation for the process of base ionization.

The ethylamine (\(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{NH}_{2}\)) reacts with water, as it is a weak base: \[ \mathrm{C}_{2} \mathrm{H}_{5} \mathrm{NH}_{2} \left(\mathrm{aq}\right) + \mathrm{H}_{2}\mathrm{O} \left(\mathrm{l}\right) \rightleftarrows\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{NH}_{3}^{+} \left(\mathrm{aq}\right) + \mathrm{OH}^{-} \left(\mathrm{aq}\right). \]

Create the equilibrium expression and solve for \(\mathrm{OH}^{-}\) concentration.

Using given \(K_{b}=6.4 \times 10^{-4}\) and initial concentrations, we can create a table to determine the equilibrium concentrations. Let \(x\) be the concentration of \(\mathrm{OH}^{-}\) at equilibrium. Initial concentrations (M): Ethylamine: \(0.050\ \mathrm{M}\) Ethylammonium ion: \(0\ \mathrm{M}\) Hydroxide ion: \(0\ \mathrm{M}\) Changes (M): Ethylamine: \(-x\) Ethylammonium ion: \(+x\) Hydroxide ion: \(+x\) Equilibrium concentrations (M): Ethylamine: \((0.050-x)\ \mathrm{M}\) Ethylammonium ion: \(x\ \mathrm{M}\) Hydroxide ion: \(x\ \mathrm{M}\) Thus, the equilibrium expression for this reaction is: \[ K_b = \frac{\left[\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{NH}_{3}^{+}\right] \left[\mathrm{OH}^{-}\right]}{\left[\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{NH}_{2}\right]}. \] Substituting the equilibrium concentrations: \[ 6.4 \times 10^{-4} = \frac{x \cdot x}{0.050 - x}. \]

Solve the equilibrium expression for \(x\) to obtain \(\mathrm{OH}^{-}\) concentration.

Assuming \(x << 0.050\), we can approximate the denominator to simply \(0.050\). Solving for x: \[ 6.4 \times 10^{-4} = \frac{x^2}{0.050} \Rightarrow x^2 = 6.4 \times 10^{-4} \times 0.050 \Rightarrow x = \sqrt{6.4\times 10^{-4}\times 0.050} = 1.8\times 10^{-2}. \] Thus, the concentration of \(\mathrm{OH}^{-}\) ions at equilibrium is \(1.8\times 10^{-2}\ \mathrm{M}\).

Calculate \(\mathrm{pOH}\) and \(\mathrm{pH}\) of the solution.

Now that we have the concentration of \(\mathrm{OH}^{-}\) ions, we can calculate the \(\mathrm{pOH}\), and subsequently, the \(\mathrm{pH}\) of the solution. Formula for \(\mathrm{pOH}\) and \(\mathrm{pH}\): \[ \mathrm{pOH}=-\log_{10}\left[\mathrm{OH}^{-}\right], \] \[ \mathrm{pH} = 14 - \mathrm{pOH}. \] So, the \(\mathrm{pOH}\) is: \[ \mathrm{pOH} = -\log_{10}\left(1.8 \times 10^{-2}\right) = 1.74. \] And the \(\mathrm{pH}\) is: \[ \mathrm{pH} = 14 - 1.74 = 12.26. \] Thus, the \(\mathrm{pH}\) of the solution is \(12.26\).

Key concepts

Weak Bases

Weak bases are substances that only partially ionize in water. Unlike strong bases, which fully dissociate, weak bases establish an equilibrium between the separated ions and the original compound. An example is ethylamine (\(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{NH}_{2} \)). Upon dissolving in water, it partially reacts to form ethylammonium ions (\(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{NH}_{3}^{+} \)) and hydroxide ions (\(\mathrm{OH}^{-} \)). This partial ionization is important for understanding how much of the base actually contributes to changing the solution's acidity or basicity.
The degree of ionization for weak bases is described by a constant known as the base ionization constant (\(K_b \)). In the case of ethylamine, \(K_b = 6.4 \times 10^{-4} \), indicating a relatively small tendency to form ions.

• Partial Ionization: Only part of the base reacts with water.
• Weak Base Example: Ethylamine (\(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{NH}_{2} \)).
• Ionization Constant: \(K_b = 6.4 \times 10^{-4} \).

Ionization Equilibrium

Ionization equilibrium refers to the state reached when a weak base, like ethylamine, reaches a balance in its ionization reaction. This occurs when the rates of the forward reaction (forming ions) and reverse reaction (reforming the base) become equal. At this point, the concentrations of all species involved in the reaction remain constant over time.
For ethylamine in water, the equilibrium is:\[\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{NH}_{2} + \mathrm{H}_{2}\mathrm{O} \rightleftarrows \mathrm{C}_{2} \mathrm{H}_{5} \mathrm{NH}_{3}^{+} + \mathrm{OH}^{-}.\]
The equation for the base ionization constant (\(K_b \)) provides a quantitative measure of this equilibrium:\[ K_b = \frac{\left[\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{NH}_{3}^{+}\right] \left[\mathrm{OH}^{-}\right]}{\left[\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{NH}_{2}\right]}.\]
By substituting values into this expression, we can find the concentration of hydroxide ions (\(\mathrm{OH}^{-} \)) at equilibrium based on the assumption that the initial change in ethylamine concentration is small.

• Equilibrium is achieved when the formation and reformation rates match.
• Resulting concentrations don't change over time.
• \(K_b \) expression helps to determine equilibrium concentrations.

pH Calculation

Calculating pH involves determining the acidity or basicity of a solution. For bases, like ethylamine, it typically starts with finding the pOH. This relates directly to the hydroxide ion concentration (\(\mathrm{OH}^{-} \)).

The pOH is calculated using:\[\mathrm{pOH} = -\log_{10}\left[\mathrm{OH}^{-}\right].\]
Once you know the pOH, you can find the pH since \(\mathrm{pH} + \mathrm{pOH} = 14 \) for aqueous solutions:\[\mathrm{pH} = 14 - \mathrm{pOH}.\]
In ethylamine's case, with an \(\mathrm{OH}^{-} \) concentration of \(1.8 \times 10^{-2} \ \mathrm{M} \), the pOH is 1.74. Therefore, the pH is calculated as:\[\mathrm{pH} = 14 - 1.74 = 12.26.\]

This solution is basic, as indicated by a pH greater than 7.

• pOH reflects the concentration of hydroxide ions.
• pH and pOH are complementary; they add up to 14.
• For bases, a high pH indicates strong basicity.