Question

An unknown acid is titrated with base, and the \(\mathrm{pH}\) is 3.64 at the point where exactly half of the acid in the original sample has been neutralized. Calculate the value of the ionization constant of the acid.

Short answer

The ionization constant, \(K_a\), is approximately \(2.29 \times 10^{-4}\).

Step-by-step solution

Write the Henderson-Hasselbalch Equation

The Henderson-Hasselbalch equation is used to relate the pH of a solution to the pKa (acid dissociation constant) and the ratio of the concentration of the deprotonated to protonated form of the acid: \[ \mathrm{pH} = \mathrm{pKa} + \log \left( \frac{[A^-]}{[HA]} \right) \] where \([A^-]\) is the concentration of the deprotonated form and \([HA]\) is the concentration of the protonated form.

Analyze the Titration Conditions

At the halfway point of a titration of a weak acid with a strong base, exactly half of the original acid has been converted to its conjugate base. Therefore, \([A^-] = [HA]\), and the log ratio in the Henderson-Hasselbalch equation becomes zero because:\[ \log \left( \frac{[A^-]}{[HA]} \right) = \log(1) = 0 \]

Simplify the Equation at Halfway Point

At the halfway point, the Henderson-Hasselbalch equation simplifies to:\[ \mathrm{pH} = \mathrm{pKa} \]This is because the log term is zero, so the pH directly equals the pKa when exactly half of the acid has been neutralized.

Calculate the Ionization Constant

Since \(\mathrm{pH} = \mathrm{pKa}\) at the halfway point, the pKa of the acid is 3.64. To find the ionization constant \(K_a\), we use the relation between pKa and \(K_a\):\[ \mathrm{pKa} = -\log(K_a) \] This implies that:\[ \log(K_a) = -3.64 \]

Solve for the Ionization Constant, \(K_a\)

Convert the log equation into its exponential form to calculate \(K_a\):\[ K_a = 10^{-3.64} \] Calculating this gives:\[ K_a \approx 2.29 \times 10^{-4} \]

Key concepts

Acid-Base Titration

An acid-base titration is a laboratory method used to determine the concentration of an acid or base by precisely adding a titrant until the chemical reaction is complete. In the case of a titration involving a weak acid and a strong base, like the one described in the exercise, the goal is to neutralize the acid. Neutralization occurs when the acid and base react in such a way that they cancel each other out, forming water and a salt.

During a titration, there are key points you should understand:

• The Start: Before any base is added, the solution is purely acidic.
• Equivalence Point: The point at which the amounts of acid and base are stoichiometrically equivalent.
• Halfway Point: This is where half of the acid is neutralized, which is crucial in calculating the pKa using the Henderson-Hasselbalch equation.

At the halfway point of the titration, the pH directly equates to the pKa, making it easier to determine the ionization constant of the acid. Understanding these stages is essential for successfully executing an acid-base titration.

Ionization Constant

The ionization constant, often denoted as \( K_a \), is a measure of how completely an acid dissociates into its ions in a solution. It indicates the strength of the acid—the higher the \( K_a \), the stronger the acid and the more it dissociates.

The ionization constant is related to the concept of chemical equilibrium. When an acid is added to water, it will split into hydrogen ions \( (H^+) \) and its conjugate base \( (A^-) \). This can be represented by the equation:
\[ HA ightleftharpoons H^+ + A^- \]
The ionization constant formula is:
\[ K_a = \frac{[H^+][A^-]}{[HA]} \]
In equilibrium conditions, this ratio describes how much of the acid remains in its undissociated form versus its dissociated form, providing insight into the acid's behavior in solution. In the exercise, we converted the pH to \( K_a \) using the relationship between pKa and the ionization constant.

pKa Calculation

Calculating the pKa is central to understanding how acidic a substance is. The pKa represents the pH at which half of the acid is dissociated, meaning there are equal concentrations of the undissociated acid \( (HA) \) and the conjugate base \( (A^-) \).

The Henderson-Hasselbalch equation is the go-to formula for this calculation:
\[ \mathrm{pH} = \mathrm{pKa} + \log \left( \frac{[A^-]}{[HA]} \right) \]
At the point where exactly half of the acid has been neutralized, \([A^-] = [HA]\), making the log term zero. Therefore, \( \mathrm{pH} = \mathrm{pKa} \). In the exercise, the pH value at this point was given as 3.64, directly informing us that the pKa is 3.64.

This direct link between pH and pKa at the halfway point simplifies the calculation of the ionization constant \( K_a \):

• Formula for pKa: \( \mathrm{pKa} = -\log(K_a) \)
• Exponential form: Solving for \( K_a \) involves converting \( \log(K_a) = -3.64 \) to \( K_a = 10^{-3.64} \).

Understanding pKa and how it relates to \( K_a \) helps clarify the strength and behavior of an acid in chemical reactions.