Question

A \(0.035 M\) solution of a weak acid (HA) has a pH of 4.88 . What is the \(K_{a}\) of the acid?

Short answer

The \( K_a \) of the acid is \ 4.99 \times 10^{-10} \.

Step-by-step solution

Determine the concentration of hydrogen ions \([H^+]\)

Use the pH value to find the concentration of hydrogen ions \([H^+]\). \([H^+] = 10^{-pH}\). For the given problem, the pH is 4.88, so: \[ [H^+] = 10^{-4.88} \]

Calculate the exact hydrogen ion concentration

Compute \[ 10^{-4.88} \] to find \([H^+]\): \[ [H^+] \approx 1.32 \times 10^{-5} \ M \]

Set up the expression for the dissociation of HA

Given the dissociation reaction \(\text{HA} \rightleftharpoons \text{H}^+ + \text{A}^-\), the initial concentration of HA is 0.035 M and it partially dissociates. So at equilibrium: \[ [H^+] = [A^-] = 1.32 \times 10^{-5} \ M \]

Calculate the concentration of undissociated HA at equilibrium

At equilibrium, the concentration of undissociated HA is: \[ [HA]_{eq} = 0.035 M - 1.32 \times 10^{-5} M \approx 0.035 M \]

Write and solve the expression for the acid dissociation constant \(K_a\)

The expression for \(K_a\) is: \[ K_a = \frac{[H^+][A^-]}{[HA]_{eq}} \] Substitute the known values into the expression: \[ K_a = \frac{(1.32 \times 10^{-5})(1.32 \times 10^{-5})}{0.035} \approx 4.99 \times 10^{-10} \]

Key concepts

Weak Acid Dissociation

In this exercise, we're dealing with a weak acid (HA), which does not fully dissociate in water. Unlike strong acids, weak acids only partially ionize. This partial ionization is represented by an equilibrium expression: \( \text{HA} \rightleftharpoons \text{H}^+ + \text{A}^- \).
The double arrow indicates that the reaction can proceed in both directions, establishing a dynamic equilibrium between the undissociated acid and the resulting ions.
Weak acids have specific dissociation constants (\( K_a \)), which quantify the extent of their ionization in water.
Understanding weak acid dissociation is crucial for calculating the equilibrium concentrations of the involved species and the acid dissociation constant.

pH Calculation

The pH is a measure of how acidic or basic a solution is, calculated using the concentration of hydrogen ions (\( [H^+] \)). For this problem, the given pH value is 4.88.
The pH of a solution is determined by the formula:

• \( \text{pH} = -\text{log}[H^+] \)

To find the concentration of hydrogen ions (\( [H^+] \)), we rearrange this formula:

• \( [H^+] = 10^{-\text{pH}} \)

Substituting the given pH of 4.88 into the equation, we get:

• \( [H^+] = 10^{-4.88} \)
• \( [H^+] \rightarrow \text{approximately } 1.32 \times 10^{-5} \text{ M} \)

This step allows us to find the hydrogen ion concentration necessary for further calculations.

Equilibrium Concentration

At equilibrium, all species (HA, H\textsuperscript{+}, and A\textsuperscript{-}) are present in specific concentrations. Given the reaction: \( \text{HA} \rightleftharpoons \text{H}^+ + \text{A}^- \),
we start with the initial concentration of the weak acid, HA, which is 0.035 M. At equilibrium, the amount dissociated into \(\text{H}^+ \text{ and } \text{A}^-\) will be equal due to the 1:1 ratio in the balanced equation.
Therefore, we can write:

• \( [\text{H}^+] = [\text{A}^-] = 1.32 \times 10^{-5} \text{ M} \)

The equilibrium concentration of undissociated HA is calculated as:

• \( [HA]_{eq} = \text{initial concentration} - [\text{H}^+] \)

Substituting the values:

• \( [HA]_{eq} = 0.035 \text{ M} - 1.32 \times 10^{-5} \text{ M} \approx 0.035 \text{ M} \)

Since the amount dissociated is very small relative to the initial concentration, the change is often negligible for weak acids.

Hydrogen Ion Concentration

The hydrogen ion concentration \( [H^+] \) is fundamental in understanding the acidity of a solution. This value directly influences the pH and is pivotal in determining the equilibrium state of a weak acid.
In our problem, we calculated \([H^+]\) using the pH value: \( [H^+] = 10^{-4.88} \approx 1.32 \times 10^{-5} \text{ M} \).
This concentration indicates the amount of hydrogen ions present in the solution at equilibrium. It's essential for setting up the equilibrium expression and calculating the dissociation constant (\( K_a \)).
The accuracy of \([H^+]\) calculation is crucial as it impacts the subsequent steps in determining the \( K_a \) and understanding the weak acid's behavior in solution.

Chemical Equilibrium

Chemical equilibrium refers to the state where the concentrations of all reactants and products remain constant over time. For weak acids, this involves the balance between dissociated and undissociated species.
The equilibrium in our weak acid problem is set up by the reaction: \( \text{HA} \rightleftharpoons \text{H}^+ + \text{A}^- \).
At equilibrium, the concentrations of \(\text{H}^+ \text{, } \text{A}^- \text{, and undissociated } \text{HA} \) remain stable. The equilibrium constant for this dissociation is represented by \( K_a \):

• \( K_a = \frac{[\text{H}^+][\text{A}^-]}{[\text{HA}]} \)

Substituting the previously determined concentrations:

• \( K_a = \frac{(1.32 \times 10^{-5})(1.32 \times 10^{-5})}{0.035} \approx 4.99 \times 10^{-10} \)

The \( K_a \) value provides insight into the acid strength and the degree of ionization. Understanding chemical equilibrium is key to predicting the behavior of weak acids in different conditions.