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Chapter 16: Problem 52

The ionization constant of a very weak acid, HA, is \(4.0 \times 10^{-9} .\) Calculate the equilibrium concentrations of \(\mathrm{H}_{3} \mathrm{O}^{+}, \mathrm{A}^{-},\) and \(\mathrm{HA}\) in a \(0.040 \mathrm{M}\) solution of the acid.

Short Answer

Expert verified
\([\text{H}_3\text{O}^+] = [\text{A}^-] = 1.26 \times 10^{-5} \text{ M} \), \([\text{HA}] = 0.040 \text{ M}\).

Step by step solution

01

Understanding the Ionization Equation

The ionization of a weak acid, HA, in water is represented by the equation: \( \text{HA} + \text{H}_2\text{O} \rightleftharpoons \text{H}_3\text{O}^+ + \text{A}^- \). The equilibrium expression based on the ionization constant \( K_a \) is \( K_a = \frac{[\text{H}_3\text{O}^+][\text{A}^-]}{[\text{HA}]} \).
02

Setting Up the Initial Concentrations

Initially, before any ionization, \([\text{HA}] = 0.040 \text{ M} \), and \( [\text{H}_3\text{O}^+] = [\text{A}^-] = 0 \text{ M} \).
03

Defining Change in Concentrations

Let the change in concentration of HA due to ionization be \( -x \). Accordingly, \([\text{H}_3\text{O}^+] = x \text{ M}\) and \([\text{A}^-] = x \text{ M}\) because they are produced in equal amounts. Thus, \([\text{HA}] = 0.040 - x \text{ M}\).
04

Substituting into the Equilibrium Expression

Substitute the equilibrium concentrations into the ionization constant expression: \( K_a = \frac{x^2}{0.040 - x} \), where \( K_a = 4.0 \times 10^{-9} \).
05

Approximating for the Very Weak Acid

Since \( K_a \) is very small, the change in \([\text{HA}]\), \( x \), is negligible. Therefore, \( 0.040 - x \approx 0.040 \). Substitute this approximation into the equation: \( 4.0 \times 10^{-9} = \frac{x^2}{0.040} \).
06

Solving for x

Rearrange the approximation to solve for \( x \): \( x^2 = (4.0 \times 10^{-9})(0.040)\). Calculate \( x \) by taking the square root: \( x = \sqrt{1.6 \times 10^{-10}} \approx 1.26 \times 10^{-5} \text{ M}\).
07

Conclusion on Equilibrium Concentrations

At equilibrium, \([\text{H}_3\text{O}^+] = 1.26 \times 10^{-5} \text{ M}\), \([\text{A}^-] = 1.26 \times 10^{-5} \text{ M}\), and \([\text{HA}] = 0.040 \text{ M} - 1.26 \times 10^{-5} \text{ M} \approx 0.040 \text{ M}\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Equilibrium Concentrations
When dealing with the ionization of a weak acid, understanding equilibrium concentrations is crucial. For the acid HA, when it dissolves in water, it partially breaks apart into ions: \( \text{H}_3\text{O}^+ \) and \( \text{A}^- \). At the start (initial state), the concentration of HA is 0.040 M, while both \([\text{H}_3\text{O}^+]\) and \([\text{A}^-]\) are 0 M.

As the reaction reaches equilibrium, HA begins to ionize, causing changes in concentrations:
  • The concentration of HA decreases by \( -x \).
  • Both \([\text{H}_3\text{O}^+]\) and \([\text{A}^-]\) increase by \( x \).
Thus, at equilibrium:
  • \([\text{H}_3\text{O}^+] = x \)
  • \([\text{A}^-] = x \)
  • \([\text{HA}] = 0.040 - x \)
For very weak acids, because their ionization constant is small, \( x \) is also small. This allows us to approximate \([\text{HA}] \approx 0.040 \text{ M}\), simplifying our calculations.
Ionization Constant
The ionization constant, \( K_a \), is a measure of the strength of an acid in solution. It tells us how well the acid can donate protons (H+ ions). For our weak acid HA, the ionization constant is very low: \( 4.0 \times 10^{-9} \).

This tiny value indicates that HA barely ionizes in water, making it a weak acid. The equilibrium expression for this ionization process is:
  • \( K_a = \frac{[\text{H}_3\text{O}^+][\text{A}^-]}{[\text{HA}]} \)
Given the initial conditions and using the approximation for the weak acid:
  • Substitute: \( K_a = \frac{x^2}{0.040} \)
  • Solve for \( x \) to determine the equilibrium concentration.
The magnitude of \( K_a \) provides insight into the extent of acid ionization, with lower values denoting weaker acids.
Acid Dissociation Equilibrium
Acid dissociation equilibrium focuses on the balance between the undissociated acid (HA) and the ions it produces (\( \text{H}_3\text{O}^+ \) and \( \text{A}^- \)). For a very weak acid, equilibrium is reached when the rate of ionization and recombination of ions are equal.

The equation representing this equilibrium is:
  • \( \text{HA} + \text{H}_2\text{O} \rightleftharpoons \text{H}_3\text{O}^+ + \text{A}^- \)
Understanding this equilibrium reveals:
  • The majority of HA remains un-ionized in the solution.
  • Only a small fraction turns into \( \text{H}_3\text{O}^+ \) and \( \text{A}^- \).
  • The equilibrium concentrations reflect how little the equilibrium shifts towards complete ionization for weak acids.
This concept helps in predicting the behavior of acid in different concentrations and conditions, guiding chemists in solution preparation and reactions.

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