**Acetic Acid Dissociation and Equilibrium Concentration Calculation** **Acetic Acid Partially Dissociates:** Acetic acid (HC₂H₃O₂) partially dissociates in water according to the following reaction: \[ \text{HC}_2\text{H}_3\text{O}_2 + \text{H}_2\text{O(l)} \rightleftharpoons \text{H}_3\text{O}^+ + \text{C}_2\text{H}_3\text{O}_2^- \] The equilibrium constant (K) for this reaction at 25°C is: \[ K = 1.8 \times 10^{-5} \] **Calculate the Equilibrium Concentrations:** To determine the equilibrium concentrations of the components based on a 0.21 M acetic acid solution, the concentration of \( \text{H}_3\text{O}^+ \) ions needs to be calculated. **Question:** What is the concentration of \( \text{H}_3\text{O}^+ \)? \[ \left[ \text{H}_3\text{O}^+ \right] = \, [ ? ] \times 10^{[?]} \, \text{M} \] **Explanation of the Calculation:** 1. **Write the Expression for the Equilibrium Constant (K):** \[ K = \frac{[\text{H}_3\text{O}^+][\text{C}_2\text{H}_3\text{O}_2^-]}{[\text{HC}_2\text{H}_3\text{O}_2]} \] 2. **Initial Concentrations:** - Initial concentration of acetic acid, \([\text{HC}_2\text{H}_3\text{O}_2]_0 = 0.21 \, \text{M}\) - Initial concentration of \( \text{H}_3\text{O}^+ = 0 \) - Initial concentration of \( \text{C}_2\text{H}_3\text{O}_2^- = 0 \) 3. **Change in Concentration During Dissociation:** Let the change in the concentration of acetic acid be \( x \). - At equilibrium
**Acetic Acid Dissociation and Equilibrium Concentration Calculation** **Acetic Acid Partially Dissociates:** Acetic acid (HC₂H₃O₂) partially dissociates in water according to the following reaction: \[ \text{HC}_2\text{H}_3\text{O}_2 + \text{H}_2\text{O(l)} \rightleftharpoons \text{H}_3\text{O}^+ + \text{C}_2\text{H}_3\text{O}_2^- \] The equilibrium constant (K) for this reaction at 25°C is: \[ K = 1.8 \times 10^{-5} \] **Calculate the Equilibrium Concentrations:** To determine the equilibrium concentrations of the components based on a 0.21 M acetic acid solution, the concentration of \( \text{H}_3\text{O}^+ \) ions needs to be calculated. **Question:** What is the concentration of \( \text{H}_3\text{O}^+ \)? \[ \left[ \text{H}_3\text{O}^+ \right] = \, [ ? ] \times 10^{[?]} \, \text{M} \] **Explanation of the Calculation:** 1. **Write the Expression for the Equilibrium Constant (K):** \[ K = \frac{[\text{H}_3\text{O}^+][\text{C}_2\text{H}_3\text{O}_2^-]}{[\text{HC}_2\text{H}_3\text{O}_2]} \] 2. **Initial Concentrations:** - Initial concentration of acetic acid, \([\text{HC}_2\text{H}_3\text{O}_2]_0 = 0.21 \, \text{M}\) - Initial concentration of \( \text{H}_3\text{O}^+ = 0 \) - Initial concentration of \( \text{C}_2\text{H}_3\text{O}_2^- = 0 \) 3. **Change in Concentration During Dissociation:** Let the change in the concentration of acetic acid be \( x \). - At equilibrium
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Chapter12: Chemical Equilibrium
Section: Chapter Questions
Problem 12.11PAE
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![**Acetic Acid Dissociation and Equilibrium Concentration Calculation**
**Acetic Acid Partially Dissociates:**
Acetic acid (HC₂H₃O₂) partially dissociates in water according to the following reaction:
\[ \text{HC}_2\text{H}_3\text{O}_2 + \text{H}_2\text{O(l)} \rightleftharpoons \text{H}_3\text{O}^+ + \text{C}_2\text{H}_3\text{O}_2^- \]
The equilibrium constant (K) for this reaction at 25°C is:
\[ K = 1.8 \times 10^{-5} \]
**Calculate the Equilibrium Concentrations:**
To determine the equilibrium concentrations of the components based on a 0.21 M acetic acid solution, the concentration of \( \text{H}_3\text{O}^+ \) ions needs to be calculated.
**Question:**
What is the concentration of \( \text{H}_3\text{O}^+ \)?
\[ \left[ \text{H}_3\text{O}^+ \right] = \, [ ? ] \times 10^{[?]} \, \text{M} \]
**Explanation of the Calculation:**
1. **Write the Expression for the Equilibrium Constant (K):**
\[ K = \frac{[\text{H}_3\text{O}^+][\text{C}_2\text{H}_3\text{O}_2^-]}{[\text{HC}_2\text{H}_3\text{O}_2]} \]
2. **Initial Concentrations:**
- Initial concentration of acetic acid, \([\text{HC}_2\text{H}_3\text{O}_2]_0 = 0.21 \, \text{M}\)
- Initial concentration of \( \text{H}_3\text{O}^+ = 0 \)
- Initial concentration of \( \text{C}_2\text{H}_3\text{O}_2^- = 0 \)
3. **Change in Concentration During Dissociation:**
Let the change in the concentration of acetic acid be \( x \).
- At equilibrium](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F9e34ac62-cb87-4fee-9bdd-4f82be5e7b79%2Fec747e1c-d790-4ec6-8c1d-0b2fea945f29%2Fnps1lfn_processed.png&w=3840&q=75)
Transcribed Image Text:**Acetic Acid Dissociation and Equilibrium Concentration Calculation**
**Acetic Acid Partially Dissociates:**
Acetic acid (HC₂H₃O₂) partially dissociates in water according to the following reaction:
\[ \text{HC}_2\text{H}_3\text{O}_2 + \text{H}_2\text{O(l)} \rightleftharpoons \text{H}_3\text{O}^+ + \text{C}_2\text{H}_3\text{O}_2^- \]
The equilibrium constant (K) for this reaction at 25°C is:
\[ K = 1.8 \times 10^{-5} \]
**Calculate the Equilibrium Concentrations:**
To determine the equilibrium concentrations of the components based on a 0.21 M acetic acid solution, the concentration of \( \text{H}_3\text{O}^+ \) ions needs to be calculated.
**Question:**
What is the concentration of \( \text{H}_3\text{O}^+ \)?
\[ \left[ \text{H}_3\text{O}^+ \right] = \, [ ? ] \times 10^{[?]} \, \text{M} \]
**Explanation of the Calculation:**
1. **Write the Expression for the Equilibrium Constant (K):**
\[ K = \frac{[\text{H}_3\text{O}^+][\text{C}_2\text{H}_3\text{O}_2^-]}{[\text{HC}_2\text{H}_3\text{O}_2]} \]
2. **Initial Concentrations:**
- Initial concentration of acetic acid, \([\text{HC}_2\text{H}_3\text{O}_2]_0 = 0.21 \, \text{M}\)
- Initial concentration of \( \text{H}_3\text{O}^+ = 0 \)
- Initial concentration of \( \text{C}_2\text{H}_3\text{O}_2^- = 0 \)
3. **Change in Concentration During Dissociation:**
Let the change in the concentration of acetic acid be \( x \).
- At equilibrium
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