Acetic acid partially dissociates according to the reaction below. HC₂H3O2 + H₂O(1) H3O+ + C₂H30₂ ⇒ K = 1.8 x 10-5 @ 25°C Calculate the equilibrium concentrations of the components based on a 0.21 M acetic acid solution. What is the concentration of H3O+? [H3O+] = [?] x 10⁰ M Exponent (yellow) Coefficient (green) Enter
Acetic acid partially dissociates according to the reaction below. HC₂H3O2 + H₂O(1) H3O+ + C₂H30₂ ⇒ K = 1.8 x 10-5 @ 25°C Calculate the equilibrium concentrations of the components based on a 0.21 M acetic acid solution. What is the concentration of H3O+? [H3O+] = [?] x 10⁰ M Exponent (yellow) Coefficient (green) Enter
Chemistry
10th Edition
ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Chapter1: Chemical Foundations
Section: Chapter Questions
Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...
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![### Acetic Acid Dissociation Equilibrium
#### Introduction:
Acetic acid partially dissociates in aqueous solution according to the following chemical equilibrium reaction:
\[ \text{HC}_2\text{H}_3\text{O}_2 + \text{H}_2\text{O} \rightleftharpoons \text{H}_3\text{O}^+ + \text{C}_2\text{H}_3\text{O}_2^- \]
The equilibrium constant, \( K \), for this reaction is given as \( 1.8 \times 10^{-5} \) at 25°C.
#### Problem Statement:
Calculate the equilibrium concentrations of the components in the solution based on an initial concentration of 0.21 M acetic acid. Specifically, find the concentration of \( \text{H}_3\text{O}^+ \).
#### Calculation:
Using the given dissociation reaction and equilibrium constant, set up the equilibrium expression:
\[ K = \frac{[\text{H}_3\text{O}^+][\text{C}_2\text{H}_3\text{O}_2^-]}{[\text{HC}_2\text{H}_3\text{O}_2]} \]
Given:
- Initial concentration of acetic acid, \([\text{HC}_2\text{H}_3\text{O}_2]_0 = 0.21 \text{ M} \)
- \[ K = 1.8 \times 10^{-5} \]
Assuming \( x \) is the change in concentration:
\[ [\text{HC}_2\text{H}_3\text{O}_2] = 0.21 - x \]
\[ [\text{H}_3\text{O}^+] = x \]
\[ [\text{C}_2\text{H}_3\text{O}_2^-] = x \]
Substitute these into the equilibrium expression:
\[ 1.8 \times 10^{-5} = \frac{x^2}{0.21 - x} \]
Solve for \( x \) to find the concentration of \( [\text{H}_3\text{O}^+] \).
#### Input:
- Coefficient (green): Enter the coefficient value from the calculated \( [\text{](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Ff93c0e8e-c503-4238-bd00-8ed92b9568ca%2Fcf7a37d3-eee4-4003-a9e5-cd150867d29a%2Frja652_processed.jpeg&w=3840&q=75)
Transcribed Image Text:### Acetic Acid Dissociation Equilibrium
#### Introduction:
Acetic acid partially dissociates in aqueous solution according to the following chemical equilibrium reaction:
\[ \text{HC}_2\text{H}_3\text{O}_2 + \text{H}_2\text{O} \rightleftharpoons \text{H}_3\text{O}^+ + \text{C}_2\text{H}_3\text{O}_2^- \]
The equilibrium constant, \( K \), for this reaction is given as \( 1.8 \times 10^{-5} \) at 25°C.
#### Problem Statement:
Calculate the equilibrium concentrations of the components in the solution based on an initial concentration of 0.21 M acetic acid. Specifically, find the concentration of \( \text{H}_3\text{O}^+ \).
#### Calculation:
Using the given dissociation reaction and equilibrium constant, set up the equilibrium expression:
\[ K = \frac{[\text{H}_3\text{O}^+][\text{C}_2\text{H}_3\text{O}_2^-]}{[\text{HC}_2\text{H}_3\text{O}_2]} \]
Given:
- Initial concentration of acetic acid, \([\text{HC}_2\text{H}_3\text{O}_2]_0 = 0.21 \text{ M} \)
- \[ K = 1.8 \times 10^{-5} \]
Assuming \( x \) is the change in concentration:
\[ [\text{HC}_2\text{H}_3\text{O}_2] = 0.21 - x \]
\[ [\text{H}_3\text{O}^+] = x \]
\[ [\text{C}_2\text{H}_3\text{O}_2^-] = x \]
Substitute these into the equilibrium expression:
\[ 1.8 \times 10^{-5} = \frac{x^2}{0.21 - x} \]
Solve for \( x \) to find the concentration of \( [\text{H}_3\text{O}^+] \).
#### Input:
- Coefficient (green): Enter the coefficient value from the calculated \( [\text{
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