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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![### Determining Equilibrium Direction in Acid-Base Reactions
**Question:** Which direction will the equilibrium lie for the following reaction?
---
**Reaction Representation:**
\( \text{HO(CH}_3\text{CO)} \) + \( \text{NaOH} \rightleftharpoons \) \( \text{CH}_3 \text{CO} \text{ONa} \) + \( \text{H}_2 \text{O} \)
- **Reactants:**
- **Acetic Acid (HO(CH₃CO))** with a pKa = 4.75
- **Sodium Hydroxide (NaOH)**
- **Products:**
- **Sodium Acetate (CH₃COONa)**
- **Water (H₂O)** with a pKa = 15.7
---
**Explanation of pKa Values:**
- **pKa of Acetic Acid = 4.75**
- Lower pKa value indicates a stronger acid.
- **pKa of Water = 15.7**
- Higher pKa value indicates a weaker acid.
**Conclusion:**
In acid-base reactions, the equilibrium tends to favor the direction that produces the weaker acid (higher pKa value). Here, the equilibrium will lie towards the products side, favoring the formation of sodium acetate and water. This is because water (pKa = 15.7) is a much weaker acid compared to acetic acid (pKa = 4.75).
By comparing the pKa values of acetic acid and water, we can determine that the equilibrium will shift towards the right, producing more of the weaker acid, water, and sodium acetate.
### Summary:
For the given reaction:
\[ \text{HO(CH}_3\text{CO)} + \text{NaOH} \rightleftharpoons \text{CH}_3 \text{CO} \text{ONa} + \text{H}_2 \text{O} \]
The equilibrium will lie to the right, favoring the formation of sodium acetate and water.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F9c611bf3-375a-4bcc-90fb-aaa1871cff40%2F93c97476-3ffb-4b3c-b76a-e39347b8f7ba%2F8hpsflc.jpeg&w=3840&q=75)
Transcribed Image Text:### Determining Equilibrium Direction in Acid-Base Reactions
**Question:** Which direction will the equilibrium lie for the following reaction?
---
**Reaction Representation:**
\( \text{HO(CH}_3\text{CO)} \) + \( \text{NaOH} \rightleftharpoons \) \( \text{CH}_3 \text{CO} \text{ONa} \) + \( \text{H}_2 \text{O} \)
- **Reactants:**
- **Acetic Acid (HO(CH₃CO))** with a pKa = 4.75
- **Sodium Hydroxide (NaOH)**
- **Products:**
- **Sodium Acetate (CH₃COONa)**
- **Water (H₂O)** with a pKa = 15.7
---
**Explanation of pKa Values:**
- **pKa of Acetic Acid = 4.75**
- Lower pKa value indicates a stronger acid.
- **pKa of Water = 15.7**
- Higher pKa value indicates a weaker acid.
**Conclusion:**
In acid-base reactions, the equilibrium tends to favor the direction that produces the weaker acid (higher pKa value). Here, the equilibrium will lie towards the products side, favoring the formation of sodium acetate and water. This is because water (pKa = 15.7) is a much weaker acid compared to acetic acid (pKa = 4.75).
By comparing the pKa values of acetic acid and water, we can determine that the equilibrium will shift towards the right, producing more of the weaker acid, water, and sodium acetate.
### Summary:
For the given reaction:
\[ \text{HO(CH}_3\text{CO)} + \text{NaOH} \rightleftharpoons \text{CH}_3 \text{CO} \text{ONa} + \text{H}_2 \text{O} \]
The equilibrium will lie to the right, favoring the formation of sodium acetate and water.
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