Chemistry for Today: General, Organic, and Biochemistry
9th Edition
ISBN:9781305960060
Author:Spencer L. Seager, Michael R. Slabaugh, Maren S. Hansen
Publisher:Spencer L. Seager, Michael R. Slabaugh, Maren S. Hansen
Chapter9: Acids, Bases, And Salts
Section: Chapter Questions
Problem 9.109E
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![### Problem Statement: Percent Ionization of Benzoic Acid
**Question:**
What is the percent ionization of a 0.100 M solution of benzoic acid?
**Given:**
- Benzoic Acid (C₆H₅COOH)
- Acid dissociation constant (Kₐ) = 6.46 × 10⁻⁵
**Solution:**
First, write the ionization reaction of benzoic acid:
\[ \text{C}_6\text{H}_5\text{COOH} \rightleftharpoons \text{C}_6\text{H}_5\text{COO}^- + \text{H}^+ \]
Using the Initial Concentration \( C = 0.100 \) M and the degree of ionization \( x \):
\[ K_a = \frac{[\text{C}_6\text{H}_5\text{COO}^-][\text{H}^+]}{[\text{C}_6\text{H}_5\text{COOH}]} \]
At equilibrium:
- \[ [\text{C}_6\text{H}_5\text{COOH}] = 0.100 - x \]
- \[ [\text{C}_6\text{H}_5\text{COO}^-] = x \]
- \[ [\text{H}^+] = x \]
Thus,
\[ 6.46 × 10⁻⁵ = \frac{x \cdot x}{0.100 - x} \]
If \( x \ll 0.100 \):
\[ 6.46 × 10⁻⁵ \approx \frac{x^2}{0.100} \]
Solving for \( x \):
\[ x^2 = 6.46 × 10⁻⁶ \]
\[ x = \sqrt{6.46 × 10⁻⁶} \]
\[ x \approx 0.00254 \]
The percent ionization is given by:
\[ \text{Percent Ionization} = \left( \frac{x}{0.100} \right) \times 100 \]
\[ \text{Percent Ionization} = \left( \frac{0.00254}{0.100} \right) \times](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F662b2d89-fcf4-479b-9295-1543bded1973%2F4f5a0db1-9cb2-479d-a47c-25c2207d1b06%2Fzc5r3uy_processed.jpeg&w=3840&q=75)
Transcribed Image Text:### Problem Statement: Percent Ionization of Benzoic Acid
**Question:**
What is the percent ionization of a 0.100 M solution of benzoic acid?
**Given:**
- Benzoic Acid (C₆H₅COOH)
- Acid dissociation constant (Kₐ) = 6.46 × 10⁻⁵
**Solution:**
First, write the ionization reaction of benzoic acid:
\[ \text{C}_6\text{H}_5\text{COOH} \rightleftharpoons \text{C}_6\text{H}_5\text{COO}^- + \text{H}^+ \]
Using the Initial Concentration \( C = 0.100 \) M and the degree of ionization \( x \):
\[ K_a = \frac{[\text{C}_6\text{H}_5\text{COO}^-][\text{H}^+]}{[\text{C}_6\text{H}_5\text{COOH}]} \]
At equilibrium:
- \[ [\text{C}_6\text{H}_5\text{COOH}] = 0.100 - x \]
- \[ [\text{C}_6\text{H}_5\text{COO}^-] = x \]
- \[ [\text{H}^+] = x \]
Thus,
\[ 6.46 × 10⁻⁵ = \frac{x \cdot x}{0.100 - x} \]
If \( x \ll 0.100 \):
\[ 6.46 × 10⁻⁵ \approx \frac{x^2}{0.100} \]
Solving for \( x \):
\[ x^2 = 6.46 × 10⁻⁶ \]
\[ x = \sqrt{6.46 × 10⁻⁶} \]
\[ x \approx 0.00254 \]
The percent ionization is given by:
\[ \text{Percent Ionization} = \left( \frac{x}{0.100} \right) \times 100 \]
\[ \text{Percent Ionization} = \left( \frac{0.00254}{0.100} \right) \times
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