A 0.100M solution of bromoacetic acid, BrCH₂COOH, is 13.2% ionized. a. Write the equation for dissociation for this acid. Be sure to include water as one of the reactants. b. Write the Ka expression for bromoacetic acid. c. What are the equilibrium concentrations of the reactants and products? d. What is the value of Ka for this acid? e. What is the pH of bromoacetic acid?
A 0.100M solution of bromoacetic acid, BrCH₂COOH, is 13.2% ionized. a. Write the equation for dissociation for this acid. Be sure to include water as one of the reactants. b. Write the Ka expression for bromoacetic acid. c. What are the equilibrium concentrations of the reactants and products? d. What is the value of Ka for this acid? e. What is the pH of bromoacetic acid?
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![### Bromoaetic Acid Ionization
**A 0.100M solution of bromoacetic acid, BrCH₂COOH, is 13.2% ionized.**
a. **Write the equation for dissociation for this acid. Be sure to include water as one of the reactants.**
**Equation:** BrCH₂COOH + H₂O ⇌ H₃O⁺ + BrCH₂COO⁻
b. **Write the \( K_a \) expression for bromoacetic acid.**
**Expression:** \( K_a = \frac{[H₃O⁺][BrCH₂COO⁻]}{[BrCH₂COOH]} \)
c. **What are the equilibrium concentrations of the reactants and products?**
- Initial concentration of BrCH₂COOH = 0.100 M
- Ionization percentage = 13.2%
**Change in concentration due to ionization:**
- \( [BrCH₂COOH]_{eq} = 0.100 - 0.100 \times 0.132 \)
- \( [H₃O⁺]_{eq} = [BrCH₂COO⁻]_{eq} = 0.100 \times 0.132 \)
d. **What is the value of \( K_a \) for this acid?**
To calculate \( K_a \), use the equilibrium concentrations:
**\( K_a \) calculation:**
- \( K_a = \frac{(0.100 \times 0.132)^2}{0.100 - 0.100 \times 0.132} \)
e. **What is the pH of bromoacetic acid?**
- **pH Calculation:**
- \( \text{pH} = -\log[H₃O⁺] \)
- \( [H₃O⁺] = 0.100 \times 0.132 \)
**Graph/Diagram Explanation:**
There are no graphs or diagrams included in this content. The focus is on the chemical equations and calculations related to the ionization and \( K_a \) value of bromoacetic acid.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F4a31084d-53de-40ac-8333-22b405fb7df5%2F275bc738-e4bb-49d7-9fcd-298d73c3080e%2Fl16x28a_processed.png&w=3840&q=75)
Transcribed Image Text:### Bromoaetic Acid Ionization
**A 0.100M solution of bromoacetic acid, BrCH₂COOH, is 13.2% ionized.**
a. **Write the equation for dissociation for this acid. Be sure to include water as one of the reactants.**
**Equation:** BrCH₂COOH + H₂O ⇌ H₃O⁺ + BrCH₂COO⁻
b. **Write the \( K_a \) expression for bromoacetic acid.**
**Expression:** \( K_a = \frac{[H₃O⁺][BrCH₂COO⁻]}{[BrCH₂COOH]} \)
c. **What are the equilibrium concentrations of the reactants and products?**
- Initial concentration of BrCH₂COOH = 0.100 M
- Ionization percentage = 13.2%
**Change in concentration due to ionization:**
- \( [BrCH₂COOH]_{eq} = 0.100 - 0.100 \times 0.132 \)
- \( [H₃O⁺]_{eq} = [BrCH₂COO⁻]_{eq} = 0.100 \times 0.132 \)
d. **What is the value of \( K_a \) for this acid?**
To calculate \( K_a \), use the equilibrium concentrations:
**\( K_a \) calculation:**
- \( K_a = \frac{(0.100 \times 0.132)^2}{0.100 - 0.100 \times 0.132} \)
e. **What is the pH of bromoacetic acid?**
- **pH Calculation:**
- \( \text{pH} = -\log[H₃O⁺] \)
- \( [H₃O⁺] = 0.100 \times 0.132 \)
**Graph/Diagram Explanation:**
There are no graphs or diagrams included in this content. The focus is on the chemical equations and calculations related to the ionization and \( K_a \) value of bromoacetic acid.
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