Introductory Chemistry: A Foundation
9th Edition
ISBN:9781337399425
Author:Steven S. Zumdahl, Donald J. DeCoste
Publisher:Steven S. Zumdahl, Donald J. DeCoste
Chapter16: Acids And Bases
Section: Chapter Questions
Problem 65AP: . The concepts of acid-base equilibria were developed in this chapter for aqueous solutions (in...
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![**Calculating Hydronium Ion Concentration in Benzoic Acid Solution**
To determine the **hydronium ion concentration** \([H_3O^+]\) of an aqueous solution of 0.532 M benzoic acid \((C_6H_5COOH)\), we use the acid dissociation constant \(K_a\).
### Given:
- **Concentration of benzoic acid**: 0.532 M
- **Acid dissociation constant \(K_a\)**: \(6.30 \times 10^{-5}\)
### Formula:
For a weak acid \((HA)\), the dissociation in water is represented by:
\[ HA \rightleftharpoons H^+ + A^- \]
Using the assumption that \([H^+] = [A^-]\):
\[ K_a = \frac{[H^+][A^-]}{[HA] - [H^+]} \approx \frac{[H^+]^2}{[HA]} \]
(since \([H^+]\) is much smaller compared to \([HA]\)),
### Calculation:
1. Use the formula to solve for \([H^+]\):
\[ [H^+] = \sqrt{K_a \times [HA]} \]
### Interactive Elements:
- **Submit Answer**: Allows students to input their calculated concentration.
- **Retry Entire Group**: Option to retry if the calculation is incorrect.
- **Attempts Remaining**: Indicates 9 attempts left for this problem.
This problem helps reinforce understanding of weak acid equilibria and provides practice with calculating ion concentrations using equilibrium constants.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F6bfb8691-d1db-4ddc-9d6d-d840f1ac88cb%2F9553a707-4185-43df-bf61-13fe8d773d25%2Fouti624_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Calculating Hydronium Ion Concentration in Benzoic Acid Solution**
To determine the **hydronium ion concentration** \([H_3O^+]\) of an aqueous solution of 0.532 M benzoic acid \((C_6H_5COOH)\), we use the acid dissociation constant \(K_a\).
### Given:
- **Concentration of benzoic acid**: 0.532 M
- **Acid dissociation constant \(K_a\)**: \(6.30 \times 10^{-5}\)
### Formula:
For a weak acid \((HA)\), the dissociation in water is represented by:
\[ HA \rightleftharpoons H^+ + A^- \]
Using the assumption that \([H^+] = [A^-]\):
\[ K_a = \frac{[H^+][A^-]}{[HA] - [H^+]} \approx \frac{[H^+]^2}{[HA]} \]
(since \([H^+]\) is much smaller compared to \([HA]\)),
### Calculation:
1. Use the formula to solve for \([H^+]\):
\[ [H^+] = \sqrt{K_a \times [HA]} \]
### Interactive Elements:
- **Submit Answer**: Allows students to input their calculated concentration.
- **Retry Entire Group**: Option to retry if the calculation is incorrect.
- **Attempts Remaining**: Indicates 9 attempts left for this problem.
This problem helps reinforce understanding of weak acid equilibria and provides practice with calculating ion concentrations using equilibrium constants.
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