**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.

Given- Concentration of benzoic acid solution = 0.532 M Ka = 6.30 x 10-5 To find- [H3O+]

Answered: The hydronium ion concentration of an…

The hydronium ion concentration of an aqueous solution of 0.532 M benzoic acid, C,H5COOH (K, = 6.30×10) is %3D [H;0"]=| M.

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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Question

**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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