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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![**Problem Statement:**
Calculate the pH of a 0.0778 M aqueous solution of formic acid (HCOOH, \( K_a = 1.8 \times 10^{-4} \)).
**Solution:**
To calculate the pH of a weak acid like formic acid, use the following steps:
1. **Set up the expression for \( K_a \):**
\[
K_a = \frac{[H^+][A^-]}{[HA]}
\]
For formic acid:
\[
K_a = \frac{[H^+][HCOO^-]}{[HCOOH]}
\]
2. **Assume initial concentration of \( H^+ \) and \( HCOO^- \) is 0, and let \( x \) be the concentration of ions formed:**
\[
[H^+] = [HCOO^-] = x, \quad [HCOOH] = 0.0778 - x
\]
3. **Substitute into the \( K_a \) expression:**
\[
1.8 \times 10^{-4} = \frac{x^2}{0.0778 - x}
\]
If \( x \) is small compared to the original concentration, \( 0.0778 - x \approx 0.0778 \).
4. **Simplify and solve for \( x \):**
\[
x^2 = (1.8 \times 10^{-4}) \times 0.0778
\]
\[
x^2 = 1.4004 \times 10^{-5}
\]
\[
x = \sqrt{1.4004 \times 10^{-5}}
\]
\[
x = 0.003743
\]
5. **Calculate pH:**
\[
\text{pH} = -\log[H^+]
\]
\[
\text{pH} = -\log(0.003743)
\]
\[
\text{pH} \approx 2.43
\]
Therefore, the pH of the solution is approximately 2.43.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F00585539-79e1-42a7-8a75-95bc6de8d9d5%2F589d9fc6-92ab-4681-8d57-6b129050702d%2F13gsn9g_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Problem Statement:**
Calculate the pH of a 0.0778 M aqueous solution of formic acid (HCOOH, \( K_a = 1.8 \times 10^{-4} \)).
**Solution:**
To calculate the pH of a weak acid like formic acid, use the following steps:
1. **Set up the expression for \( K_a \):**
\[
K_a = \frac{[H^+][A^-]}{[HA]}
\]
For formic acid:
\[
K_a = \frac{[H^+][HCOO^-]}{[HCOOH]}
\]
2. **Assume initial concentration of \( H^+ \) and \( HCOO^- \) is 0, and let \( x \) be the concentration of ions formed:**
\[
[H^+] = [HCOO^-] = x, \quad [HCOOH] = 0.0778 - x
\]
3. **Substitute into the \( K_a \) expression:**
\[
1.8 \times 10^{-4} = \frac{x^2}{0.0778 - x}
\]
If \( x \) is small compared to the original concentration, \( 0.0778 - x \approx 0.0778 \).
4. **Simplify and solve for \( x \):**
\[
x^2 = (1.8 \times 10^{-4}) \times 0.0778
\]
\[
x^2 = 1.4004 \times 10^{-5}
\]
\[
x = \sqrt{1.4004 \times 10^{-5}}
\]
\[
x = 0.003743
\]
5. **Calculate pH:**
\[
\text{pH} = -\log[H^+]
\]
\[
\text{pH} = -\log(0.003743)
\]
\[
\text{pH} \approx 2.43
\]
Therefore, the pH of the solution is approximately 2.43.
Expert Solution
Step 1
The value of Ka for the formic acid is 1.8 × 10-4
Step by step
Solved in 3 steps with 2 images
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