8. What is the pH of a buffer solution made with 0.53M benzoic acid with Ka = 6.5 x 105 and 0.478M sodium benzoate? 9. What is the [H*] for the buffer solution made in #8?
8. What is the pH of a buffer solution made with 0.53M benzoic acid with Ka = 6.5 x 105 and 0.478M sodium benzoate? 9. What is the [H*] for the buffer solution made in #8?
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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![**Buffer Solution pH Calculation**
**Question 8:**
What is the pH of a buffer solution made with 0.53M benzoic acid with \( K_a = 6.5 \times 10^{-5} \) and 0.478M sodium benzoate?
**Question 9:**
What is the \([H^+]\) for the buffer solution made in #8?
In this set of questions, the goal is to determine the pH of a buffer solution and the concentration of hydrogen ions, \([H^+]\), in that solution. For these calculations, the concentrations of the weak acid (benzoic acid) and its conjugate base (sodium benzoate) are given, as well as the acid dissociation constant, \(K_a\), for benzoic acid.
### Steps to Solve:
1. **Apply the Henderson-Hasselbalch Equation** for pH calculation:
\[
\text{pH} = \text{p}K_a + \log\left(\frac{[\text{A}^-]}{[\text{HA}]}\right)
\]
Where:
- \(\text{p}K_a = -\log(K_a)\)
- \([\text{A}^-]\) is the concentration of the conjugate base (sodium benzoate)
- \([\text{HA}]\) is the concentration of the weak acid (benzoic acid)
Given:
- \(K_a = 6.5 \times 10^{-5}\)
- \([\text{HA}] = 0.53M\)
- \([\text{A}^-] = 0.478M\)
Calculate \(\text{p}K_a\):
\[
\text{p}K_a = -\log(6.5 \times 10^{-5}) \approx 4.19
\]
Plug into the Henderson-Hasselbalch Equation:
\[
\text{pH} = 4.19 + \log\left(\frac{0.478}{0.53}\right)
\]
Calculate the ratio:
\[
\frac{0.478}{0.53} \approx 0.902
\]
Find the log of](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fae62e410-6976-4570-b87e-1788a8ca26be%2Ffbbcef5d-d060-43bd-8037-cf5370123da7%2Ftcwq6im_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Buffer Solution pH Calculation**
**Question 8:**
What is the pH of a buffer solution made with 0.53M benzoic acid with \( K_a = 6.5 \times 10^{-5} \) and 0.478M sodium benzoate?
**Question 9:**
What is the \([H^+]\) for the buffer solution made in #8?
In this set of questions, the goal is to determine the pH of a buffer solution and the concentration of hydrogen ions, \([H^+]\), in that solution. For these calculations, the concentrations of the weak acid (benzoic acid) and its conjugate base (sodium benzoate) are given, as well as the acid dissociation constant, \(K_a\), for benzoic acid.
### Steps to Solve:
1. **Apply the Henderson-Hasselbalch Equation** for pH calculation:
\[
\text{pH} = \text{p}K_a + \log\left(\frac{[\text{A}^-]}{[\text{HA}]}\right)
\]
Where:
- \(\text{p}K_a = -\log(K_a)\)
- \([\text{A}^-]\) is the concentration of the conjugate base (sodium benzoate)
- \([\text{HA}]\) is the concentration of the weak acid (benzoic acid)
Given:
- \(K_a = 6.5 \times 10^{-5}\)
- \([\text{HA}] = 0.53M\)
- \([\text{A}^-] = 0.478M\)
Calculate \(\text{p}K_a\):
\[
\text{p}K_a = -\log(6.5 \times 10^{-5}) \approx 4.19
\]
Plug into the Henderson-Hasselbalch Equation:
\[
\text{pH} = 4.19 + \log\left(\frac{0.478}{0.53}\right)
\]
Calculate the ratio:
\[
\frac{0.478}{0.53} \approx 0.902
\]
Find the log of
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