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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Question
![**Question:**
What is the pH of a 1.0 L buffer made with 0.300 mol of HF (Ka = 6.8 × 10⁻⁴) and 0.200 mol of NaF to which 0.080 mol of HCl were added?
**Explanation:**
This question involves calculating the pH of a buffer solution. The buffer is composed of a weak acid (HF) and its conjugate base (NaF). The addition of HCl affects the concentrations of these components.
To calculate the pH after adding HCl:
1. **Determine Initial Concentrations:**
- HF: 0.300 M
- NaF: 0.200 M
- HCl added: 0.080 M
2. **Reaction with added HCl:**
HCl will react with NaF (F⁻), converting some of the F⁻ back to HF:
\[
\text{F}^- + \text{HCl} \rightarrow \text{HF} + \text{Cl}^-
\]
New concentrations:
- F⁻: 0.200 M - 0.080 M = 0.120 M
- HF: 0.300 M + 0.080 M = 0.380 M
3. **Use the Henderson-Hasselbalch equation:**
\[
\text{pH} = \text{pKa} + \log \left( \frac{[\text{A}^-]}{[\text{HA}]} \right)
\]
Where \(\text{pKa} = -\log(\text{Ka})\).
4. **Calculate pKa:**
\[
\text{pKa} = -\log(6.8 \times 10^{-4}) \approx 3.17
\]
5. **Calculate pH:**
\[
\text{pH} = 3.17 + \log \left( \frac{0.120}{0.380} \right) \approx 3.17 + (-0.50) = 2.67
\]
Therefore, the pH of the buffer solution after adding HCl is approximately 2.67.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F055086f6-4d26-404b-b10d-4e78791428d4%2F7d2d1376-5d85-4110-b976-3b0242b50798%2Fl68j834_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Question:**
What is the pH of a 1.0 L buffer made with 0.300 mol of HF (Ka = 6.8 × 10⁻⁴) and 0.200 mol of NaF to which 0.080 mol of HCl were added?
**Explanation:**
This question involves calculating the pH of a buffer solution. The buffer is composed of a weak acid (HF) and its conjugate base (NaF). The addition of HCl affects the concentrations of these components.
To calculate the pH after adding HCl:
1. **Determine Initial Concentrations:**
- HF: 0.300 M
- NaF: 0.200 M
- HCl added: 0.080 M
2. **Reaction with added HCl:**
HCl will react with NaF (F⁻), converting some of the F⁻ back to HF:
\[
\text{F}^- + \text{HCl} \rightarrow \text{HF} + \text{Cl}^-
\]
New concentrations:
- F⁻: 0.200 M - 0.080 M = 0.120 M
- HF: 0.300 M + 0.080 M = 0.380 M
3. **Use the Henderson-Hasselbalch equation:**
\[
\text{pH} = \text{pKa} + \log \left( \frac{[\text{A}^-]}{[\text{HA}]} \right)
\]
Where \(\text{pKa} = -\log(\text{Ka})\).
4. **Calculate pKa:**
\[
\text{pKa} = -\log(6.8 \times 10^{-4}) \approx 3.17
\]
5. **Calculate pH:**
\[
\text{pH} = 3.17 + \log \left( \frac{0.120}{0.380} \right) \approx 3.17 + (-0.50) = 2.67
\]
Therefore, the pH of the buffer solution after adding HCl is approximately 2.67.
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