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 pH Calculation Example
#### Problem Statement:
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 0.070 mol of HCl were added?
#### Explanation:
We can calculate the pH of the buffer solution using the Henderson-Hasselbalch equation:
\[ \text{pH} = \text{p}K_a + \log \left( \frac{[\text{A}^-]}{[\text{HA}]} \right) \]
Where:
- \(\text{p}K_a\) is the negative logarithm of the acid dissociation constant (\(K_a\)).
- \([\text{A}^-]\) is the concentration of the conjugate base (F⁻ from NaF).
- \([\text{HA}]\) is the concentration of the weak acid (HF).
**Steps to solve the problem:**
1. **Calculate \(\text{p}K_a\):**
\[ \text{p}K_a = -\log(6.8 \times 10^{-4}) \]
2. **Determine the changes in molarity due to the addition of HCl:**
- Upon adding 0.070 mol of HCl to the buffer, HCl will react completely with the conjugate base NaF.
- \( 0.070 \text{ mol} \) HCl will reduce the amount of \( \text{A}^- (F^-) \) and increase the amount of \(\text{HA (HF)}\).
Initial moles before the reaction:
- HF: 0.300 mol
- NaF: 0.200 mol
Change in moles after addition of HCl:
- HF: 0.300 mol + 0.070 mol = 0.370 mol
- NaF: 0.200 mol - 0.070 mol = 0.130 mol
3. **Apply the Henderson-Hasselbalch equation:**
\[ \text{pH} = \text{p}K_a + \log \left( \frac{0.130}{0.370} \right) \]
#### Diagram Explanation:
The image contains a](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fabf3f71a-8f38-4387-af5c-ba72dda07b09%2F40767cda-4c4a-44d6-ac60-3e29cd18fb2b%2F4cdjxaa_processed.png&w=3840&q=75)
Transcribed Image Text:### Buffer pH Calculation Example
#### Problem Statement:
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 0.070 mol of HCl were added?
#### Explanation:
We can calculate the pH of the buffer solution using the Henderson-Hasselbalch equation:
\[ \text{pH} = \text{p}K_a + \log \left( \frac{[\text{A}^-]}{[\text{HA}]} \right) \]
Where:
- \(\text{p}K_a\) is the negative logarithm of the acid dissociation constant (\(K_a\)).
- \([\text{A}^-]\) is the concentration of the conjugate base (F⁻ from NaF).
- \([\text{HA}]\) is the concentration of the weak acid (HF).
**Steps to solve the problem:**
1. **Calculate \(\text{p}K_a\):**
\[ \text{p}K_a = -\log(6.8 \times 10^{-4}) \]
2. **Determine the changes in molarity due to the addition of HCl:**
- Upon adding 0.070 mol of HCl to the buffer, HCl will react completely with the conjugate base NaF.
- \( 0.070 \text{ mol} \) HCl will reduce the amount of \( \text{A}^- (F^-) \) and increase the amount of \(\text{HA (HF)}\).
Initial moles before the reaction:
- HF: 0.300 mol
- NaF: 0.200 mol
Change in moles after addition of HCl:
- HF: 0.300 mol + 0.070 mol = 0.370 mol
- NaF: 0.200 mol - 0.070 mol = 0.130 mol
3. **Apply the Henderson-Hasselbalch equation:**
\[ \text{pH} = \text{p}K_a + \log \left( \frac{0.130}{0.370} \right) \]
#### Diagram Explanation:
The image contains a
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