A 1.50 L buffer solution is 0.250 M in HF and 0.250 M in NaF. Calculate the pH of the solution after the addition of 0.100 moles of solid NaOH. Assume no volume change upon the addition of base. The K, for HF is 3.5 x 10-4. 3.82 3.22 3.69 4.46 3.09
A 1.50 L buffer solution is 0.250 M in HF and 0.250 M in NaF. Calculate the pH of the solution after the addition of 0.100 moles of solid NaOH. Assume no volume change upon the addition of base. The K, for HF is 3.5 x 10-4. 3.82 3.22 3.69 4.46 3.09
Chemistry
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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:**
A 1.50 L buffer solution is 0.250 M in HF and 0.250 M in NaF. Calculate the pH of the solution after the addition of 0.100 moles of solid NaOH. Assume no volume change upon the addition of base. The \(K_a\) for HF is \(3.5 \times 10^{-4}\).
**Multiple Choice Options:**
- ○ 3.82
- ○ 3.22
- ○ 3.69
- ○ 4.46
- ○ 3.09
**Explanation:**
This problem involves calculating the pH of a buffer solution after the addition of a strong base. The buffer consists of a weak acid (HF) and its conjugate base (NaF). The addition of NaOH introduces OH⁻ ions, which will react with HF, changing the concentrations of both HF and NaF. You can use the Henderson-Hasselbalch equation to find the new pH:
\[ \text{pH} = \text{pKa} + \log \left(\frac{[\text{A}^-]}{[\text{HA}]}\right) \]
Where:
- \([\text{A}^-]\) is the concentration of the conjugate base (NaF).
- \([\text{HA}]\) is the concentration of the weak acid (HF).
- \(\text{pKa} = -\log(K_a)\).](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F1c6de790-3967-4822-b83d-f877aa472903%2F82652766-8cf3-4aa5-b771-b833686aeb3c%2Fsr5wgh_processed.png&w=3840&q=75)
Transcribed Image Text:**Problem Statement:**
A 1.50 L buffer solution is 0.250 M in HF and 0.250 M in NaF. Calculate the pH of the solution after the addition of 0.100 moles of solid NaOH. Assume no volume change upon the addition of base. The \(K_a\) for HF is \(3.5 \times 10^{-4}\).
**Multiple Choice Options:**
- ○ 3.82
- ○ 3.22
- ○ 3.69
- ○ 4.46
- ○ 3.09
**Explanation:**
This problem involves calculating the pH of a buffer solution after the addition of a strong base. The buffer consists of a weak acid (HF) and its conjugate base (NaF). The addition of NaOH introduces OH⁻ ions, which will react with HF, changing the concentrations of both HF and NaF. You can use the Henderson-Hasselbalch equation to find the new pH:
\[ \text{pH} = \text{pKa} + \log \left(\frac{[\text{A}^-]}{[\text{HA}]}\right) \]
Where:
- \([\text{A}^-]\) is the concentration of the conjugate base (NaF).
- \([\text{HA}]\) is the concentration of the weak acid (HF).
- \(\text{pKa} = -\log(K_a)\).
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