What mass (in grams) of sodium acetate (CH3COONa) must be added to 100.00 mL of a 0.115 M solution of acetic acid in order to prepare a buffer solution with a pH of 4.70? The K, of acetic acid is 1.8 x 105. Mass of sodium acetate =
What mass (in grams) of sodium acetate (CH3COONa) must be added to 100.00 mL of a 0.115 M solution of acetic acid in order to prepare a buffer solution with a pH of 4.70? The K, of acetic acid is 1.8 x 105. Mass of sodium acetate =
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
Chapter15: Acid-base Equilibria
Section: Chapter Questions
Problem 3RQ: One of the most challenging parts of solving acidbase problems is writing out the correct equation....
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![### Buffer Solution Calculation
To prepare a buffer solution, it is essential to calculate the required mass of a solute. Here we are tasked with finding the mass of sodium acetate (CH₃COONa) needed to prepare a buffer solution.
#### Problem:
What mass (in grams) of sodium acetate (CH₃COONa) must be added to 100.00 mL of a 0.115 M solution of acetic acid in order to prepare a buffer solution with a pH of 4.70? The \( \text{K}_\text{a} \) of acetic acid is \( 1.8 \times 10^{-5} \).
**Mass of sodium acetate =** \( \_\_\_\_\_\_\_\_\_ \text{g} \)
---
#### Explanation:
To solve this problem, we can use the Henderson-Hasselbalch equation, which is given by:
\[ \text{pH} = \text{p}K_\text{a} + \log{\left(\frac{[\text{A}^-]}{[\text{HA}]}\right)} \]
Where:
- \( \text{pH} \) is the desired pH of the buffer solution.
- \( \text{p}K_\text{a} \) is the negative logarithm of the acid dissociation constant (Kₐ).
- \( [\text{A}^-] \) is the concentration of the conjugate base (sodium acetate).
- \( [\text{HA}] \) is the concentration of the acid (acetic acid).
Given the values:
- Desired pH = 4.70
- \( \text{K}_\text{a} = 1.8 \times 10^{-5} \)
- Volume of the acetic acid solution = 100.00 mL
- Molarity of the acetic acid solution = 0.115 M
First, calculate the pKₐ:
\[ \text{p}K_\text{a} = -\log(1.8 \times 10^{-5}) \approx 4.74 \]
By substituting into the Henderson-Hasselbalch equation:
\[ 4.70 = 4.74 + \log{\left(\frac{[\text{A}^-]}{0.115}\right)}](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F209d3457-b98a-429a-9f03-d4f3f1cae73c%2Fd796f118-8a3b-4a45-850d-f103c0b85f63%2F8tnh1m4_processed.jpeg&w=3840&q=75)
Transcribed Image Text:### Buffer Solution Calculation
To prepare a buffer solution, it is essential to calculate the required mass of a solute. Here we are tasked with finding the mass of sodium acetate (CH₃COONa) needed to prepare a buffer solution.
#### Problem:
What mass (in grams) of sodium acetate (CH₃COONa) must be added to 100.00 mL of a 0.115 M solution of acetic acid in order to prepare a buffer solution with a pH of 4.70? The \( \text{K}_\text{a} \) of acetic acid is \( 1.8 \times 10^{-5} \).
**Mass of sodium acetate =** \( \_\_\_\_\_\_\_\_\_ \text{g} \)
---
#### Explanation:
To solve this problem, we can use the Henderson-Hasselbalch equation, which is given by:
\[ \text{pH} = \text{p}K_\text{a} + \log{\left(\frac{[\text{A}^-]}{[\text{HA}]}\right)} \]
Where:
- \( \text{pH} \) is the desired pH of the buffer solution.
- \( \text{p}K_\text{a} \) is the negative logarithm of the acid dissociation constant (Kₐ).
- \( [\text{A}^-] \) is the concentration of the conjugate base (sodium acetate).
- \( [\text{HA}] \) is the concentration of the acid (acetic acid).
Given the values:
- Desired pH = 4.70
- \( \text{K}_\text{a} = 1.8 \times 10^{-5} \)
- Volume of the acetic acid solution = 100.00 mL
- Molarity of the acetic acid solution = 0.115 M
First, calculate the pKₐ:
\[ \text{p}K_\text{a} = -\log(1.8 \times 10^{-5}) \approx 4.74 \]
By substituting into the Henderson-Hasselbalch equation:
\[ 4.70 = 4.74 + \log{\left(\frac{[\text{A}^-]}{0.115}\right)}
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