### Buffer Solution CalculationTo 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 MFirst, 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)}

Given data contains, Volume of solution = 100 mL = 0.100 L [CH3COOH] = 0.115 M pH of the solution =…

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 =

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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

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Concept explainers

Acid Base Titration

An acid-base titration is the method of quantitative analysis for determining the concentration of an acid and base by exactly neutralizing it with a standard solution of base or acid having known concentration.

Question

### 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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