1) Calculate the pH of a solution prepared by dissolving 1.90 g of sodium acetate, CH3COONA, in 85.0 mL of 0.20 M acetic acid, CH3COOH(aq). Assume the volume change upon dissolving the sodium acetate is negligible. K, of CH3COOH is 1.75 x 10-5. pH =
1) Calculate the pH of a solution prepared by dissolving 1.90 g of sodium acetate, CH3COONA, in 85.0 mL of 0.20 M acetic acid, CH3COOH(aq). Assume the volume change upon dissolving the sodium acetate is negligible. K, of CH3COOH is 1.75 x 10-5. pH =
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ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
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Chapter1: Chemical Foundations
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![### Problem 1: Solution pH Calculation
**Objective:** Calculate the pH of a solution prepared by dissolving 1.90 g of sodium acetate, \( \text{CH}_3\text{COONa} \), in 85.0 mL of 0.20 M acetic acid, \( \text{CH}_3\text{COOH (aq)} \). Assume the volume change upon dissolving the sodium acetate is negligible. The \( K_a \) of \( \text{CH}_3\text{COOH} \) is \( 1.75 \times 10^{-5} \).
**Given Data:**
- Mass of sodium acetate (\( \text{CH}_3\text{COONa} \)): 1.90 g
- Volume of acetic acid solution: 85.0 mL
- Molarity of acetic acid solution: 0.20 M
- \( K_a \) of acetic acid: \( 1.75 \times 10^{-5} \)
**Steps to Calculate pH:**
1. **Convert mass of sodium acetate to moles:**
- Molar mass of \( \text{CH}_3\text{COONa} \): 82.03 g/mol
\[
\text{Moles of } \text{CH}_3\text{COONa} = \frac{1.90 \text{ g}}{82.03 \text{ g/mol}} = 0.02316 \text{ moles}
\]
2. **Convert volume of acetic acid solution from mL to L:**
\[
\text{Volume (L)} = 85.0 \text{ mL} \times \frac{1 \text{ L}}{1000 \text{ mL}} = 0.085 \text{ L}
\]
3. **Calculate the initial concentration of acetic acid:**
\[
[\text{CH}_3\text{COOH}]_0 = 0.20 \text{ M}
\]
4. **Calculate the concentration of sodium acetate:**
\[
[\text{CH}_3\text{COONa}] = \frac{0.02316 \text{ moles}}{0.085 \text{](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F209d3457-b98a-429a-9f03-d4f3f1cae73c%2F779d2da1-3275-4bef-a3da-aa8caa9ebfaa%2Fpt1ngwk_processed.jpeg&w=3840&q=75)
Transcribed Image Text:### Problem 1: Solution pH Calculation
**Objective:** Calculate the pH of a solution prepared by dissolving 1.90 g of sodium acetate, \( \text{CH}_3\text{COONa} \), in 85.0 mL of 0.20 M acetic acid, \( \text{CH}_3\text{COOH (aq)} \). Assume the volume change upon dissolving the sodium acetate is negligible. The \( K_a \) of \( \text{CH}_3\text{COOH} \) is \( 1.75 \times 10^{-5} \).
**Given Data:**
- Mass of sodium acetate (\( \text{CH}_3\text{COONa} \)): 1.90 g
- Volume of acetic acid solution: 85.0 mL
- Molarity of acetic acid solution: 0.20 M
- \( K_a \) of acetic acid: \( 1.75 \times 10^{-5} \)
**Steps to Calculate pH:**
1. **Convert mass of sodium acetate to moles:**
- Molar mass of \( \text{CH}_3\text{COONa} \): 82.03 g/mol
\[
\text{Moles of } \text{CH}_3\text{COONa} = \frac{1.90 \text{ g}}{82.03 \text{ g/mol}} = 0.02316 \text{ moles}
\]
2. **Convert volume of acetic acid solution from mL to L:**
\[
\text{Volume (L)} = 85.0 \text{ mL} \times \frac{1 \text{ L}}{1000 \text{ mL}} = 0.085 \text{ L}
\]
3. **Calculate the initial concentration of acetic acid:**
\[
[\text{CH}_3\text{COOH}]_0 = 0.20 \text{ M}
\]
4. **Calculate the concentration of sodium acetate:**
\[
[\text{CH}_3\text{COONa}] = \frac{0.02316 \text{ moles}}{0.085 \text{
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