A buffer solution is prepared by dissolving 1.000 g of sodium acetate (CH;COONA) into 100.00 mL of a 0.100 M solution of acetic acid. Then 2.20 mL of a 10.00 M solution of hydrochloric acid is added to the acetic acid/sodium acetate buffer solution. The K, of acetic acid is 1.8 x 105. What is the pH of the solution after adding HCI? pH =|0.9035 %3D
A buffer solution is prepared by dissolving 1.000 g of sodium acetate (CH;COONA) into 100.00 mL of a 0.100 M solution of acetic acid. Then 2.20 mL of a 10.00 M solution of hydrochloric acid is added to the acetic acid/sodium acetate buffer solution. The K, of acetic acid is 1.8 x 105. What is the pH of the solution after adding HCI? pH =|0.9035 %3D
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
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Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...
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![### Buffer Solution Preparation and pH Calculation
A buffer solution is prepared by dissolving 1.000 g of sodium acetate (CH₃COONa) into 100.00 mL of a 0.100 M solution of acetic acid. Then 2.20 mL of a 10.00 M solution of hydrochloric acid is added to the acetic acid/sodium acetate buffer solution. The \( K_a \) of acetic acid is \( 1.8 \times 10^{-5} \).
#### Problem Statement
**What is the pH of the solution after adding HCl?**
#### Calculation Result
**pH = 0.9035**
[Image Description]
The image shows a problem related to the preparation of a buffer solution and the calculation of its pH after the addition of hydrochloric acid. A formula or a calculated value \(0.9035\) is provided as the pH of the resulting solution. There are no graphs or diagrams present in the image.
#### Step-by-step Solution:
1. **Calculate the moles of sodium acetate:**
\[
\text{Molar mass of sodium acetate (CH₃COONa)} = 82.03 \, \text{g/mol}
\]
\[
\text{Moles of sodium acetate} = \frac{1.000 \, \text{g}}{82.03 \, \text{g/mol}} = 0.0122 \, \text{mol}
\]
2. **Moles of acetic acid:**
\[
\text{Molarity of acetic acid solution} = 0.100 \, \text{M}
\]
\[
\text{Volume of acetic acid solution} = 100.00 \, \text{mL} = 0.100 \, \text{L}
\]
\[
\text{Moles of acetic acid} = 0.100 \, \text{M} \times 0.100 \, \text{L} = 0.0100 \, \text{mol}
\]
3. **Calculate the moles of HCl added:**
\[
\text{Molarity of HCl solution} = 10.00 \, \](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F209d3457-b98a-429a-9f03-d4f3f1cae73c%2F18052a58-70d5-4f46-bf47-8a35e823b522%2Fdjfvv1_processed.jpeg&w=3840&q=75)
Transcribed Image Text:### Buffer Solution Preparation and pH Calculation
A buffer solution is prepared by dissolving 1.000 g of sodium acetate (CH₃COONa) into 100.00 mL of a 0.100 M solution of acetic acid. Then 2.20 mL of a 10.00 M solution of hydrochloric acid is added to the acetic acid/sodium acetate buffer solution. The \( K_a \) of acetic acid is \( 1.8 \times 10^{-5} \).
#### Problem Statement
**What is the pH of the solution after adding HCl?**
#### Calculation Result
**pH = 0.9035**
[Image Description]
The image shows a problem related to the preparation of a buffer solution and the calculation of its pH after the addition of hydrochloric acid. A formula or a calculated value \(0.9035\) is provided as the pH of the resulting solution. There are no graphs or diagrams present in the image.
#### Step-by-step Solution:
1. **Calculate the moles of sodium acetate:**
\[
\text{Molar mass of sodium acetate (CH₃COONa)} = 82.03 \, \text{g/mol}
\]
\[
\text{Moles of sodium acetate} = \frac{1.000 \, \text{g}}{82.03 \, \text{g/mol}} = 0.0122 \, \text{mol}
\]
2. **Moles of acetic acid:**
\[
\text{Molarity of acetic acid solution} = 0.100 \, \text{M}
\]
\[
\text{Volume of acetic acid solution} = 100.00 \, \text{mL} = 0.100 \, \text{L}
\]
\[
\text{Moles of acetic acid} = 0.100 \, \text{M} \times 0.100 \, \text{L} = 0.0100 \, \text{mol}
\]
3. **Calculate the moles of HCl added:**
\[
\text{Molarity of HCl solution} = 10.00 \, \
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