1. What is the pH of a buffer prepared by mixing 20.0 g of acetic acid (HC₂H302) and 20.0 g of sodium acetate (NaC₂H302) in enough water to form 2.00L of solution. The Ka of acetic acid is 1.82 x 10-5.
1. What is the pH of a buffer prepared by mixing 20.0 g of acetic acid (HC₂H302) and 20.0 g of sodium acetate (NaC₂H302) in enough water to form 2.00L of solution. The Ka of acetic acid is 1.82 x 10-5.
Chemistry
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Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
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![**Buffer pH Calculation**
**Question:**
1. What is the pH of a buffer prepared by mixing 20.0 g of acetic acid (HC₂H₃O₂) and 20.0 g of sodium acetate (NaC₂H₃O₂) in enough water to form 2.00 L of solution? The Kₐ of acetic acid is 1.82 × 10⁻⁵.
**Description:**
This problem involves calculating the pH of a buffer solution. A buffer solution consists of a weak acid, acetic acid (HC₂H₃O₂), and its conjugate base, sodium acetate (NaC₂H₃O₂). To solve this, you need to use the Henderson-Hasselbalch equation, which relates the pH of a buffer to the concentration of acid and its conjugate base:
\[ \text{pH} = \text{pK}_a + \log \left( \frac{[\text{A}^-]}{[\text{HA}]} \right) \]
Where:
- \([\text{A}^-]\) is the concentration of the conjugate base (sodium acetate).
- \([\text{HA}]\) is the concentration of the weak acid (acetic acid).
- \(\text{pK}_a = -\log(\text{K}_a)\).
To find the concentrations:
1. Convert grams of acetic acid and sodium acetate to moles.
2. Divide the moles by the volume of the solution in liters (2.00 L) to find molarity.
Calculate pH using the Henderson-Hasselbalch equation after determining the concentrations and \(\text{pK}_a\).](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F5b8356a3-8ffd-4b0d-99e3-2fe5523dae0d%2F381ba5d2-207e-4552-b9dd-9415c16604f9%2Fzhuvojd_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Buffer pH Calculation**
**Question:**
1. What is the pH of a buffer prepared by mixing 20.0 g of acetic acid (HC₂H₃O₂) and 20.0 g of sodium acetate (NaC₂H₃O₂) in enough water to form 2.00 L of solution? The Kₐ of acetic acid is 1.82 × 10⁻⁵.
**Description:**
This problem involves calculating the pH of a buffer solution. A buffer solution consists of a weak acid, acetic acid (HC₂H₃O₂), and its conjugate base, sodium acetate (NaC₂H₃O₂). To solve this, you need to use the Henderson-Hasselbalch equation, which relates the pH of a buffer to the concentration of acid and its conjugate base:
\[ \text{pH} = \text{pK}_a + \log \left( \frac{[\text{A}^-]}{[\text{HA}]} \right) \]
Where:
- \([\text{A}^-]\) is the concentration of the conjugate base (sodium acetate).
- \([\text{HA}]\) is the concentration of the weak acid (acetic acid).
- \(\text{pK}_a = -\log(\text{K}_a)\).
To find the concentrations:
1. Convert grams of acetic acid and sodium acetate to moles.
2. Divide the moles by the volume of the solution in liters (2.00 L) to find molarity.
Calculate pH using the Henderson-Hasselbalch equation after determining the concentrations and \(\text{pK}_a\).
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