A buffer is made up of 2.0 mL of 0.1 M HC2H3O2 and 20.0 mL of 0.1 M N a C2H3O2. The pH of the buffer solution is 4.77. What is the pKą? HINT : Use pH = pKa + log ([B minus]/[HB]) or, if concentrations are the same, pH = pKa + log (Volume of B minus/ Volume of HB)
A buffer is made up of 2.0 mL of 0.1 M HC2H3O2 and 20.0 mL of 0.1 M N a C2H3O2. The pH of the buffer solution is 4.77. What is the pKą? HINT : Use pH = pKa + log ([B minus]/[HB]) or, if concentrations are the same, pH = pKa + log (Volume of B minus/ Volume of HB)
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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
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![**Buffer Solution Preparation**
**Problem Statement:**
A buffer is made up of 2.0 mL of 0.1 M HC₂H₃O₂ and 20.0 mL of 0.1 M NaC₂H₃O₂. The pH of the buffer solution is 4.77. What is the pKₐ?
**Hint:**
Use the Henderson-Hasselbalch equation:
\[ \text{pH} = \text{p}K_a + \log \left(\frac{[\text{B}^-]}{[\text{HB}]}\right) \]
Alternatively, if the concentrations are the same:
\[ \text{pH} = \text{p}K_a + \log \left(\frac{\text{Volume of B}^{-}}{\text{Volume of HB}}\right) \]
**Detailed Explanation:**
- **HC₂H₃O₂** (acetic acid) is the weak acid (HB).
- **NaC₂H₃O₂** (sodium acetate) provides the conjugate base (\[ \text{B}^- \]).
To find the pKₐ of the weak acid (HB), you can rearrange the Henderson-Hasselbalch equation:
\[ \text{p}K_a = \text{pH} - \log \left(\frac{[\text{B}^-]}{[\text{HB}]}\right) \]
Since the molar concentrations of HC₂H₃O₂ and NaC₂H₃O₂ are both 0.1 M and their volumes are given, if you convert to the volume approach:
\[ \text{p}K_a = \text{pH} - \log \left(\frac{\text{Volume of B}^{-}}{\text{Volume of HB}}\right) \]
Using the provided volumes:
- Volume of HB (HC₂H₃O₂) = 2.0 mL
- Volume of B⁻ (NaC₂H₃O₂) = 20.0 mL
So,
\[ \text{p}K_a = 4.77 - \log \left(\frac{20.0}{2.0}\right) \]
Calculate the ratio:
\[ \frac{20.0}{2.0} = 10](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F8c8d8a42-8de4-4466-9c46-ae253053e314%2F4e8f1b50-99c1-46d7-84f3-2eebc71a97dc%2Frxbjroh.png&w=3840&q=75)
Transcribed Image Text:**Buffer Solution Preparation**
**Problem Statement:**
A buffer is made up of 2.0 mL of 0.1 M HC₂H₃O₂ and 20.0 mL of 0.1 M NaC₂H₃O₂. The pH of the buffer solution is 4.77. What is the pKₐ?
**Hint:**
Use the Henderson-Hasselbalch equation:
\[ \text{pH} = \text{p}K_a + \log \left(\frac{[\text{B}^-]}{[\text{HB}]}\right) \]
Alternatively, if the concentrations are the same:
\[ \text{pH} = \text{p}K_a + \log \left(\frac{\text{Volume of B}^{-}}{\text{Volume of HB}}\right) \]
**Detailed Explanation:**
- **HC₂H₃O₂** (acetic acid) is the weak acid (HB).
- **NaC₂H₃O₂** (sodium acetate) provides the conjugate base (\[ \text{B}^- \]).
To find the pKₐ of the weak acid (HB), you can rearrange the Henderson-Hasselbalch equation:
\[ \text{p}K_a = \text{pH} - \log \left(\frac{[\text{B}^-]}{[\text{HB}]}\right) \]
Since the molar concentrations of HC₂H₃O₂ and NaC₂H₃O₂ are both 0.1 M and their volumes are given, if you convert to the volume approach:
\[ \text{p}K_a = \text{pH} - \log \left(\frac{\text{Volume of B}^{-}}{\text{Volume of HB}}\right) \]
Using the provided volumes:
- Volume of HB (HC₂H₃O₂) = 2.0 mL
- Volume of B⁻ (NaC₂H₃O₂) = 20.0 mL
So,
\[ \text{p}K_a = 4.77 - \log \left(\frac{20.0}{2.0}\right) \]
Calculate the ratio:
\[ \frac{20.0}{2.0} = 10
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