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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**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
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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