A buffer is made up of 15.0 mL of 0.1 M of HC2H3O2 and 15.0 mL of 0.1 M N a C2H3O2. The solution is diluted with 30.0 mL of water. The pH of the buffer solution is 4.72. What is the pKg? HINT : Use pH = pKa + log ([B minus]/[HB]) or, if concentrations are the same, pH = pką + log (Volume of B minus/ Volume of HB)

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I know how to solve for the pka. But i'm not sure if 30 mL diluted water makes a difference? or Do i solve for pka the same way? 

### Determining the pKa of a Buffer Solution

A buffer is made up of 15.0 mL of 0.1 M of HC₂H₃O₂ and 15.0 mL of 0.1 M NaC₂H₃O₂. The solution is diluted with 30.0 mL of water. The pH of the buffer solution is 4.72. What is the pKa?

#### HINT: 
Use the Henderson-Hasselbalch equation:

\[ \text{pH} = \text{p}K_a + \log \left( \frac{[\text{B}^{\minus}]}{[\text{HB}]} \right) \]

Alternatively, if the concentrations are the same:

\[ \text{pH} = \text{p}K_a + \log \left( \frac{\text{Volume of B}^{\minus}}{\text{Volume of HB}} \right) \]

By using the provided values and the hint, one can calculate the pKa of the buffer solution.
Transcribed Image Text:### Determining the pKa of a Buffer Solution A buffer is made up of 15.0 mL of 0.1 M of HC₂H₃O₂ and 15.0 mL of 0.1 M NaC₂H₃O₂. The solution is diluted with 30.0 mL of water. The pH of the buffer solution is 4.72. What is the pKa? #### HINT: Use the Henderson-Hasselbalch equation: \[ \text{pH} = \text{p}K_a + \log \left( \frac{[\text{B}^{\minus}]}{[\text{HB}]} \right) \] Alternatively, if the concentrations are the same: \[ \text{pH} = \text{p}K_a + \log \left( \frac{\text{Volume of B}^{\minus}}{\text{Volume of HB}} \right) \] By using the provided values and the hint, one can calculate the pKa of the buffer solution.
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