An analytical chemist is titrating 92.3 mL of a 0.6600M solution of ethylamine (C₂H5NH₂) with a 0.2000M solution of HNO3. The pK, of ethylamine is 3.19. Calculate the pH of the base solution after the chemist has added 83.0 mL of the HNO3 solution to it. Note for advanced students: you may assume the final volume equals the initial volume of the solution plus the volume of HNO3 solution added. Round your answer to 2 decimal places. pH = d A M

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**Calculating the pH of a Base Solution: A Practical Example**

An analytical chemist is titrating 92.3 mL of a 0.6600 M solution of ethylamine (C₂H₅NH₂) with a 0.2000 M solution of HNO₃. The \( pK_b \) of ethylamine is 3.19. Calculate the pH of the base solution after the chemist has added 83.0 mL of the HNO₃ solution to it.

### Note for Advanced Students
You may assume the final volume equals the initial volume of the solution plus the volume of HNO₃ solution added.

### Titration Procedure:
1. Initial volume of ethylamine solution: 92.3 mL
2. Concentration of ethylamine solution: 0.6600 M
3. Volume of HNO₃ solution added: 83.0 mL
4. Concentration of HNO₃ solution: 0.2000 M
5. \( pK_b \) of ethylamine: 3.19

### Calculation Steps:
1. Use the provided volumes and concentrations to determine the moles of ethylamine and HNO₃.
2. Apply stoichiometry to find the remaining moles of ethylamine or the moles of the resulting product.
3. Calculate the pH using the appropriate equilibrium expressions and the provided \( pK_b \).

### Final Volume:
Final Volume = Initial Volume of ethylamine solution + Volume of HNO₃ solution added
\[ V_{final} = 92.3 \text{ mL} + 83.0 \text{ mL} \]

### Detailed Solution:
After collecting all necessary data and performing the calculations:
\[ \text{pH} = \boxed{\box
Transcribed Image Text:**Calculating the pH of a Base Solution: A Practical Example** An analytical chemist is titrating 92.3 mL of a 0.6600 M solution of ethylamine (C₂H₅NH₂) with a 0.2000 M solution of HNO₃. The \( pK_b \) of ethylamine is 3.19. Calculate the pH of the base solution after the chemist has added 83.0 mL of the HNO₃ solution to it. ### Note for Advanced Students You may assume the final volume equals the initial volume of the solution plus the volume of HNO₃ solution added. ### Titration Procedure: 1. Initial volume of ethylamine solution: 92.3 mL 2. Concentration of ethylamine solution: 0.6600 M 3. Volume of HNO₃ solution added: 83.0 mL 4. Concentration of HNO₃ solution: 0.2000 M 5. \( pK_b \) of ethylamine: 3.19 ### Calculation Steps: 1. Use the provided volumes and concentrations to determine the moles of ethylamine and HNO₃. 2. Apply stoichiometry to find the remaining moles of ethylamine or the moles of the resulting product. 3. Calculate the pH using the appropriate equilibrium expressions and the provided \( pK_b \). ### Final Volume: Final Volume = Initial Volume of ethylamine solution + Volume of HNO₃ solution added \[ V_{final} = 92.3 \text{ mL} + 83.0 \text{ mL} \] ### Detailed Solution: After collecting all necessary data and performing the calculations: \[ \text{pH} = \boxed{\box
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