A solution is prepared at 25 °C that is initially 0.077M in trimethylamine ((CH³)¸N), a weak base with K₁=7.4×10¯ª, and 0.50M in trimethylammonium chloride ((CH3)NHCI Calculate the pH of the solution. Round your answer to 2 decimal places.
A solution is prepared at 25 °C that is initially 0.077M in trimethylamine ((CH³)¸N), a weak base with K₁=7.4×10¯ª, and 0.50M in trimethylammonium chloride ((CH3)NHCI Calculate the pH of the solution. Round your answer to 2 decimal places.
Chemistry
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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
Chapter1: Chemical Foundations
Section: Chapter Questions
Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...
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### Acids and Bases: Calculating the pH of a Buffer
A solution is prepared at 25°C that is initially 0.077 M in trimethylamine \(( (CH_3)_3N )\), a weak base with \( K_b = 7.4 \times 10^{-4} \), and 0.50 M in trimethylammonium chloride \(( (CH_3)_3NHCl )\). Calculate the pH of the solution. Round your answer to 2 decimal places.
#### Calculation:
To solve for the pH of this buffer solution, you can use the Henderson-Hasselbalch equation for a basic buffer:
\[ \text{pH} = 14 - \left( pK_b + \log \frac{[ \text{Base} ]}{[ \text{Acid} ]} \right) \]
Here:
- Base = trimethylamine, \((CH_3)_3N\)
- Acid = trimethylammonium chloride, \((CH_3)_3NHCl\)
- \( K_b = 7.4 \times 10^{-4} \)
First, calculate \(pK_b\):
\[ pK_b = -\log K_b \]
\[ pK_b = -\log (7.4 \times 10^{-4}) \]
\[ pK_b = 3.13 \]
Next, apply the concentrations into the Henderson-Hasselbalch equation:
\[ \text{pOH} = pK_b + \log \frac{[ \text{Base} ]}{[ \text{Acid} ]} \]
\[ \text{pOH} = 3.13 + \log \left( \frac{0.077}{0.50} \right) \]
\[ \text{pOH} = 3.13 + \log (0.154) \]
\[ \text{pOH} = 3.13 + (-0.812) \]
\[ \text{pOH} = 2.318 \]
Finally, convert pOH to pH:
\[ \text{pH} = 14 - \text{pOH} \]
\[ \text{pH} = 14 - 2.318 \]
\[ \text{pH} = 11.68 \]
Hence,](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fb9aefb8e-0e61-429a-9e9e-42e0be9cb154%2Fc0cbcbac-38c6-4b1f-ab6d-b4933ca6b312%2Fhxu44e8_processed.jpeg&w=3840&q=75)
Transcribed Image Text:---
### Acids and Bases: Calculating the pH of a Buffer
A solution is prepared at 25°C that is initially 0.077 M in trimethylamine \(( (CH_3)_3N )\), a weak base with \( K_b = 7.4 \times 10^{-4} \), and 0.50 M in trimethylammonium chloride \(( (CH_3)_3NHCl )\). Calculate the pH of the solution. Round your answer to 2 decimal places.
#### Calculation:
To solve for the pH of this buffer solution, you can use the Henderson-Hasselbalch equation for a basic buffer:
\[ \text{pH} = 14 - \left( pK_b + \log \frac{[ \text{Base} ]}{[ \text{Acid} ]} \right) \]
Here:
- Base = trimethylamine, \((CH_3)_3N\)
- Acid = trimethylammonium chloride, \((CH_3)_3NHCl\)
- \( K_b = 7.4 \times 10^{-4} \)
First, calculate \(pK_b\):
\[ pK_b = -\log K_b \]
\[ pK_b = -\log (7.4 \times 10^{-4}) \]
\[ pK_b = 3.13 \]
Next, apply the concentrations into the Henderson-Hasselbalch equation:
\[ \text{pOH} = pK_b + \log \frac{[ \text{Base} ]}{[ \text{Acid} ]} \]
\[ \text{pOH} = 3.13 + \log \left( \frac{0.077}{0.50} \right) \]
\[ \text{pOH} = 3.13 + \log (0.154) \]
\[ \text{pOH} = 3.13 + (-0.812) \]
\[ \text{pOH} = 2.318 \]
Finally, convert pOH to pH:
\[ \text{pH} = 14 - \text{pOH} \]
\[ \text{pH} = 14 - 2.318 \]
\[ \text{pH} = 11.68 \]
Hence,
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