Chemistry
10th Edition
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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![### Post-lab Question #2:
Using the \( K_b \) for \( \text{NH}_3 \) (from Appendix G: \( K_b = 1.8 \times 10^{-5} \)), calculate the \( K_a \) for the \( \text{NH}_4^+ \) ion.
#### Answer Choices:
- \( 5.6 \times 10^{-10} \)
- \( 1.8 \times 10^{-10} \)
- \( 5.6 \times 10^{-5} \)
- \( 1.8 \times 10^{-5} \)
To calculate the \( K_a \) of the \( \text{NH}_4^+ \) ion from the provided \( K_b \) of \( \text{NH}_3 \), use the relationship between \( K_a \) and \( K_b \) for a conjugate acid-base pair, which is given by the equation:
\[ K_a \times K_b = K_w \]
where \( K_w \) is the ion-product constant for water (\( K_w = 1.0 \times 10^{-14} \) at 25°C).
Rearrange the equation to solve for \( K_a \):
\[ K_a = \frac{K_w}{K_b} \]
Substitute the given \( K_b \) value:
\[ K_a = \frac{1.0 \times 10^{-14}}{1.8 \times 10^{-5}} \]
Perform the calculation:
\[ K_a = \frac{1.0 \times 10^{-14}}{1.8 \times 10^{-5}} = 5.6 \times 10^{-10} \]
So, the correct answer is:
- \( 5.6 \times 10^{-10} \)](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F6a70e2d2-d641-435a-b212-d4266cff05fc%2Fab115e60-db97-42b6-8281-fc6e1e77d34f%2Fn3qgb3h_processed.png&w=3840&q=75)
Transcribed Image Text:### Post-lab Question #2:
Using the \( K_b \) for \( \text{NH}_3 \) (from Appendix G: \( K_b = 1.8 \times 10^{-5} \)), calculate the \( K_a \) for the \( \text{NH}_4^+ \) ion.
#### Answer Choices:
- \( 5.6 \times 10^{-10} \)
- \( 1.8 \times 10^{-10} \)
- \( 5.6 \times 10^{-5} \)
- \( 1.8 \times 10^{-5} \)
To calculate the \( K_a \) of the \( \text{NH}_4^+ \) ion from the provided \( K_b \) of \( \text{NH}_3 \), use the relationship between \( K_a \) and \( K_b \) for a conjugate acid-base pair, which is given by the equation:
\[ K_a \times K_b = K_w \]
where \( K_w \) is the ion-product constant for water (\( K_w = 1.0 \times 10^{-14} \) at 25°C).
Rearrange the equation to solve for \( K_a \):
\[ K_a = \frac{K_w}{K_b} \]
Substitute the given \( K_b \) value:
\[ K_a = \frac{1.0 \times 10^{-14}}{1.8 \times 10^{-5}} \]
Perform the calculation:
\[ K_a = \frac{1.0 \times 10^{-14}}{1.8 \times 10^{-5}} = 5.6 \times 10^{-10} \]
So, the correct answer is:
- \( 5.6 \times 10^{-10} \)
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