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Chemistry
10th Edition
ISBN: 9781305957404
Author: Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher: Cengage Learning
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![**Question:**
Calculate the pH of a 0.10 M ammonia (NH₃) solution. The base dissociation constant (K_b) for ammonia is 1.8 x 10⁻⁵.
**Your Answer:**
*Input Box:* [ ]
**Answer Explanation:**
To solve this problem, follow these steps:
1. Write the equilibrium expression for the dissociation of ammonia (NH₃):
- NH₃ + H₂O ⇌ NH₄⁺ + OH⁻
2. Set up an expression for the base dissociation constant (K_b):
- \( K_b = \frac{{[NH_4^+][OH^-]}}{{[NH_3]}} \)
3. Use the initial concentrations and change in concentration (x) to determine the equilibrium concentrations:
- Initial: [NH₃] = 0.10 M, [NH₄⁺] = 0, [OH⁻] = 0
- Change: [NH₃] = -x, [NH₄⁺] = +x, [OH⁻] = +x
- Equilibrium: [NH₃] = 0.10 - x, [NH₄⁺] = x, [OH⁻] = x
4. Substitute the equilibrium concentrations back into the K_b expression:
- \( 1.8 \times 10^{-5} = \frac{{x^2}}{{0.10 - x}} \)
5. Solve for x, which represents [OH⁻].
6. Calculate the pOH:
- pOH = -log[OH⁻]
7. Finally, convert pOH to pH:
- pH = 14 - pOH
This will provide the pH of the 0.10 M ammonia solution.](https://content.bartleby.com/qna-images/question/b9b814ac-1a5f-4504-b547-b7326372e811/c9939e8b-475f-45c5-a6cd-bc7e73064f2a/g8k8ith_thumbnail.jpeg)
Transcribed Image Text:**Question:**
Calculate the pH of a 0.10 M ammonia (NH₃) solution. The base dissociation constant (K_b) for ammonia is 1.8 x 10⁻⁵.
**Your Answer:**
*Input Box:* [ ]
**Answer Explanation:**
To solve this problem, follow these steps:
1. Write the equilibrium expression for the dissociation of ammonia (NH₃):
- NH₃ + H₂O ⇌ NH₄⁺ + OH⁻
2. Set up an expression for the base dissociation constant (K_b):
- \( K_b = \frac{{[NH_4^+][OH^-]}}{{[NH_3]}} \)
3. Use the initial concentrations and change in concentration (x) to determine the equilibrium concentrations:
- Initial: [NH₃] = 0.10 M, [NH₄⁺] = 0, [OH⁻] = 0
- Change: [NH₃] = -x, [NH₄⁺] = +x, [OH⁻] = +x
- Equilibrium: [NH₃] = 0.10 - x, [NH₄⁺] = x, [OH⁻] = x
4. Substitute the equilibrium concentrations back into the K_b expression:
- \( 1.8 \times 10^{-5} = \frac{{x^2}}{{0.10 - x}} \)
5. Solve for x, which represents [OH⁻].
6. Calculate the pOH:
- pOH = -log[OH⁻]
7. Finally, convert pOH to pH:
- pH = 14 - pOH
This will provide the pH of the 0.10 M ammonia solution.
Expert Solution
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Step 1
Ammonia reacts with water and dissociates into ammonium ion hydroxide ions.
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