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The pH of a 0.10 M ammonia NH3 solution (K = 1.8 x 10-5) is: Your Answer: Answer

The pH of a 0.10 M ammonia NH3 solution (K = 1.8 x 10-5) is: Your Answer: Answer

Chemistry
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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**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.
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.
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Ammonia reacts with water and dissociates into ammonium ion hydroxide ions. 

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