What is the percent ionization of a 0.75 M solution of ammonia? Kb(NH,) = 1.8 × 10-5 [?] %

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
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### Problem Statement

**What is the percent ionization of a 0.75 M solution of ammonia?**

**Given:**
\[ K_b(\text{NH}_3) = 1.8 \times 10^{-5} \]

### Solution Input

\[ 
\text{Percent Ionization} = \boxed{ \ ] \% 
\]

### User Interaction

- **Input Box**: Below the problem statement, there is an input box labeled “% Ionization” where the user can enter their answer.
- **Submit Button**: There is an "Enter" button next to the input box to submit the answer for evaluation.

---

This question involves calculating the percent ionization of ammonia in a solution. 

Percent ionization is found using the formula:

\[ \text{Percent Ionization} = \left( \frac{[\text{OH}^-]_{\text{equilibrium}}}{[\text{NH}_3]_{\text{initial}}} \right) \times 100\]

Where:
- \([\text{OH}^-]_{\text{equilibrium}}\) is the concentration of hydroxide ions at equilibrium.
- \([\text{NH}_3]_{\text{initial}}\) is the initial concentration of ammonia.

This calculation typically involves setting up an ICE (Initial, Change, Equilibrium) table and using the base dissociation constant \(K_b\) to find the equilibrium concentration of ions in the solution.
Transcribed Image Text:### Problem Statement **What is the percent ionization of a 0.75 M solution of ammonia?** **Given:** \[ K_b(\text{NH}_3) = 1.8 \times 10^{-5} \] ### Solution Input \[ \text{Percent Ionization} = \boxed{ \ ] \% \] ### User Interaction - **Input Box**: Below the problem statement, there is an input box labeled “% Ionization” where the user can enter their answer. - **Submit Button**: There is an "Enter" button next to the input box to submit the answer for evaluation. --- This question involves calculating the percent ionization of ammonia in a solution. Percent ionization is found using the formula: \[ \text{Percent Ionization} = \left( \frac{[\text{OH}^-]_{\text{equilibrium}}}{[\text{NH}_3]_{\text{initial}}} \right) \times 100\] Where: - \([\text{OH}^-]_{\text{equilibrium}}\) is the concentration of hydroxide ions at equilibrium. - \([\text{NH}_3]_{\text{initial}}\) is the initial concentration of ammonia. This calculation typically involves setting up an ICE (Initial, Change, Equilibrium) table and using the base dissociation constant \(K_b\) to find the equilibrium concentration of ions in the solution.
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