Chemistry
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ISBN: 9781305957404
Author: Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher: Cengage Learning
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![### Calculating the pH of a Solution Containing Ammonia and Ammonium Chloride
**Problem Statement:**
A solution is prepared at \(25 \, ^\circ \mathrm{C}\) that is initially \(0.16\, M\) in ammonia \((\mathrm{NH}_3)\), a weak base with \(K_b = 1.8 \times 10^{-5}\), and \(0.011\, M\) in ammonium chloride \((\mathrm{NH}_4\mathrm{Cl})\). Calculate the pH of the solution. Round your answer to 2 decimal places.
**Solution:**
To calculate the pH of the solution containing ammonia and ammonium chloride, follow these steps:
1. **Determine the concentration of \(\mathrm{OH}^-\)** using the given \(K_b\):
- Ammonia (\(\mathrm{NH}_3\)) is a weak base, which partially dissociates in water:
\[
\mathrm{NH}_3 + \mathrm{H}_2\mathrm{O} \rightleftharpoons \mathrm{NH}_4^+ + \mathrm{OH}^-
\]
- Using the given \(K_b\) value:
\[
K_b = \frac{[\mathrm{NH}_4^+][\mathrm{OH}^-]}{[\mathrm{NH}_3]}
\]
Given \( [\mathrm{NH}_4^+] \) from \(\mathrm{NH}_4\mathrm{Cl} = 0.011\, M\) and initial \([\mathrm{NH}_3] = 0.16\, M\).
2. **Set up the equilibrium expression:**
\[
1.8 \times 10^{-5} = \frac{(0.011 + x)x}{0.16 - x}
\]
Because \(K_b\) is very small, \(x\), the amount dissociated, will be very small relative to the initial concentrations; thus, we can approximate \(0.011 + x \approx 0.011\) and \(0.16 - x \approx 0.16\).
3. **Simplify the equation:**
\[
1.8 \times 10^{-5}](https://content.bartleby.com/qna-images/question/868a318f-0624-4ed5-b0cc-007ff8576a9b/5f93ccd7-8ddb-453d-91d7-414f55d05498/zj04bqt_processed.png)
Transcribed Image Text:### Calculating the pH of a Solution Containing Ammonia and Ammonium Chloride
**Problem Statement:**
A solution is prepared at \(25 \, ^\circ \mathrm{C}\) that is initially \(0.16\, M\) in ammonia \((\mathrm{NH}_3)\), a weak base with \(K_b = 1.8 \times 10^{-5}\), and \(0.011\, M\) in ammonium chloride \((\mathrm{NH}_4\mathrm{Cl})\). Calculate the pH of the solution. Round your answer to 2 decimal places.
**Solution:**
To calculate the pH of the solution containing ammonia and ammonium chloride, follow these steps:
1. **Determine the concentration of \(\mathrm{OH}^-\)** using the given \(K_b\):
- Ammonia (\(\mathrm{NH}_3\)) is a weak base, which partially dissociates in water:
\[
\mathrm{NH}_3 + \mathrm{H}_2\mathrm{O} \rightleftharpoons \mathrm{NH}_4^+ + \mathrm{OH}^-
\]
- Using the given \(K_b\) value:
\[
K_b = \frac{[\mathrm{NH}_4^+][\mathrm{OH}^-]}{[\mathrm{NH}_3]}
\]
Given \( [\mathrm{NH}_4^+] \) from \(\mathrm{NH}_4\mathrm{Cl} = 0.011\, M\) and initial \([\mathrm{NH}_3] = 0.16\, M\).
2. **Set up the equilibrium expression:**
\[
1.8 \times 10^{-5} = \frac{(0.011 + x)x}{0.16 - x}
\]
Because \(K_b\) is very small, \(x\), the amount dissociated, will be very small relative to the initial concentrations; thus, we can approximate \(0.011 + x \approx 0.011\) and \(0.16 - x \approx 0.16\).
3. **Simplify the equation:**
\[
1.8 \times 10^{-5}
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