Calculate the pH of a buffer solution that is 0.339 M in NH3 (ammonia) and 0.143 M in NH4CI. pH =

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### Calculating the pH of a Buffer Solution

#### Problem Statement:
Calculate the pH of a buffer solution that is **0.339 M** in **NH₃ (ammonia)** and **0.143 M** in **NH₄Cl**.

#### Solution:
To calculate the pH of a buffer solution, you can use the Henderson-Hasselbalch equation:

\[ \text{pH} = \text{p}K_{\text{a}} + \log \left( \frac{[\text{Base}]}{[\text{Acid}]} \right) \]

In this problem:
- The base (\( \text{NH}_3 \)) concentration (\( [\text{Base}] \)) is **0.339 M**
- The conjugate acid (\( \text{NH}_4^+ \), provided by \( \text{NH}_4\text{Cl} \)) concentration (\( [\text{Acid}] \)) is **0.143 M**

Here, the \( \text{Kb} \) for ammonia (\( \text{NH}_3 \)) can be used to find the \( \text{pKa} \) of its conjugate acid, ammonia (\( \text{NH}_4^+ \)).

\[ \text{p}K_{\text{a}} = 14 - \text{p}K_{\text{b}} \]

Assuming \( \text{p}K_{\text{b}} \) of ammonia is approximately 4.75:

\[ \text{p}K_{\text{a}} = 14 - 4.75 = 9.25 \]

So, the Henderson-Hasselbalch equation becomes:

\[ \text{pH} = 9.25 + \log \left( \frac{0.339}{0.143} \right) \]

#### Graphs and Diagrams:
In this example, no graphs or diagrams were provided. However, if there were, the explanation would include descriptions of axes, data points, and any significant trends or patterns observed within the graph or diagram.

You can input these values into the equation and calculate the log term to find the pH of the buffer solution. The final pH will be determined numerically.
Transcribed Image Text:### Calculating the pH of a Buffer Solution #### Problem Statement: Calculate the pH of a buffer solution that is **0.339 M** in **NH₃ (ammonia)** and **0.143 M** in **NH₄Cl**. #### Solution: To calculate the pH of a buffer solution, you can use the Henderson-Hasselbalch equation: \[ \text{pH} = \text{p}K_{\text{a}} + \log \left( \frac{[\text{Base}]}{[\text{Acid}]} \right) \] In this problem: - The base (\( \text{NH}_3 \)) concentration (\( [\text{Base}] \)) is **0.339 M** - The conjugate acid (\( \text{NH}_4^+ \), provided by \( \text{NH}_4\text{Cl} \)) concentration (\( [\text{Acid}] \)) is **0.143 M** Here, the \( \text{Kb} \) for ammonia (\( \text{NH}_3 \)) can be used to find the \( \text{pKa} \) of its conjugate acid, ammonia (\( \text{NH}_4^+ \)). \[ \text{p}K_{\text{a}} = 14 - \text{p}K_{\text{b}} \] Assuming \( \text{p}K_{\text{b}} \) of ammonia is approximately 4.75: \[ \text{p}K_{\text{a}} = 14 - 4.75 = 9.25 \] So, the Henderson-Hasselbalch equation becomes: \[ \text{pH} = 9.25 + \log \left( \frac{0.339}{0.143} \right) \] #### Graphs and Diagrams: In this example, no graphs or diagrams were provided. However, if there were, the explanation would include descriptions of axes, data points, and any significant trends or patterns observed within the graph or diagram. You can input these values into the equation and calculate the log term to find the pH of the buffer solution. The final pH will be determined numerically.
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