To calculate the pH of the buffer, we use the Henderson-Hasselbalch equation:

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

Information provided:

• $[ \text{Base} ] = 0.22 \, \text{M}$ (ammonia, $\text{NH}_3$)
• $[ \text{Acid} ] = 0.24 \, \text{M}$ (ammonium ion, $\text{NH}_4^+$)
• $\text{pKa} = 9.24$

Step-by-step solution:

1. Substitute the values into the Henderson-Hasselbalch equation: $\text{pH} = 9.24 + \log \left( \frac{0.22}{0.24} \right)$

2. Calculate the ratio $\frac{[ \text{Base} ]}{[ \text{Acid} ]}$: $\frac{0.22}{0.24} = 0.9167$

3. Take the logarithm of the ratio: $\log(0.9167) \approx -0.037$

4. Add this to the pKa: $\text{pH} = 9.24 + (-0.037)$

5. Final pH: $\text{pH} = 9.20$

Final Answer:

The pH of the buffer is 9.20.

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Do you want more details or have any questions? Here are 5 related questions for further exploration:

1. What happens to the buffer pH if the concentration of $\text{NH}_3$ is increased to 0.30 M?
2. How does the buffer resist changes in pH when a small amount of acid is added?
3. What would the pH be if equal concentrations of $\text{NH}_3$ and $\text{NH}_4^+$ were used?
4. How is the buffer capacity of this solution determined?
5. What is the significance of the pKa value in determining the strength of the buffer?

Tip: Always ensure the ratio of base to acid in a buffer stays close to 1 for optimal buffering capacity!