Consider the titration of a 25.0-mL sample of 0.175 M CH3NH2 with 0.150 M HBr. Determine each quantity: b. the volume of added acid required to reach the equivalence point, d. the pH at one-half of the equivalence point, f. the pH after adding 5.0 mL of acid beyond the equivalence point.

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1. b. Volume of added acid required to reach the equivalence point:
2. 1. Write the balanced chemical equation for the reaction: \( \text{CH}_3\text{NH}_2 + \text{HBr} \rightarrow \text{CH}_3\text{NH}_3^+ + \text{Br}^- \).
3. 2. Calculate the moles of \( \text{CH}_3\text{NH}_2 \) in the initial solution: \( \text{moles} = \text{volume (L)} \times \text{molarity} \).
4. 3. Use the stoichiometry of the reaction to determine the moles of \( \text{HBr} \) needed, which is equal to the moles of \( \text{CH}_3\text{NH}_2 \) since the reaction is 1:1.
5. 4. Calculate the volume of \( \text{HBr} \) solution required using its molarity: \( \text{volume (L)} = \frac{\text{moles of HBr}}{\text{molarity of HBr}} \).
6. d. pH at one-half of the equivalence point:
7. 1. At one-half of the equivalence point, the concentration of \( \text{CH}_3\text{NH}_2 \) equals the concentration of \( \text{CH}_3\text{NH}_3^+ \).
8. 2. Use the Henderson-Hasselbalch equation: \( \text{pH} = \text{pK}_a + \log\left(\frac{[\text{base}]}{[\text{acid}]}\right) \).
9. 3. Since \([\text{base}] = [\text{acid}]\), \( \log\left(\frac{[\text{base}]}{[\text{acid}]}\right) = 0 \), so \( \text{pH} = \text{pK}_a \).
10. 4. Calculate \( \text{pK}_a \) from \( \text{pK}_w = 14 \) and \( \text{pK}_b \) of \( \text{CH}_3\text{NH}_2 \): \( \text{pK}_a = 14 - \text{pK}_b \).
11. f. pH after adding 5.0 mL of acid beyond the equivalence point:
12. 1. Calculate the total volume of \( \text{HBr} \) added: \( \text{volume at equivalence point} + 5.0 \text{ mL} \).
13. 2. Determine the moles of excess \( \text{HBr} \) added beyond the equivalence point.
14. 3. Calculate the concentration of excess \( \text{H}^+ \) ions in the total solution volume.
15. 4. Use the concentration of \( \text{H}^+ \) to find the pH: \( \text{pH} = -\log[\text{H}^+] \).