Consider the titration of 100.0 mL of 0.100 M CH₃NH₂ with 0.250 M HNO₃ at the equivalence point. What would be the pH of the solution at the equivalence point? The K_b of CH₃NH₂ is 4.4 × 10⁻⁴.

• A. 8.10
• B. 11.79
• C. 2.21
• D. 5.90

Answer: (D)

Verified step by step guidance

1. Start by understanding that at the equivalence point in a titration, the amount of acid equals the amount of base. Here, CH3NH2 (methylamine) is a weak base and HNO3 is a strong acid.
2. Calculate the moles of CH3NH2 initially present using the formula: \( \text{moles} = \text{concentration} \times \text{volume} \). For CH3NH2, \( \text{moles} = 0.100 \, \text{M} \times 0.100 \, \text{L} = 0.010 \, \text{moles} \).
3. At the equivalence point, all CH3NH2 is converted to its conjugate acid, CH3NH3^+. The concentration of CH3NH3^+ can be found by dividing the moles of CH3NH2 by the total volume of the solution after titration. Calculate the total volume: \( 100.0 \, \text{mL} + \frac{0.010 \, \text{moles}}{0.250 \, \text{M}} \).
4. Use the Kb of CH3NH2 to find the Ka of CH3NH3^+ using the relation: \( K_w = K_a \times K_b \), where \( K_w = 1.0 \times 10^{-14} \). Solve for \( K_a \) using \( K_a = \frac{K_w}{K_b} \).
5. Finally, calculate the pH using the formula for the dissociation of the conjugate acid: \( [H^+] = \sqrt{K_a \times [CH3NH3^+]} \). Then, find the pH using \( \text{pH} = -\log[H^+] \).