A 25.0-mL sample of 0.10 M C₂H₃NH₂(ethylamine) is titrated with 0.15 M HCl. What is the pH of the solution after 9.00 mL of acid have been added to the amine? K_b = 6.5 × 10⁻⁴ A)11.08 B)10.88 C)10.74 D)10.55 E)10.49

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Quizplus · Accepted Answer

After adding 9.00 mL of 0.15 M HCl to 25.0 mL of 0.10 M ethylamine, the moles of HCl added are 0.00135 mol, and the moles of ethylamine initially present are 0.0025 mol. The reaction between HCl and ethylamine forms C2H3NH3+. The remaining moles of ethylamine are 0.00115 mol, and the moles of C2H3NH3+ formed are 0.00135 mol. Using the Henderson-Hasselbalch equation for the buffer solution, pH = pKa + log([base]/[acid]), where pKa = 14 - pKb = 10.187, the pH is calculated to be approximately 10.74.

A 250-ML Sample of 0