A chemist titrates 240.0 mL of a 0.7318 M ethylamine (C_2 H_5NH_2) solution with 0.3280 M HCl solution at 25 degree C. Calculate the pH at equivalence. The p^K_ b of ethylamine is 3.19. Round your answer to 2 decimal places. Note for advanced students: you may assume the total volume of the solution equals the initial volume plus the volume of HCl solution added.

To calculate the pH at the equivalence point, we first determine the moles of ethylamine, C2H5NH2, and then calculate the moles of HCl needed to reach the equivalence point.

Step 1: Calculate moles of ethylamine (base) using its concentration and volume. \[ \text{moles of ethylamine} = Molarity_{\text{ethylamine}} \times Volume_{\text{ethylamine}} \] \[ \text{moles of ethylamine} = 0.7318\ M \times 0.2400\ L = 0.1756\ mol \]

Step 2: Since we are at the equivalence point, the moles of HCl (acid) added equals the moles of ethylamine because one mole of HCl will react with one mole of ethylamine. \[ \text{moles of HCl} = \text{moles of ethylamine} = 0.1756\ mol \]

Step 3: Calculate the volume of 0.3280 M HCl solution required to reach the equivalence point. \[ Volume_{\text{HCl}} = \frac{\text{moles of HCl}}{Molarity_{\text{HCl}}} \] \[ Volume_{\text{HCl}} = \frac{0.1756\ mol}{0.3280\ M} = 0.5354\ L \]

At the equivalence point, ethylamine has fully reacted with HCl to form the ethylammonium ion, C2H5NH3+. We calculate the pH using the pKb of ethylamine and the Kw (ion product constant of water, which is 1.00 x 10^-14 at 25 degrees Celsius).

Step 4: Since we're dealing with the conjugate acid of a weak base, we need the pKa of the conjugate acid. pKa and pKb are related by the following equation: \[ pKa + pKb = pKw = 14 \] \[ pKa = 14 - pKb = 14 - 3.19 = 10.81 \]

Step 5: To find the pH, convert pKa to Ka and use the formula for the pH of a solution of a weak acid at the equivalence point. \[ Ka = 10^{-pKa} = 10^{-10.81} \]

At the equivalence point in the titration of a weak base with a strong acid, the solution contains the conjugate acid of the base and will be acidic.

Step 6: Calculate the concentration of the ethylammonium ion in the solution after adding the volume of HCl solution. \[ [C_2H_5NH_3^+] = \frac{\text{moles of C_2H_5NH_3^+}}{\text{Total Volume of the solution}} \] \[ [C_2H_5NH_3^+] = \frac{0.1756\ mol}{0.2400\ L + 0.5354\ L} = \frac{0.1756\ mol}{0.7754\ L} = 0.2265\ M \]

Step 7: Use the Henderson-Hasselbalch equation to calculate the pH of the solution. Since we are at equivalence, the concentration of the acid equals the concentration of its conjugate base. \[ pH = pKa + \log\left(\frac{[\text{base}]}{[\text{acid}]}\right) \] In our case, since the acid is the conjugate acid of a weak base, the base concentration can be considered zero at equivalence, and we can simplify the equation to: \[ pH = pKa + \log(1) = pKa = 10.81 \]

Thus, the pH at the equivalence point of this titration is 10.81 (Note: Normally, we would use the Henderson-Hasselbalch equation in such problems, but because we're interested only in the conjugate acid at this point, we go directly to pKa.)