A chemist titrates 50.0mL of a 0.4246M propionic acid HC2H5CO2 solution with 0.6989M NaOH solution at 25°C. Calculate the pH at equivalence. The pKa of propionic acid is 4.89.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 NaOH solution added.

To solve this problem, we need to follow these steps:

1. Calculate the moles of propionic acid (HC2H5CO2) in the solution. We can do this by multiplying the volume of the solution (in liters) by the molarity of the solution.

   Moles of HC2H5CO2 = 0.0500 L * 0.4246 mol/L = 0.02123 mol

2. At equivalence point, all the propionic acid has reacted with the sodium hydroxide (NaOH). Therefore, the moles of NaOH added would be equal to the moles of propionic acid.

   Moles of NaOH = 0.02123 mol

3. The reaction between propionic acid and NaOH produces propionate ion (C2H5CO2-) and water. At the equivalence point, all the propionic acid has been converted to propionate ion.

4. The propionate ion is the conjugate base of the propionic acid and can react with water to produce OH- ions. This reaction can be represented as follows:

   C2H5CO2- + H2O -> HC2H5CO2 + OH-

5. The pKa of the propionic acid is given as 4.89. We can use this to calculate the pKb of the propionate ion using the following relationship:

   pKb = 14 - pKa = 14 - 4.89 = 9.11

6. The Kb of the propionate ion can be calculated from the pKb using the following relationship:

   Kb = 10^-pKb = 10^-9.11

7. At the equivalence point, the concentration of the propionate ion is equal to the moles of propionate ion divided by the total volume of the solution. The total volume of the solution is the sum of the initial volume of the propionic acid solution and the volume of the NaOH solution added.

   [C2H5CO2-] = moles of C2H5CO2- / total volume

   We don't have the volume of NaOH solution added, but we know that the moles of NaOH added is equal to the moles of propionic acid, which is 0.02123 mol. Therefore, we can calculate the volume of NaOH solution added as follows:

   Volume of NaOH = moles of NaOH / molarity of NaOH = 0.02123 mol / 0.6989 mol/L = 0.0304 L

   Therefore, the total volume of the solution is 0.0500 L + 0.0304 L = 0.0804 L

   Therefore, [C2H5CO2-] = 0.02123 mol / 0.0804 L = 0.264 mol/L

8. We can now use the Kb and the concentration of the propionate ion to calculate the concentration of OH- ions using the following relationship:

   Kb = [OH-][HC2H5CO2] / [C2H5CO2-]

   Since the concentration of HC2H5CO2 is very small compared to the concentration of C2H5CO2-, we can assume that [OH-] = [HC2H5CO2]. Therefore, we can simplify the above equation as follows:

   Kb = [OH-]^2 / [C2H5CO2-]

   Therefore, [OH-] = sqrt(Kb * [C2H5CO2-]) = sqrt((10^-9.11) * 0.264) = 1.63 * 10^-5 M

9. We can now calculate the pOH of the solution using the following relationship:

   pOH = -log[OH-] = -log(1.63 * 10^-5) = 4.79

10. Finally, we can calculate the pH of the solution using the following relationship:

    pH = 14 - pOH = 14 - 4.79 = 9.21

Therefore, the pH at the equivalence point is 9.21.