A 70.0 mL solution of 0.165 M potassium alaninate (H2NC2H5CO2K) is titrated with 0.165 M HCl. The pKa values for the amino acid alanine are 2.344 (pKa1) and 9.868 (pKa2), which correspond to the carboxylic acid and amino groups, respectively.
Calculate pH of second equivalence point.
To calculate the pH of the second equivalence point, we need to determine the volume of HCl required to reach this point.
The second equivalence point occurs when all the alanine has been fully neutralized. This means that both the carboxylic acid group and the amino group of alanine have reacted with HCl. Since alanine is a diprotic amino acid, there are two equivalence points: one when the carboxylic acid group is neutralized and the second when the amino group is neutralized.
To find the volume of HCl required for the second equivalence point, we need to consider the stoichiometry of the reaction. Each mole of alanine reacts with two moles of HCl, one for each acidic proton. Therefore, the mole ratio of alanine to HCl is 1:2.
Given:
Initial volume of potassium alaninate solution (V1) = 70.0 mL
Concentration of potassium alaninate solution (C1) = 0.165 M
Concentration of HCl solution (C2) = 0.165 M
Using the equation: V1 * C1 = V2 * C2
V2 = (V1 * C1) / C2
V2 = (70.0 mL * 0.165 M) / 0.165 M
V2 = 70.0 mL
At the second equivalence point, the volume of HCl added is equal to the initial volume of the potassium alaninate solution (70.0 mL).
Next, we need to convert the remaining amount of alanine in the solution into moles. Since the alanine molecule is diprotic, the moles of alanine can be calculated as follows:
Moles of alanine = (70.0 mL * 0.165 M) / 1000 mL/L
Now, let's calculate the pOH at the second equivalence point. The pOH can be determined by considering the hydroxide ion concentration, which is equal to the concentration of alanine still present in the solution.
pOH = -log[OH-]
OH- = moles of alanine / total volume of solution
Total volume of solution = volume of potassium alaninate solution + volume of HCl added
Total volume of solution = 70.0 mL + 70.0 mL = 140.0 mL = 0.140 L
pOH = -log[(70.0 mL * 0.165 M) / (0.140 L)]
Finally, the pH at the second equivalence point can be calculated by subtracting the pOH from 14 (pH + pOH = 14):
pH = 14 - pOH
Substituting the calculated pOH value in the equation above will give you the pH of the second equivalence point in the titration.